Mathematics Today In - 2017 - 02
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10
76
19
Vol. XXXV
No. 2
February 2017
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contents 8
8
41
Maths Musing Problem Set - 170 23
Competition edge
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mathematics today | February‘17
7
M
aths Musing was started in January 2003 issue of Mathematics Today with the suggestion of Shri Mahabir Singh. The aim of Maths Musing is to augment the chances of bright students seeking admission into IITs with additional study material. During the last 10 years there have been several changes in JEE pattern. To suit these changes Maths Musing also adopted the new pattern by changing the style of problems. Some of the Maths Musing problems have been adapted in JEE benefitting thousand of our readers. It is heartening that we receive solutions of Maths Musing problems from all over India. Maths Musing has been receiving tremendous response from candidates preparing for JEE and teachers coaching them. We do hope that students will continue to use Maths Musing to boost up their ranks in JEE Main and Advanced.
jee main
1. If the tangent at the point (a, b) on the curve x3 + y3 = a3 + b3 meets the curve again at the point (p, q), then (a) ap + bq + ab = 0 (b) bp + aq + ab = 0 (c) ap + bq – pq = 0 (d) bp + aq – pq = 0 an +1 3 + 1 1 − 3 an , n ∈ N. 2. Let = bn +1 3 − 1 1 + 3 bn If a1 = b1 = 1, then a22 = (a) 231 (b) 232 (c) 233 (d) 333 3. In a triangle ABC, if tan A : tan B : tan C = 2 : 3 : 4, ∆ then 2 = R 9 6 18 6 36 2 18 3 (a) (b) (c) (d) 35 35 35 35 3 −i 4. If z = , then (i101 + z101)103 = 2 (a) z (b) z (c) iz (d) iz 5. Let m be the number of 5-element subsets that can be chosen from the set of the first 15 natural numbers so that at least two of the five numbers are consecutive. The sum of the digits of m is (a) 11 (b) 12 (c) 13 (d) 14 jee advanced 4 6 6 6. Let A = 1 3 2 . A value of l, such that −1 −5 −2 AX = lX, where X is a non-zero column vector, is (a) 0 (b) 1 (c) 2 (d) 3 comprehension The parametric equations of a parabola are x = u cos a⋅t, 1 y = u sin a⋅t − gt2, t is a parameter. 2 7. Its latus rectum is 2u2 2 2u2 sin a cos2 a (a) (b) g g 8
mathematics today | FeBruary‘17
2 u2 cos2 a (c) u sin2 a (d) 2g 2g 8. Its focus is u2 u2 (a) sin 2a, cos 2a 2g 2g
u2
−u 2
cos 2a (b) sin 2a, 2g 2g
u2 −u 2 2 cos2 a (c) sin a, 2g 2g u2
−u 2
2 sin2 a (d) cos a, 2g 2g integer match 9. I is the incentre of the triangle ABC. AI, BI, CI when produced meet the opposite sides at D, E, F respectively. Then the value of AI BI CI AI BI CI ⋅ ⋅ − + + is ID IE IF ID IE IF matrix match 10. List-I contains S and List-II gives last digit of S. List-I List-II
P. S = Q. S = R. S = S. S = (a) (b) (c) (d)
P 4 4 2 1
11
∑ (2n − 1)2
1.
0
∑ (2n − 1)3
2.
1
∑ (2n − 1)2 (−1)n
3.
5
∑ (2n − 1)3(−1)n−1
4.
8
n =1 10
n =1 18 n =1 15
n =1
Q 2 2 1 2
R 1 3 4 3
S 3 1 3 4
See Solution Set of Maths Musing 169 on page no. 89
Exam Dates OFFLINE : 2nd April ONLINE : 8th & 9th April
1. Th ree numbers are chosen at random from 3 20 21 2 (a) (b) (c) (d) 1, 2, ..., 30. What is the probability that they will 5 25 25 5 form a G.P.? 7. If two independent events A and B are such 10 12 19 20 3 8 (a) (b) (c) (d) that P ( A BC ) and P ( AC B) and 4060 4060 4060 4060 25 25 1 2. If n balls are distributed into m boxes, so that each P ( A) , the value of P(A) + P(B) = 2 ball is equally likely to fall in any box, the probability 8 4 that a specified box will contain r balls is (a) (b) n nr n nr 5 5 Pr (m 1) Cr (m 1) (a) (b) 7 n n Cm Cm (c) (d) None of these 5 n nr n nr Cr (m 1) Cr (m 1) 8. If two events A and B are such that P(AC) = 0.3, (c) (d) n m nm P(B) = 0.4, P(ABC) = 0.5, then P(B|A BC) is 3. Th e probability that at least one of the eventsA and 4 1 1 1 (a) (b) (c) (d) B occur is 0.4. If A and B be occur simultaneously 5 2 4 5 with probability 0.1, then probability of AC or BC is 9. Th e independent probabilities thatA, B, C can equal to 1 1 1 (a) 1.2 (b) 1.4 (c) 1.5 (d) 1.8 solve the problem are , , respectively. Th e 2 3 4 4. In an attempt to land an unmanned rocket on the probability that just two of them only solve the moon, the probability of a successful landing is 0.4 problem is 2 3 and the probability of monitoring system giving the (a) (b) 3 correct information of landing is 0.9 in either case. Th e 4 1 probability of a successful landing, it being known that (c) (d) None of these the monitoring system indicated it correctly 4 10. A bag A contains 2 white and 3 red balls and bag B 6 9 4 2 (a) (b) (c) (d) contains 4 white and 5 red balls. One ball is drawn 7 10 25 5 at random from one of the bags and is found to be 5. Each of N + 1 identical urns marked 0, 1, 2, ..., N th red. Th e probability that it was drawn from bagB is contains N balls so that the i urn contains i black and N – i white balls (0 i N). An urn is chosen 25 8 13 25 (a) (b) (c) (d) at random and n balls are drawn from it one after 51 17 27 52 another with replacement. If all the n balls turn out 11. Th e set of equations to be black, the probability that the next ball will x − y + (cos)z = 0 also black [assume that N is large] is 3x + y + 2z = 0 n n n 1 n 1 (cos)x + y + 2z = 0 (a) (b) (c) (d) n2 n 1 n2 n 1 0 < 2, has non−trivial solutions (a) for no value of and 6. Th e probability that a teacher will give a surprise (b) for all values of and test during any class meeting is 3/5. If a student is (c) for all values of and only two values of absent on two days, the probability that he will miss (d) for only one value of and all values of . at least one test is 10
MATHEMATICS TODAY | FEBRUARY ‘17
12. Orthogonal trajectories of family of parabolas y2 = 4a(x + a) where ‘a’ is an arbitrary constant is (a) ax2 = 4cy (b) x2 + y2 = a2 −
x 2a
(c) y = ce (d) axy = c2, where c is a constant. 13. If
5z1 2z1 + 3z2 is purely imaginary, then = 7 z2 2z1 − 3z2
(a) 5/7
(b) 7/5
(c) 25/29 (d) 1
14. The mean daily profit made by a shopkeeper in a month of 30 days was ` 350. If the mean profit for the first twenty days was ` 400, then the mean profit for the last 10 days would be (a) ` 200 (b) ` 250 (c) ` 800 (d) ` 300 15. Consider points A(3, 4) and B(7, 13). If P be a point on the line y = x such that PA + PB is minimum, then coordinates of P are 13 13 12 12 (a) , (b) , 7 7 7 7 31 31 (c) , (d) (0, 0) 7 7 16. lim
log(2 + x 2 ) − log(2 − x 2 ) x2 (b) 2
x →0
(a) –1
17. The sum of the series is equal to (a) 1
(b) 0
= k , the value of k is
(c) 1 3 12 ⋅ 22
+
(d) 0 5 22 ⋅ 32
(c) –1
+
7 32 ⋅ 42
+ ...∞
(d) 2
18. If f : R → R, f (x + y) = f (x) · f (y). If f (1) = 1, then n
∑ f (k) = k =1
(a) n (c) 1
(b) n(n + 1)(2n + 1)/6 (d) n(n + 1)/2
sin(a + 1)x + sin x , for x c , for 19. If f (x ) = 2 1/2 1/2 (x + bx ) − x , forr bx 3/2
x <0 x =0 x >0
is continuous at x = 0, then a = (a) 3/2 (b) –3/2 (c) 1/4 (d) –1/4
12
mathematics today | February ‘17
20. If f (x) = xn, then the value of
2
2 2 n f ′(1) f ′′(1) f (1) { f (1)} + + + .... + 1! 2 ! n ! is given by (a) 2nCn (b) 2nCn – 1 2n (c) Cn + 1 (d) 2nC1 2
21. For the three events A, B and C, P(exactly one of the events A or B occurs) = P (exactly one of the events B or C occurs) = P(exactly one of the events C or A occurs) = p and P(all the three events occurs simultaneously) = p2, where 0 < p < 1/2. Then the probability of at least one of the three events A, B and C occurring is 1 1 2 (a) (3 p + 2 p ) (b) ( p + 3 p2 ) 2 4 1 1 2 (c) ( p + 3 p2 ) (d) (3 p + 2 p ) 4 2 22. The area of the smaller region bounded by the curves x2 + y2 = 5 and y2 = 4x is π 1 π 1 + (a) (b) − 4 6 4 6 1 5 −1 2 1 5 −1 2 (c) 2 − sin (d) 2 + sin 3 2 3 2 5 5 x + a2 ab ac 23. If a, b, c are real, then f (x ) = is decreasing in 2 (a) − (a2 + b2 + c 2 ), 3
ab
x + b2
bc
ac
bc
x + c2
0
2 (b) 0, (a2 + b2 + c 2 ) 3 a 2 + b2 + c 2 (c) , 0 (d) None of these 3
24. If P(n) : n2 + n is an odd integer. It is seen that truth of P(n) ⇒ the truth of P(n + 1). Therefore, P(n) is true for all (a) n > 1 (b) n (c) n > 2 (d) none of these 2 2 2 25. If y = cos x + cos x + cos x + ...... to ∞ ,
then (a)
dy is dx − sin x 2 x(2 y − 1)
(b)
−2 x sin x 2 2y −1
(c)
− sin x 2y −1
26. Area lying between the curves y2 = 4x and y = 2x is (a) 2/3 (b) 1/3 (c) 1/4 (d) none of these 27. If f(x) is a continuous function satisfying f(x) f(1/x) = f(x) + f(1/x) and f (1) > 0, then lim f ( x) x →1
is equal to (a) 2 (c) 3
(b) 1 (d) none of these
28. For any integer n, the integral π
∫e
cos 2 x
0
cos 3 x( 2n + 1)x dx has the value
(a) π (c) 0
(b) 1 (d) none of these
29. Let E and F be two independent events. The probability that both E and F happen is 1/12 and the probability that neither E nor F happens is 1/2. Then (a) P(E) = 1/3, P(F) = 1/4 (b) P(E) = 1/2, P(F) = 1/6 (c) P(E) = 1/6, P(F) = 1/2 (d) P(E) = 1/8, P(F) = 1/3 30. If nCr denotes the number of combinations of n things taken r at a time, then the expression n
n −1
C0 + ∑
n+ k
k =0
(a) (c)
2nC 2nC
n–1 n
5/2 4/3 5/3 5/4
(d) none of these
Ck +1 equals (b) 2nCn + 1 (d) nCn – 1 SOLUTIONS
1. (c): Let A be the event that the numbers of the triplets form a G.P. Since 3 numbers can be selected from 30 numbers in 30C ways, therefore total number of triplets = 30C 3 3 Now we count the triplets (arranged in increasing order) whose terms form a G.P. by listing them as follows. Common ratio Triplet 2 {(i, 2i, 4i), 1 ≤ i ≤ 7} 3 {(i, 3i, 9i), 1 ≤ i ≤ 3} 4 (1, 4, 16) 5 (1, 5, 25) 3/2 (4, 6, 9), (8, 12, 18), (12, 18, 27)
(4, 10, 25) (9, 12, 16) (9, 15, 25) (16, 20, 25)
Thus among 30C3 triplets, there are 19 triplets, whose terms form a G.P. 19 19 \ Required probability P ( A) = 30 = C3 4060
2. (c): The total number of ways of distributing n balls into m boxes is mn which are assumed to be equally likely. Since r balls which should go to the specified box can be chosen in n C r ways and for any such way the remaining (n – r) balls can be distributed to the (m – 1) boxes in (m – 1)n – r ways. n \ The number of favourable cases = (m − 1)n−r r n n−r r (m − 1) \ P ( A) = mn [where A is the event that a specified box will contain r balls] 3. \ or or or
(c): Given that P(A ∪ B) = 0.4 and P(A ∩ B) = 0.1 P(A) + P(B) – P(A ∩ B) = 0.4 P(A) + P(B) = 0.4 + 0.1 = 0.5 1 – P(AC) + 1 – P(BC) = 0.5 P(AC) + P(BC) = 2 – 0.5 = 1.5
4. (d) : Let A denote the event of a successful landing of the rocket, B1 denote the event of the monitoring system indicating it correctly and B2 denote the event of its indicating unsuccessful landing. \ P(A) = 0.4, P(B1 |A) = 0.9, P(B2|AC) = 0.9 P (B1 | A) P ( A) The required probability = P(A|B1) = P (B1 ) Now, P(B1) = P(A ∩ B1) + P(AC ∩ B1) = P(B1|A)P(A) + P(B1|AC) P(AC) = P(B1|A)P(A) + [1 – P(B1C|AC)] P(AC ) = P(B1|A) P(A) + [1 – P(B2|AC)] P(AC ) = 0.9 × 0.4 + (1 – 0.9) × 0.6 = 0.42 0. 9 × 0. 4 6 Hence the required probability = = 0.42 7 5. (a) : Let Ai be the event of choosing ith urn and B the event of choosing n black balls in succession and C the event of drawing (n + 1)th ball as black. The required probability is mathematics today | February‘17
13
P (B ∩ C ) P ( B)
P (C | B) =
1 3 4 As P ( A) > , we must have P ( A) = and P (B) = 2 5 5 7 \ P ( A) + P (B) = 5
N
Now, P (B) = ∑ P ( Ai ) ⋅ P (B |Ai) i =0
N
=∑ i =0
1
n
1 i n = ∫ x dx when N → ∞ N +1 n 0
Similarly, N
1 i N +1 N
P (B ∩ C ) = ∑ i =0
n +1
1
=∫x
n +1
dx when N → ∞
=
i ∑ n i =0 N
1
n +1
i ∑ n
n
i =0
=
∫x 0
1
=
n +1
dx if n is large
∫ x dx n
0
n +1 = n+2 6. (d) : Required probability = 1 – P (no test is missed) = 1 – P (no test on his first day of absence and no test of his second day of absence) = 1 – P (no test of his first day of absence) × P (no test of his second day of absence) 2 2 21 = 1− = 5 5 25 7. (c): Given that P ( A ∩ BC ) = 8 1 , P ( A) > , 25 2 Let P(A) = x, P(B) = y
3 25
and P ( AC ∩ B) =
\ We are given that P(A ∩ BC) = P(A) – P(A ∩ B) 3 = x − P ( A)P (B) 25 3 = x − xy [Q A and B are independent] 25 Also, P(AC ∩ B) = P(B) – P(A ∩ B) 8 1 ⇒ = y − xy. Hence y = x + 25 5 1 2 \ Solving x and y we get x = , y = 5 5 3 4 x = and y = or 5 5 14
=
0
Hence the required probability N
8. (a) : P(AC) = 0.3, P(B) = 0.4, P(ABC) = 0.5 \ P(A ∪ BC) = P(A) + P(BC) – P(A ∩ BC) = 0.7 + 0.6 – 0.5 = 0.8 Now, P(B |A ∪ BC)
mathematics today | February ‘17
=
P[B ∩ ( A ∪ BC )] P ( A ∪ BC ) P ( A ∩ B)
P ( A ∪ BC )
=
=
P[(B ∩ A) ∪ (B ∩ BC )] P ( A ∪ BC )
P ( A) − P ( A ∩ BC ) P ( A ∪ BC )
0. 7 − 0 . 5 0 . 2 1 = = 0.8 0.8 4
1 1 1 9. (c): Given that P ( A) = , P (B) = , P (C ) = 2 3 4 \ Probability that only two of them can solve the problem = P(A ∩ B ∩ CC) + P(A ∩ BC ∩ C) + P(AC ∩ B ∩ C) 1 1 3 1 2 1 1 1 1 6 1 = ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ = = 2 3 4 2 3 4 2 3 4 24 4 1 10. (a) : P (E1 ) = P (E2 ) = 2 3 5 and P (R |E1) = , P (R |E2 ) = 5 9 By Bayes' theorem P (E2 ) ⋅ P (R | E2 ) P (E2 | R) = P (E1 ) ⋅ P (R | E1 ) + P (E2 ) ⋅ P (R | E2 ) 1 5 5 ⋅ 25 2 9 = 9 = = 1 3 1 5 3 5 27 + 25 ⋅ + ⋅ + 2 5 2 9 5 9 25 P (E2 | R) = 52 11. (a) : Determinant of coefficients λ −1 cos θ 1 2 = cos θ − cos2 θ + 6 = 3 2 cos θ 1 and this is positive for all θ since |cosθ| ≤ 1. The only solution is therefore the trivial solution. dy dy 2a = 4a ⇒ = dx dx y For orthogonal trajectory, 12. (c): 2 y ⋅
mathematics today | February‘17
15
−
dx 2a = ⇒ dy y
dy 1 ∫ y = ∫ − 2a dx x
− x ⇒ ln y = − + k ⇒ y = ce 2a 2a
5z1 z 7K = Ki (K ∈ R), then 1 = i 7 z2 z2 5 z 2 1 + 3 14 K i + 3 2z1 + 3z2 z2 =1 = 5 = Consider 14 K 2z1 − 3z2 z1 i−3 2 −3 5 z2 14. (b) : 30 × 350 = 20 × 400 + 10 × x ⇒ 10x = 10500 – 8000 ⇒ 10x = 2500 ⇒ x = 250 13. (d) : Let
15. (c): Consider a point A′, the image of A through y=x
19. (b) : f (0 − 0) = lim f (0 − h) h→0
(a + 1 + 1)(−h) sin(a + 1)(−h) + sin(−h) = lim =a+2 = lim h→0 ( − h) h→0 0−h f (0 + 0) = lim f (0 + h) = lim h→0
(h + bh2 )1/2 − h1/2
h→0
bh b1/2 1 + − 1 =1 2 = lim 1/2 h→0 2 bh ⋅ h
bh3/2
\ The given function is continuous f (0 – 0) = f (0 + 0). 1 3 \ a+2= , a = − 2 2 f r (1) n = Cr 20. (a) : f (x ) = x n ⇒ r! n
∑ (n Cr )2 = 2nCn \ Coordinates of A′ = (4, 3) r =0 [Notice that A and B lie to 21. (a) : P( A ∪ B ∪ C ) = ∑ P( A) − P( AB) − P(BC ) − P( AC ) + P( A ∩ B ∩ C ) the same side with respect P( A ∪ B ∪ C ) = ∑ P( A) − P( AB) − P(BC ) − P( AC ) + P( A ∩ B ∩ C ) to y = x]. P ( A) − P ( AB) + P (B) − P (BC ) + P (B) Then PA = PA′ 1 Thus, PA + PB is minimum, = − P ( AB) + P (C ) − P (BC ) + P ( A) − P ( AC ) + P ( ABC ) 2 if PA′ + PB is minimum, if P, A′, B are collinear. + P (C ) − P ( AC ) 13 − 3 (x − 4) ⇒ 3 y − 10 x + 31 = 0 Now, AB is y − 3 = 1 7−4 = {P ( AB ) + P ( AB) + P ( AC ) + P (BC ) + P (BC ) + P ( AC )} 2 31 31 It intersects y = x at , , which is the required + P(ABC) 7 7 1 1 point P. = (3 p) + p2 = (3 p + 2 p2 ) 2 2 1 16. (c): lim 2 [log(2 + x 2 ) − log(2 − x 2 )] x →0 x 2 − x2 1 2 + x2 log log − 2 x →0 x 2 2
= lim
1/ x 2 1/ x 2 2 2 x x = lim log 1 + − log 1 − x →0 2 2 1 1 = log e1/2 − log e −1/2 = + = 1 2 2 17. (a) : The given series is, ∞
(2r + 1)
∞
1
∞
1
∑ r 2 (r + 1)2 = ∑ r 2 − ∑ (r + 1)2 = 1 r =1
r =1
r =1
18. (a) : f(1) = 1 ⇒ f(2) = 1 ⇒ f(3) = 1 f(r) = 1 ∀ r = 1, 2, ...., n n
∑ f (k) = n k =1
16
mathematics today | February ‘17
WEST BENGAL at
mathematics today | February‘17
17
(2x)2 = 4x ⇒ 4x2 = 4x ⇒ 4x2 – 4x = 0 ⇒ 4x(x – 1) = 0 ⇒ x = 0, 1 1
∫ (2
Required area =
22. (d) :
x − 2 x )dx
0
1
4 1 2 = 2 ⋅ x 3/ 2 − x 2 = − 1 = 3 3 3 0 Area =
2
∫
−2
5 − y2 −
27. (a) : Since f is continuous function so,
y 2 dy 4
lim f ( x ) = f (1).
x →1
Put x = 1 in the given equation we have (f (1))2 = 2f (1), so f(1) = 0 or 2. Since f (1) > 0, so f (1) = 2.
2
2 1 5 2 = 2 ∫ 5 − y 2 − y dy = 2 + sin −1 5 3 2 4 0
28. (c)
23. (a) :
1
0
f ′(x ) = ab x + b2 ac
bc
0
x + a2 ab bc + 0 1 x + c2
ac +
ac 0
bc x + c 2 x + a2
ab
ac
ab 0
x + b2 0
bc 1
= (x + b2)(x + c2) – b2c2 + (x + a2)(x + c2) – a2c2 + (x + a2)(x + b2) – a2b2 2 2 2 2 = 3x + 2x(a + b + c ) f (x) will be decreasing when f ′(x) < 0 ⇒ 3x2 + 2x(a2 + b2 + c2) < 0 2 ⇒ x ∈ − (a2 + b2 + c 2 ), 0 3 24. (d) : The sufficient condition for the statement : P(n) to be true for n ≥ a is (i) truth of P(n) ⇒ truth of P(n + 1) (ii) P(a) is true. Hence answer is none of these. 2 2 2 25. (b) : y = cos x + y ⇒ y − y = cos x dy dy \ 2y − = − sin x 2 ⋅ 2 x dx dx dy 2 x sin x 2 ⇒ =− dx 2y −1
1 12 1 1 (1 − x )(1 − y ) = , or 1 − (x + y ) + xy = 2 2 1 1 7 7 1 x+y= + = x + y = , xy = ; 2 12 12 12 12 1 1 P (E ) = and P (F ) = 3 4 1 1 or, P (E ) = and P (F ) = 4 3 29. (a) : Let P(E) = x, P(F) = y, xy =
n −1
n 30. (c): C0 + ∑
n+ k
= ( n + 2C 2 +
3)
k =0
+ n + 2C3 + .... + n + n – 1Cn = (n + 1C1 + n + 1C2) + n + 2C3 + .... + 2n – 1Cn Similarly,
n + 2C
2n – 1C
18
JEE MAIN
n–1 +
2n – 1C
2n – 1C
n
=
n 2nC
n
1st April to 30th April (Online) 2nd April (Offline) 8th & 9th April (Online)
VITEEE
5th April to 16th April (Online)
NATA
16th April
WBJEE
23rd April
Karnataka CET
mathematics today | February ‘17
+ .... +
Exam Dates 2017 SRMJEEE
Kerala PET
26. (b) :
Ck +1 = (n C0 + nC1 ) + n+1C2
24th April (Physics & Chemistry) 25th April (Mathematics) 2nd May (Biology & Mathematics) 3rd May (Physics & Chemistry)
BITSAT
16th May to 30th May (Online)
JEE Advanced
21st May
SECTION-1 Single OPTiOn CORReCT TyPe
1. Consider the two statements. S 1 : The number of solutions to the equation ex = x2 is 1. S 2 : The number of solutions to the equation ex = x3 is 1. (a) S1 is true, S2 is false. (b) S1 is true, S2 is true. (c) S1 is false, S2 is true. (d) S1 is false, S2 is also false.
2. The probability that a random chosen divisor of 3039 is a multiple of 3029 is 11 (a) 40
3
21 (b) 40
3
11 (c) 40
5
21 (d) 40
5
1 (c) 2 (d) 4 3 4. In a certain lottery, 7 balls are drawn at random from n balls numbered 1 through n. If the probability that no pair of consecutive numbers is drawn equals the probability of drawing exactly one pair of consecutive numbers, then n = (a) 50 (b) 60 (c) 54 (d) 64 8 27
(b)
7. An ellipse has foci at (9, 20) and (49, 55) in the xy plane and is tangent to x-axis. The length of its major axis is (a) 70 (b) 75 (c) 80 (d) 85 8. A rectangular box has faces parallel to co-ordinate planes. If two of its vertices are (1, –1, 0) and (2, 3, 6) then volume of the box is (a) 20 (b) 22 (c) 24 (d) 26 9. The number of distinct real number pairs (a, b) such that a + b ∈ integers and a2 + b2 = 2 is/are (a) 10 (b) 6 (c) 2 (d) 4
3. 4 persons are playing a game of toy-gun shooting. At exactly midnight, each person randomly chooses one of the other three and shoots him. The probability that exactly 2 persons are shot is (a)
6. Let A, B, C, D be distinct points on a circle with centre O. If there exists non-zero real numbers x and y such that | x OA + y OB | = | x OB + y OC | = | x OC + y OD | = | x OD + y OA |, then ABCD is (a) a square (b) a trapezium (c) a rectangle (d) none of these
4 9
5. If B is an idempotent matrix satisfying the condition (I – aB)–1 = I – 3B, where I is unit matrix of same order as B and a ∈ R then 2a = (a) 1 (b) 2 (c) –1 (d) 3
2p 10. Let n ≥ 3 be an integer and z = cis . Consider n the sets A = {1, z, z2, ....., zn –1} and B = {1, 1 + z, 1 + z + z2, ..., 1 + z + z2 + ... + zn – 1}. The number of elements in the set (A ∩ B) when n is even is (a) 0 (b) 1 (c) 2 (d) 3 11. Let a and b be positive real numbers such that a[a] = 17 and b[b] = 11 then (a – b) = (a) 3/7 (b) 4/7 (c) 7/12 (d) 5/12 ([·] denotes greatest integer function] 12. Let a function f(x), x ≠ 0 be such that 1 1 f (x) + f = f (x) ⋅ f then f (x) can be x x (a) 1 – x4 + x3 (c)
p 2 tan −1 | x |
(b)
| x | −x +1 2x (d) 1 + λ log | x |
By : Tapas Kr. Yogi, Mob : 9533632105. mathematics today | february ‘17
19
13. Assume that a (> 1) is a root of the equation x3 – x – 1 = 0 then (a) 1 (b) 2
3
3a2 − 4a + 3 3a2 + 4a + 2 = (c) 2a (d) a1/3
14. The maximum and minimum of the function 6sinx cosy + 2sinx siny + 3cosx is (a) 7 and –7 (b) 5 and –5 (c) 6 + 13 and 6 − 13 (d) 5 + 2 and 5 − 2 15. For a real number a, let p
I(a) = ∫ log(1 − 2a cos x + a2 ) dx then I(a) = 0
1 I(a2 ) 2 (d) I(–a2)
(a) I(a2)
(b)
(c) 2I(a2)
16. The minimum of the expression 4 2 16 x + 2 − 2 (1 + cos q)x + (1 + sin q) x x + (3 + 2cosq + 2sinq) for x > 0 and q ∈ [0, 2p] is (a) 2 2 − 1 (b) 2 2 + 3 (c) 2 2 + 1 (d) none of these 17. The area of the region contained by all the points (x, y) such that x2 + y2 ≤ 100 and sin(x + y) ≥ 0 is (a) 10p (b) 25p (c) 50p (d) 100p SECTION-2 COmPRehenSiOn TyPe
Passage-1 Let a be a fixed real number, a ∈(0, p) and Ir =
a
∫
1 − r cos u
2 −a 1 − 2r cos u + r
18. I1 = (a) 0
(b) a
du (c) 2a
(d) –a
19. lim Ir = + r →1
(a) a – p (b) a + p (c) a
(d) –a
20. lim Ir = − r →1
(a) a – p (b) a + p (c) a
(d) –a
Passage-2
Let r1 < 0 < r2 < r3 be the real roots of 8x 3 − 6x + 3 = 0 and let s 1 < 0 < s 2 < s 3 b e t he re a l ro ots of 8x3 – 6x + 1 = 0, then 20
mathematics today | february ‘17
21. 4r22 – 4r24 = (a) r12 (b) r22 22. r12 + s22 = (a) 1
(b) 2
(c) r32
(d) r32 – r12
(c) 4
(d) 8
SOLUTIONS
1. (a) : For x < 0, e x = x 2 has one solution and ex = x3 has no solution. x For x > 0, consider the function f (x) = log x log x − 1 \ f ′(x) = (log x)2 So, for x > e, f(x) increases and for x < e, f (x) decreases and moreover x = 1 is an asymptote. So, graph of f (x) is
Now, y = 2 does not intersect the graph but y = 3 meets the graph at 2 different points. Hence, in total ex = x2 intersects once and ex = x3 intersects twice. 2. (a) : 3039 = 239 × 339 × 539. So, 40 × 40 × 40 divisors in total. 2 a 3 b 5 c to be a multiple of 30 29 , we must have a ∈ [29, 39], b ∈ [29, 39], c ∈ [29, 39] i.e. 11 choices for each a, b, c. 3 11 × 11 × 11 11 = Hence, required probability = 40 × 40 × 40 40 3. (a) : Let the 4 persons were A, B, C, D. First, we choose which two were shot = 4C2. Let C, D were shot. Then C D and A → C or D and B → C or D. Any person shooting any other person has probability = 1/3. So, probability that C, D were shot 1 1 2 2 4 = × × × = 3 3 3 3 81 [for C, D] [for A, B]
Hence, probability that exactly 2 persons were shot 4 8 = 4C2 × = 81 27 4. (c) : There are n – 6C7 ways to draw 7 balls so that no two balls are consecutive and (n – 6) × n – 7C5 ways to draw 7 balls so that there is exactly one pair of consecutive balls. Hence, n – 6C7 = (n – 6) × n – 7C5 gives n = 54
5. (d) : From given relation, I = (I – 3B)(I – aB) i.e., I = I – aB – 3B + 3aB = I + 2aB – 3B Hence, 2a = 3 6. (a) : Squaring, | x OA + y OB |2 = | x OB + y OC |2 ⇒ x 2OA ⋅ OA + y 2OB ⋅ OB + 2xy OA ⋅ OB = x 2OB ⋅ OB + y 2OC ⋅ OC + 2xy OB ⋅ OC i.e., x 2r 2 + y 2r 2 + 2xy OA ⋅ OB = x 2r 2 + y 2r 2 + 2xy OB ⋅ OC Hence, ABCD is a square. 7. (d) : Let F1 and F2 be the foci and P be the point of tangency. Let F1′ be the image of F1 in x-axis then F1P + F2P = F1′P + F2P = F1′F2 = 2a = (40)2 + (75)2 = 85 8. (c) : Vertices are (1, –1, 0) and (2, 3, 6). Notice that these must be diagonal vertices. Making (1, –1, 0) as origin, we have (2, 3, 6) as (1, 4, 6). Hence, dimensions of the box are 6 × 4 × 1 i.e. Volume = 24 cubic units
12. (2) : Rearranging the given fractional equation, we have 1 1 , f −1 = x f (x) − 1 p which is satisfied by only f (x) = , out of the 2 tan −1 x four options. 13. (b) : Since, a3 – a – 1 = 0, we have 3
3a2 − 4a = 3 3a2 − 4a − (a3 − a − 1) = 3 (1 − a)3 = 1 − a
and 3 3a2 + 4a + 2 + a3 − a − 1 = 3 (1 + a)3 = 1 + a Hence, given expression = 2 14. (a) : Using Cauchy-Schwarz inequality \ (x1y1 + x2y2 + x3y3)2 ≤ (x12 + x22 + x32) (y12 + y22 + y32) We have, (6sinx cosy + 2sinx siny + 3cosx)2 ≤ (62 + 22 + 32) ((sinxcosy)2 + (sinx siny)2 + cos2x) i.e. (6sinx cosy + 2sinx siny + 3cosx)2 ≤ 49 Hence, maximum = 7 and minimum = –7
9. (b) : Using, R.M.S ≥ A.M., we have |a + b| ≤ 2 So, a + b ∈ {–2, –1, 0, 1, 2} and a2 + b2 = 2 gives 1± 3 1 3 , (a, b) = (1, –1), (–1, 1), , 2 2 −1 ± 3 −1 3 , 2 2 10. (c) : Clearly, 1 ∈ A ∩ B. Let w ∈ A ∩ B, w ≠ 1 As an element of B, for some k = 1, 2, 3, ... (n – 1) 1 − z k +1 1− z For w ∈ A, we have |w| = 1 w = 1 + z + z2 + .... + zk =
(k + 1)p 1 − z k +1 p = 1 ⇒ sin = sin n 1− z n i.e. k = n – 2 1 − 1 / z −1 = So, w = 1− z z −1 So, A ∩ B = 1, z ⇒
{ }
11. (c) : a[a] = 17 ⇒ a ∈ (4, 5). So, [a] = 4 11 17 and a = . Similarly, b[b] = 11 ⇒ b = 3 4 7 Hence, a − b = 12 mathematics today | february ‘17
21
15. (b) : We have, I(a) = I(–a) by putting x = p – y. Now, we have (1 – 2acosx + a2)(1 + 2acosx + a2) = 1 – 2a2 cos2x + a4
Putting tan(u/2) = t and simplifying Jr =
1+ r a tan −1 tan 1− r 2 |1 − r | 4
p
So, I(a) + I(−a) = ∫ log(1 − 2a2 cos 2x + a 4 ) dx
So, Ir = a +
0
=
1 2p 2 4 ∫ log(1 − 2a cos y + a ) dy, ( y = 2x) 2 0
= a − 2⋅
Putting y = 2p – t we have
∫ log(1 − 2a
2
4 a (1 − r 2 ) 1+ r tan −1 tan ⋅ 2 2 1− r 2 |1 − r |
1+ r a So, lim Ir = a − 2 tan −1 tan + r −1 2 r →1 r →1
1 1 2p = I(a2 ) + ∫ log(1 − 2a2 cos y + a 4 ) dy 2 2 p
2p
2
p =a−p 2
a −1 1 + r tan 20. (b) : lim− Ir = a + 2 tan 1− r 2
cos y + a 4 ) dy = I(a2)
r →1
p
= a + 2⋅
1 1 Hence, I(a) + I(−a) = I(a2 ) + I(a2 ) 2 2
3 2 p 2p 7p 8p 13p 14p ⇒ q= , , , , , 9 9 9 9 9 9
sin 3q =
16. (d) : The given expression can be rearranged as 2
4 which is square of distance between point x, x on hyperbola xy = 4 and point (1 + cosq, 1 + sinq) on a circle centre (1, 1) and radius 1. Clearly the minimum distance occurs at (2, 2) on the hyperbola 1 1 and 1 + , 1+ on the circle. 2 2 2
1 So, minimum distance = 2 2 − 1 + = 2 −1 2 17. (c) : Using the symmetry, required area of the region 1 is × 100p = 50p sq. units 2 a a 1 1 − cos u 18. (b) : I1 = ∫ du = ∫ du = a 2 2 − 2 cos u −a −a 1 19. (a) : For r > 0, Ir = a + (1 − r 2 ) ⋅ 2 (1 − r 2) Ir = a + ⋅ J r (let) 2 Jr = 22
a
∫
du
−a 1 − 2r cos u + r
a
∫
du
−a 1 − 2r cos u + r
2
mathematics today | february ‘17
p =a+p 2
21. (c) : Putting x = sinq in 8x 3 − 6x + 3 = 0 , we have
1 So, I(a) = I(−a) = I(a2 ) 2 4 [x − (1 + cos q)]2 + − (1 + sin q) x
r →1
2
Hence, r1 = sin
in [0, 2p]
13p 4p p 2p = − sin , r2 = sin , r3 = sin 9 9 9 9
Similarly, putting x = cosq in 8x3 – 6x + 1 = 0, 1 we have, cos 3q = − 2 8p p So, s1 = cos = − cos , and 9 9 4p 2p s2 = cos , s3 = cos 9 9 p p 2p 2 Hence, 4r22 − 4r24 = 4 sin2 − 4 sin 4 = sin2 = r3 9 9 9 4p 4p 22. (a) : r12 + s22 = sin2 + cos2 =1 9 9
MPP-8 CLASS XII 1. 6. 11. 15. 20.
(c) 2. (d) (c) 7. (a,b,c) (a,b,c,d) (d) 16. (d) (3)
3. 8. 12. 17.
ANSWER KEY (d) (a,b,c) (a,b) (2)
4. 9. 13. 18.
(a) (c,d) (b,c) (1)
5. 10. 14. 19.
(b) (a,b) (a) (1)
Y U ASK
WE ANSWER Do you have a question that you just can’t get answered? Use the vast expertise of our mtg team to get to the bottom of the question. From the serious to the silly, the controversial to the trivial, the team will tackle the questions, easy and tough. the best questions and their solutions will be printed in this column each month.
1. The base of a D is divided into three equal parts. If t1, t2, t3 be the tangents of the angles subtended by these parts at the opposite vertex, prove that: 1 1 1 1 1 (Yogesh, Delhi) t + t t + t = 4 1 + 2 2 3 1 2 t2 A Ans. Let the points P and Q divide the side BC in three equal parts α γ β such that BP = PQ = QC = x Also let, ∠BAP = a, ∠PAQ = b, θ ∠QAC = g and ∠AQC = q B P Q C From question, tan a = t1, tan b = t2, tan g = t3. Applying, m : n rule in triangle ABC we get, (2x + x) cot q = 2x cot (a + b) – x cot g ...(1) From DAPC, we get (x + x) cot q = x cot b – x cot g ...(2) Dividing (1) by (2), we get 3 2 cot (a + b) − cot g = cot b − cot g 2 4 (cot a ⋅ cot b − 1) or 3 cot b – cot g = cot b + cot a or 4 + 4 cot2 b = cot2 b + cot a ⋅ cot b + cot b ⋅ cot g + cot g ⋅ cot a or 4(1 + cot2 b) = (cot b + cot a)(cot b + cot g) 1 1 1 1 t or 4 1 + 2 = + + t2 t1 t2 t2 t3 2. If a regular polygon of n sides has the circumradius R and inradius r then prove that each side of the polygon is equal to π 2(R + r)tan . (Raman, Gujarat) 2n Ans. Let A1A2 be a side and O be the centre. Let OB ⊥ A1A2. Clearly, OA1 = R and OB = r. 2π π Also, ∠A1OA2 = and ∠A1OB = . n n
r π = cos ∠A1OB = cos ...(1) R n AB π π and 1 = tan , i.e., A1B = r tan n n OB π \ A1A2 = 2A1B = 2r tan ...(2) n π π r Now, 2(R + r) tan =2 + r tan , (from (1)) π 2n 2n cos n π π 1 + cos 1 − cos q 1 − cos q n ⋅ n tan = = 2r π π 2 sin q cos sin n n π π 1 − cos2 sin2 n = 2r ⋅ n = 2r ⋅ π π π π cos ⋅ sin cos ⋅ sin n n n n π = 2r tan = A1A2, (from (2)). n 3. In how many ways can two distinct subsets of the set A of k(k ≥ 3) elements be selected so that they have exactly two common elements? (Akhil, Assam) Ans. Let the two subsets be called P and Q. The elements for the two sets will be selected as follows: (i) 2 elements out of k elements for both the sets. This can be done in kC2 ways. (ii) r elements for the subset P from k – 2 elements and any number of elements for Q from the remaining k – 2 – r elements. Here r can vary from 0 to k – 2. For a fixed r, the number of selections = k – 2Cr ⋅ 2k – 2 – r, (because the number of selections of any number of things from n things is 2n.) If r varies from 0 to k – 2, the total number of selections In DA1OB,
k−2
= ∑
r =0
k−2
Cr ⋅ 2
k−2−r
− 1,
excluding the case when both the subsets are equal having only the two common elements. But every pair of P, Q is appearing twice like {a1, a2, a3}, {a1, a2, a4, a5} and {a1, a2, a4, a5}, {a1, a2, a3}. Hence, the required number of ways k−2 1 k−2 k−2−r Cr ⋅ 2 − 1 = k C2 × ∑ 2 r = 0 1 k(k − 1) k – 2 = ⋅ ⋅ [( C0 ⋅ 2k – 2 + k – 2C1 ⋅ 2k – 3 2 2 + k – 2C2 ⋅ 2k – 4 + ... + k – 2Ck – 2) – 1] k(k − 1) k(k − 1) k − 2 k−2 ⋅ {(2 + 1) − 1} = ⋅ (3 − 1). = 4 4 mathematics today | February‘17
23
*ALOK KUMAR, B.Tech, IIT Kanpur single option correct type
3p/ 4
∫
e p/ 4dx
=k
p /2
∫
sec xdx ,
1. If f(x) = 0 is a cubic equation with positive and distinct roots a, b, g such that b is the H.M. of the roots of f ′(x) = 0. Then a, b, g are in (a) A.P. (b) G.P. (c) H. P. (d) none of these
7. If
2. All the roots of the equation 11z10 + 10iz9 + 10iz – 11 = 0 lie (a) inside |z| = 1 (b) on |z| = 1 (c) outside |z| = 1 (d) none of these
8. A line meets the co-ordinate axes at A and B. A circle is circumscribed about DOAB. If the distances from A and B of the tangent to the circle at the origin be m and n respectively, then diameter of the circle is (a) m(m + n) (b) m + n (c) n(m + n) (d) m2 + n2
3. Let y = f(x) be a continuous and differentiable curve. The normals at (1, f (1)), (2, f (2)) and (3, f (3)) make angles p/3, p/4 and p/6 with positive x-axis respectively. Then value of 2
3
1
2
∫ ( f (x) + xf ′(x)) dx + ∫ f ′(x) f ′′(x) dx
(a) f ′(2) + f (1) + 1 (c) f (3) – f (2) + 1 4.
lim
(b) 2f (2) – f(1) + 1 (d) none of these
sin(p cos2(tan(sin x)))
x →0
(a) p (c) p/2
x
2
is equal to
is equal to
(b) p/4 (d) none of these
5. If a = x + y + z + w and b = (xy + yz + zw + wx + wy + xz), then which of the following statement(s), is/are true? (a) 8a2 ≥ 3b (b) 3a2 ≥ 8b 2 (c) a b ≥ 27 (d) none of these ^ ^ ^ ^ ^ 6. If a = i + j + k , a ⋅ b = 1 and a × b = j − k , then b is equal to ^ ^ ^ ^ (a) 2 i (b) i − j + k (c) i^
^
^
(d) 2 j − k
− p/ 4 (e
x
+ e p/ 4 )(sin x + cos x)
then the value of k is 1 1 1 1 (a) (b) (c) (d) − 2 2 2 2 2
9. If I1 =
101
dx
∫
−100 (5 + 2x − 2x 101
)(1 + e 2− 4x )
∫
2
| x |, for 0 < | x | ≤ 3 10. Let f (x) = , then at x = 0 f(x) x =0 1, for has (a) a local maximum (b) no local maximum (c) a local minimum (d) no extremum 15 5 3 11. If x 2 + 9 y 2 + 25z 2 = xyz + + , then x, y, z x y z are in (a) A.P. (b) G.P. (c) A.G.P. (d) H.P. 12. The set of values of ‘r’ for which 23 Cr + 2 ⋅ 23Cr +1+ 23 Cr +2 ≥ 25C15 contains (a) 3 elements (b) 4 elements (c) 5 elements (d) 6 elements
He trains IIT and Olympiad aspirants.
mathematics today | February ‘17
and
I , then 1 is I 2 −100 5 + 2x − 2x (a) 1/2 (b) 2 (c) 0 (d) none of these I2 =
dx
2
* Alok Kumar is a winner of INDIAN NATIONAL MATHEMATICS OLYMPIAD (INMO-91). 24
− p /2
13. In an equilateral triangle inradius(r), circumradius (R) and exradius (r1) are in (a) A.P. (b) G.P. (c) H.P. (d) none of these 14. If tanA, tanB are the roots of the quadratic equation abx2 – c2x + ab = 0 where a, b, c are sides of the DABC then sin2A + sin2B + sin2C is (a) 1 (b) 2 (c) 3 (d) 3/2 15. The real function f (x) = cos −1 x 2 + 3x + 1 + cos −1 x 2 + 3x is defined on the set (a) {0, 3} (b) (0, 3) (c) {0, –3} (d) (–3, 0) 16. If a function satisfies f (x + 1) + f (x − 1) = 2 f (x) , then the period of f(x) can be (a) 2 (b) 4 (c) 6 (d) 8 17. If (1 + x)n = nC0 + nC1x + nC2x2 + .... + nCn xn where nC , nC , nC , ... are binomial coefficients. Then the 0 1 2 value of 2(C0 + C3 + C6 + ...) + (C1 + C4 + C7 + ...)(1 + w) + (C2 + C5 + C8 + ...)(1 + w2), where w is the cube root of unity and n is a multiple of 3, is equal to (a) 2n + 1 (b) 2n – 1 + 1 n + 1 (c) 2 –1 (d) 2n – 1 18. The chords of contact of the pair of tangents drawn from each point on the line 2x + y = 4 to x2 + y2 = 1 pass through the point (a) (1, 2) (b) (1/2, 1/4) (c) (2, 4) (d) none of these 19. If
a bc
−2=
b c , where a, b, c > 0, then family + c b
of lines ax + b y + c = 0 passes through the fixed point given by (a) (1, 1) (b) (1, –2) (c) (–1, 2) (d) (–1, 1) 20. If f(x) is continuous such that |f(x)| ≤ 1 " x ∈ R and e f (x) − e f (x) , then range of g(x) is e f (x ) + e f (x ) e2 + 1 (a) [0, 1] (b) 0, 2 e − 1 g (x) =
e2 − 1 (c) 0, 2 e + 1
1 − e2 (d) , 0 2 1 + e
21. Let three non-collinear points A, B and C have position vector a , b and c , then the vector a × b + b × c + c × a is
(a) parallel to plane containing a, b and c (b) perpendicular to the plane formed with the vectors AB and AC (c) perpendicular to the plane containing a, b and c (d) none of these 22. 271 should be split into how many parts so as to maximize their product? (a) 99 (b) 10 (c) 105 (d) none of these more than one option correct type
p and cos a + cos b = 1 , then 2 p (a) a + b ≥ (b) cos(a + b) ≤ 0 2 p (c) a + b ≤ (d) cos(a + b) ≥ 0 2
23. If 0 ≤ a, b ≤
24. If lim
ae 2x − b cos 2x + ce −2x − x sin x
= 1 and x2 f (t) = (a + b)t2 + (a – b)t + c, then (a) a + b + c = 1 (b) a + b + c = 2 (c) f (1) = 3/4 (d) f (1) = 1 x →0
25. If the conics whose equations are S1 : (sin2q)x2 + (2htanq)xy + (cos2q)y2 + 32x + 16y + 19 = 0 S2 : (cos2q)x2 + (2h′cotq)xy + (sin2q)y2 + 16x + 32y + 19 = 0 p intersect in four concyclic points, where q ∈ 0, , 2 then the correct statement(s) can be (a) l = 0 (b) l = 0 (c) q = p/4 (d) none of these 26. ABCD is a square of side 1 unit. P and Q lie on the side AB and R lies on the side CD. The possible values for the circumradius of triangle PQR is ? (a) 0.5 (b) 0.6 (c) 0.7 (d) none of these ^ ^ ^ ^ ^ ^ 27. If a = i + j + k, a ⋅ b = 2 and a × b = 2 i + j − 3 k, then ^ ^ ^ (a) a + b = 5 i − 4 j + 2 k ^ ^ ^ (b) a + b = 3 i + 2 k (c) b = 2 ^i − ^j + k ^ (d) b = ^i − 2 ^j − 3 k 28. The solution of x1/3 + (2x – 3)1/3 = [3(x – 1)]1/3 is (a) 0 (b) 3/2 (c) 1 (d) none of these mathematics today | February ‘17
25
29. Let a1, a2, a3, ..., an are in G.P. such that 3a1 + 7a2 + 3a3 – 4a5 = 0, then common ratio of G.P. can be (a) 2 (b) 3/2 (c) 5/2 (d) –1/2 30. The diagonals of a square are along the pair represented by 2x2 – 3xy – 2y2 = 0. If (2, 1) is the vertex of the square, then the other vertices are (a) (–1, 2) (b) (1, –2) (c) (–2, –1)(d) (1, 2) x ∫ (5 + | 1 − t |) dt , if 31. Let f (x) = 0 5x + 1, if function is (a) continuous at x = 2 (b) differentiable at x = 2 (c) discontinuous at x = 2 (d) not differentiable at x = 2
x >2
then the
x ≤2
32. The length of other diagonal BD is 10 3
(b) 2
(c)
20 3
(d) none of these
33. The length of side AB is equal to (a)
4 58 3 58 2 58 5 58 (b) (c) (d) 9 9 3 9
34. The length of BC is equal to 4 10 2 10 8 10 (a) (b) (c) (d) none of these 3 3 3 matriX match type
35. Match the following. Column I sin 1 sin 5 − (A) sin 2 sin 6 (B)
3 9 tan − 2 4
Column II P.
positive
Q. negative
e x − 1 (C) lim (where [⋅] denotes R. 1 x →0 x the greatest integer function) S. d o e s n ot exist 26
Column I
Column II 24 (A) The area enclosed between the P. 5 2 curves |x| + |y| = 2 and x = y (in sq. units) is 7 (B) The maximum value of the function Q. 3 7 f (x) = 3 sin x − 4 cos x − 3 will be given by 16 (C) The length of common chord of R. 3 two circles of radii 3 and 4 units which intersect orthogonally is
comprehension type Let ABCD be a parallelogram whose diagonals equations are AC : x + 2y = 3; BD : 2x + y = 3. If length of diagonal AC = 4 units and area of ABCD = 8 sq. units.
(a)
36. Match the following.
mathematics today | February ‘17
(D) The length of chord intercepted by S. the parabola y2 = 4(x + 1) passing through its focus and inclined at 60° with positive x-axis is
8 3
integer type
37. If the normal to the curve x = t – 1, y = 3t2 – 6 at the point (1, 6) make intercepts a and b on x and y-axes a +12b respectively, then the value of is______ 146 38. Let S(n) denotes sum of first n terms of an A.P., n f (r ) then the value of S = lim ∑ , where n→∞ r =−n n S(3n) , is ______ f (n) = S(2n) − S(n) 39. If f (u, v, w, x) = u2 + v2 + w2 + x2 – 2(u + v + w + x) + 10 with u, w ∈ [–3, 3] and v, x ∈ [1, 3] then 1 max ⋅ f (u, v, w, x) is 23 40. If a = 2, b = 3, and c = 4 then 2 2 2 a −b + b −c + c −a cannot exceed _____ 87 solutions
1. (b) : f(x) = (x – a)(x – b)(x – g) ⇒ f ′(x) = 3x2 – 2x(a + b + g) + ab + bg + ga 2a1b1 ⇒ b= (where a1, b1 are the roots of f ′(x) = 0) a1 + b1 ⇒ b=
2(ab + bg + ga) ⇒ 2(a + b + g)
b2 = ga
11 − 10iz 2. (b) : z (11z + 10i) = 11 − 10iz ⇒ z = 11z + 10i i z + 11 10 ⇒ | z9 | = 11z + 10i 9
9
⇒ | 11i + 10z |2 − | 11z + 10i |2 = 21(1− | z |2 ) 9
⇒ if | z | < 1, then | z | > 1 (not possible)
Also a ⋅ b = 1 ⇒ a + b + g = 1 ⇒ b+1+b+b=1 ⇒ b=0 \ a = 1, g = 0 ^ \ b=i 7. (c) : I =
3p/ 4
⇒ |z| = 1 3. (b) : So,
=
∫ xf ′(x) dx
1
2 = xf (x)|12 − ∫ 1
2
3
1
2
f (x) dx
∫ ( f (x) + xf ′(x))dx + ∫ f ′(x) f ′′(x) dx =xf
I=
2 3 2 ( f ′(x)) (x) 1 + 2 2
1 = 2 f (2) − 1 f (1) + [( f ′(3))2 − ( f ′(2))2] 2 1 = 2 f (2) − f (1) + [3 − 1] = 2 f (2) − f (1) + 1 2 4. (a) : lim
(
sin p (1 − sin2 ( tan(sin x))) x
x →0
2
)
sin ( p sin (tan(sin x))) p sin ( tan(sin x)) x →0 p sin2 ( tan(sin x)) tan2(sin x) 2
= lim
2
tan2 (sin x ) sin2 x 2 = p 2 sin x x
2 − p/2 (e + 1)cos t
2 − p/2 (et + 1)cos t
Adding, 2I = \ I=
1 p /2
∫
2 − p /2
p /2
1
∫
2 2 − p /2
sec tdt
sec xdx
a b 8. (b) : Circumcentre of DOAB ≡ , 2 2 a 2 b2 + 4 4
Circumradius =
Equation of circle x2 + y2 – ax – by = 0 \ Equation of tangent at origin ax + by = 0 AL =
a2
⇒ 2I1 =
, BM =
a 2 + b2
(0, b) B m M L
b2
a 2 + b2
n
B(a, 0)
,
AL + BM = m + n = a2 + b2 = diameter of circle.
⇒ I1 =
101
dx
∫
−100 (5 + 2x − 2x
101
2
)(1 + e 2− 4x )
dx
∫
2 2 − 4(1− x) ) −100 (5 + 2(1 − x) − 2(1 − x) )(1 + e 101
I 1 = I2 ⇒ 1 = I2 2 −100 5 + 2x − 2x
∫
^
⇒ b – g = 0, a – g = 1, a – b = 1 ⇒ b = g, a = 1 + g, a = 1 + b,
dt
t
et dt
^ ^ ^ 6. (c) : Let b = a i + b j + g k, ^
∫
∫
9. (a) : I1 =
^
1 p /2
1 p /2
5. (b) : Since, 3(x2 + y2 + z2 + w2) – 2(xy + yz + zx + zw + wx + wy) = (x – y)2 + (x – z)2 + (x – w)2 + (y – z)2 + (y – w)2 + (z – w)2 ≥ 0 2 2 ⇒ 3Sx2 – 2Sxy ≥ 0 ⇒ Sx ≥ Sxy 3 2 2 2 Now, (x + y + z + w) = Sx + 2Sxy ≥ Sxy + 2Sxy 3 8 8 ⇒ (x + y + z + w)2 ≥ Sxy ⇒ a2 ≥ b 3 3
i j k ^ ^ ^ ^ a ×b = j −k ⇒ 1 1 1 = j −k a b g
p 2(e x − p/4 + 1)cos x − 4
− p/ 4
and if | z | > 1 ⇒ | z 9 | < 1 (not possible) 2
dx
∫
dx
2
(0, 1)
10. (a) : �3
3
x2
9y2
11. (d) : We have, + + 25z2 = 15yz + 5zx + 3xy 2 2 2 ⇒ (x) + (3y) + (5z) – (x)(3y) – (3y)(5z) – (x)(5z) = 0 mathematics today | February ‘17
27
f (x + 8) = –f (x + 4) = f (x) " x 1 = 0f(x) is periodic with period 8. 2(x)2 + 2(3 y)2 + 2(5z)2 − 2(x)(3 y) − 2(3 y)(5z) − 2(x)(5z)\ 2 1 17. (d) : (1 + w)n = C0 + C1w + C2w2 + ... + Cnwn 2 2 2 2(x) + 2(3 y) + 2(5z) − 2(x)(3 y) − 2(3 y)(5z) − 2(x)(5z) = 0 2 (1 + 1)n = C0 + C1 + C2 + ... + Cn 1 (1 + w)n + (1 + 1)n = 2C0 + C1(1 + w) + C2(1 + w2) ⇒ (x − 3 y)2 + (3 y − 5z)2 + (x − 5z)2 = 0 2 + C3(1 + w3) + C4(1 + w) + C5(1 + w2) + C6(1 + w3) ⇒ x – 3y = 0, 3y – 5z = 0, x – 5z = 0 + ... + Cn(1 + wn) 1 1 5 1 1 1 = , = and = ⇒ = 2(C0 + C3 + C6 + ....) + (C1 + C4 + C7 + ...)(1 + w) x 3y 3y z 5z x + (C2 + C5 + C8 + ...)(1 + w2) = –w2n + 2n 1 1 1 5 2 + = + = ⇒ ⇒ 2n – 1 (Q n is a multiple of 3, wn = 1) x z 3y 3y y 18. (b) : Let (h, k) be any point on the given line 1 1 1 ⇒\ , , are in A.P. ⇒ x, y, z are in H.P. \ 2h + k = 4 and the chord of contact hx + ky = 1 x y z hx + (4 – 2h)y = 1 ⇒ (4y – 1) + h(x – 2y) = 0, p + lq = 0, it 12. (d) : 23Cr + 2 · 23Cr + 1 + 23Cr + 2 = 24Cr + 1 + 24Cr + 2 passes through the intersection of p = 0, q = 0 or = 25Cr + 2 ≥ 25C15 (1/2, 1/4). ⇒
\ (r + 2) can be 10, 11, 12, 13, 14 and 15. So, 6 elements. 13. (a) : In an equilateral triangle r = R/2 A B C Also ex-radius r1 = 4R sin cos cos 2 2 2 1 3 3 3 = 4R . . = R 2 2 2 2 ⇒ r, R, r1, are in A.P. c2 and tan A·tan B = 1 ab tan A ⋅ tan B = 1 ⇒ tan A = cot B ⇒ A = 90° – B ⇒ A + B = 90° ⇒ C = 90° a b ⇒ sin A = , sin B = c c 14. (b) : Here, tan A + tan B =
a 2 b2 \ sin2 A + sin2 B + sin2 C = 2 + 2 + 1 = 2 c c 2 15. (c) : Obviously, 0 ≤ x + 3x + 1 ≤ 1 and 0 ≤ x2 + 3x ≤ 1 ⇒ x2 + 3x = 0 ⇒ x = 0, –3 16. (d) : By replacing x = x + 1 and x = x – 1, we get ...(1) f (x + 2) + f (x) = 2 f (x + 1) ...(2) f (x) + f (x − 2) = 2 f (x − 1) Adding (1) & (2) gives, f (x + 2) + f (x − 2) + 2 f (x) = 2 [ f (x + 1) + f (x − 1)] = 2 2 f (x) \ f (x + 2) + f (x – 2) = 0 On replacing x by x + 2 we get f(x + 4) + f(x) = 0 Finally replacing x by x + 4, we get f (x + 8) + f (x + 4) = 0, 28
mathematics today | February ‘17
19. (d) :
a
bc
−2=
b c + ⇒ a = b + c + 2 bc c b
⇒ a = ( b + c )2 ⇒ ( a − b − c )( a + b + c ) = 0 ⇒ a − b − c = 0, a + b + c ≠ 0 (Q a, b, c > 0) Comparing with ax + b y + c = 0 Hence x = –1, y = 1. e f (x) − e| f (x)| 20. (d) : We have, g (x) = f (x) | f (x)| , −1 ≤ f (x) ≤ 1 e +e For 0 ≤ f(x) ≤ 1, g(x) = 0 For –1 ≤ f (x) < 0, e f (x ) − e − f (x ) e 2 f (x ) − 1 2 g (x) = f (x) − f (x) = 2 f (x) = 1 − 2 f (x ) e +e e +1 e +1 1 − e2 , 0 For –1 ≤ f (x) < 0, g (x) ∈ 2 1 + e 1 − e2 For –1 ≤ f (x) < 1, g (x) ∈ , 0 . 2 1 + e 21. (c) : Let r be the position vector of any point on the plane of a, b and c ⇒ (r − a) ⋅ (b − a) × (c − a) = 0 ⇒ (r − a) ⋅ (a × b + b × c + c × a) = 0 Thus a × b + b × c + c × a is perpendicular to r − a and thus is perpendicular to a, b and c . 22. (d) 23. (a, b) : 0 ≤ a, b ≤ p and cos a + cos b = 1 2 a −b 1 a +b cos = ⇒ cos 2 2 2
mathematics today | February ‘17
29
1 a −b a +b and cos ≥ ⇒ atleast one of cos 2 2 2 p ⇒ atleast one of a + b and a − b ≥ 2 p p but a − b ≥ and a + b ≥ and cos(a + b) ≤ 0 2 2 24. (a, c) : lim
ae 2x − b cos 2x + ce −2x − x sin x x2
x →0
=1
(2x)2 (2x)4 (2x)2 a 1 + 2x + + ... − b 1 + + − ... 2! 2! 4! (2x)2 +c 1 − 2x + − ... 2!
⇒ lim
x
x →0
⇒ a–b+c=0 2a – 2c = 0 2a – 2b + 2c = 2 ⇒ a = c and b = 2a ⇒
a + 2a + a = 1 ⇒ a =
2
\
2
b=
=2
3 1 1 3t − t + 1 Now, f (t ) = t 2 − t + = 4 4 4 4 25. (b, c) : Curve through the intersection of S1 and S2 is given by S1 + lS2 = 0 ⇒ x2(sin2 q + l cos2 q) + 2(h tan q – lh′ cot q)xy
+ (cos2 q + l sin2 q)y2 + (32 + 16l)x + (16 + 32l)y + 19(1 + l) = 0 The above equation will represent a circle if sin2 q + l cos2 q = cos2 q + l sin2 q ⇒ sin2 q – l sin2 q = cos2 q – l cos2 q ⇒ (1 – l)sin2 q = (1 – l)cos2 q p ⇒ (1 – l)(sin2 q – cos2q) = 0 ⇒ l = 1 or q = 4 h tan q – lh′ cot q = 0 ⇒ h tan q = lh′ cot q which is satisfied if l = 1 and q = p/4 ⇒ h = h′ 26. (a, b, c) : Let O be the circumcentre. Then OP + OR ≥ PR ≥ AD = 1, so the radius is at least 1/2. P, Q, R always lie inside or on the circle through A, B, C, D 1 1 which has radius , so the radius is at most . 2 2
30
mathematics today | February ‘17
28. (a, b, c) : a1/3 + b1/3 = (a + b)1/3 is true, when either a = 0 or b = 0 or a + b = 0 3 ⇒ x = 0, , 1 2 ⇒ 7(a1 + a2 + a3) = 4(a1 + a3 + a5)
1 2
27. (b, c) : We have (a × b ) × a = (a ⋅ a) b − (b ⋅ a) a = | a |2 b − 2a
(4 ^i − 5 ^j + k^) + 2(^i + ^j + k^) ^ ^ ^ \ b= = 2 i − j + k. 3
29. (b, d) : Given 3a1 + 7a2 + 3a3 – 4a5 = 0
...(1) ...(2) ...(3) 1 =c 4
(a × b ) × a + 2a \ b= | a |2 ^ ^ ^ Now, (a × b ) × a = 4 i − 5 j + k and | a | = 3
⇒ 7(1 + r + r2) = 4(1 + r2 + r4)
⇒ 7 = 4(r2 – r + 1) ⇒ 4r2 – 4r + 1 = 4 ⇒ (2r – 1)2 = 4 3 1 r= ,− 2 2 30. (a, b, c) : Since the diagonals intersect at origin and are at right angles. Let B and D be the points adjacent to A. ⇒ 2r – 1 = ±2 ⇒
2
D O A
2
Also OA = 2 + 1 = 5
(0, 0) B
(2, 1)
Let affix of B and D are z2 and z1 respectively. z p \ arg 2 = 2+i 2
⇒ z2 = (2 + i)i = –1 + 2i ⇒ B(–1, 2) p z Also, arg 4 = − 2+i 2 ⇒ z4 = (2 + i)(–i) = 1 – 2i ⇒ D(1, –2) x
1
0
0
x
31. (c, d) : x > 2∫ (5 + | 1 − t |) dt = ∫ (6 − t )dt + ∫ (4 + t ) dt = 1+ 4 +
2
1
x 2
x2 1 + 4x + , x > 2 ⇒ f (x) = 2 5x + 1, x ≤2 4 + x, x > 2 f ′(x) = x ≤2 5, f(2+) = f(2–) = 11 continuous at x = 2 f ′(2+) ≠ f ′(2–) ⇒ not differentiable at x = 2
A
1 − +2 3 3 32. (c) : tan θ = 2 = , sin θ = , 1+1 4 5 1 20 area of ∆CPB = × PC × PB sin θ = 2 ⇒ BD = 2 3 AP 2 + PB2 − AB2 33. (a) : cos(π − θ) = 2AP ⋅ PB 100 2 4 4 + 9 − AB 2 58 ⇒ AB = ⇒ − = 10 5 3 2×2× 3 PC 2 + PB2 − BC 2 34. (a) : ln ∆CPB, cos θ = 2PC ⋅ PB ⇒ BC =
2 10 3
35. (A)-(Q), (B)-(P), (C)-(S) sin1 sin 5 sin x (A) − , Take f (x) = sin 2 sin 6 sin(x + 1) sin(x + 1)cos x − cos(x + 1)sin x sin1 f ′(x) = = 2 >0 2 sin (x + 1) sin (x + 1) ⇒ f(x) is increasing ⇒ f(1) < f(5). (B) Take f (x) = tanx – x2 f ′(x) = sec2x – 2x, f ′′(x) = 2sec2x tanx – 2 f ′(1) > 0 and f ′′(x) > 0 for x > 1 3 9 π ⇒ f (x) is increasing in 1, ⇒ tan > 2 2 4 e x − 1 (C) lim does not exist as left hand and right x →0 x hand limits are not equal. 36. (A) - (Q), (B)-(S), (C)-(P), (D)-(R) 1 1 (A) Required area = 2 (2 + 1) × 1 − ∫ x 2dx 2 0 3 1 7 = 2 − = sq. units 2 3 3 7 8 (B) 5 − = 3 3 (C) Length of chord =
2r1r2
r12 + r22
(D) Parabola is y2 = 4(x + 1)
=
2 × 3 × 4 24 = 5 5
... (i), focus is (0, 0) x −0 y −0 = =r Equation of AB is 1/ 2 3 /2
60° (1, 0) B
Substituting parametric coordinates in (1) 2
2 3 r 3r r = + , − 2r − 4 = 0 4 1 2 2 4
Length of AB = PA − PB =
(PA + PB )2 − 4PAPB
2
16 16 8 = −4× = 3 3 3 37. (1) : Given point is corresponding to t = 2 and dy 1 = 6t ⇒ slope of normal at t = 2 is − 12 dx 1 \ Equation of normal is y − 6 = − (x − 1) 12 73 ⇒ a = 73, b = ⇒ a + 12b = 146 12 38. (6) : Let a be the first term of A.P. and d be the common difference. 3n S(3n) S(3n) = {2a + (3n − 1) d}, = 3 = f (n) 2 S(2n) − S(n) . 1 n 2n + 1 \ S = lim = 6 ∑ f (r) = lim 3 n n→∞ n r =−n n→∞ 39. (2) : f(u, v, w, x) = (u – 1)2 + (v – 1)2 + (w – 1)2 + (x – 1)2 + 6 Clearly f is max when u = w = –3 and v = x = 3 ⇒ fmax =(–3 – 1)2 + (–3 – 1)2 + (3 – 1)2 + (3 – 1)2 + 6 = 46 2 40. (1) : | a − b |2 + | b − c |2 + c − a = 2(a2 + b2 + c 2) −2(a ⋅ b + b ⋅ c + c ⋅ a) = 2(4 + 9 + 16) − 2(a ⋅ b + b ⋅ c + c ⋅ a) = 58 − 2(a ⋅ b + b ⋅ c + c ⋅ a) Now (a + b + c )2 ≥ 0 ⇒ a2 + b2 + c 2 ≥ −2(a ⋅ b + b ⋅ c + c ⋅ a) ⇒ − 2(a ⋅ b + b ⋅ c + c ⋅ a) ≤ 29 ⇒ | a − b |2 + | b − c |2 + | c − a |2 ≤ 87 mathematics today | February ‘17
31
Series-7 The entire syllabus of Mathematics of WB-JEE is being divided in to eight units, on each unit there will be a Mock Test Paper (MTP) which will be published in the subsequent issues. The syllabus for module break-up is given bellow: Unit No.
Topic
Syllabus In Details
UNIT NO. 7
Co-ordinate Geometry-3D
Direction ratios and direction cosines. Angle between two intersecting lines. Skew lines, the shortest distance between them and its equation. Equation of a line and a plane in different forms, intersection of a line and a plane, coplanar lines. Differential Rolle’s and Lagrange’s Mean value theorem theorems, Applications of derivatives: Rate of calculus change of quantities, monotonic-increasing and decreasing functions, Maxima and minima of functions of one variable, tangents and normals. Integral calculus Integral as an anti-derivative. Fundamental integrals involving algebraic, trigonometric, exponential and logarithmic functions. Integration by substitution, by parts and by partial fractions. Integration using trigonometric identities. evaluation of simple integrals of type: dx dx dx dx (px + q)dx dx dx ,∫ ,∫ 2 ∫ 2 2 ,∫ 2 2 ,∫ 2 2 ,∫ 2 2, ∫ 2 2 x ±a a −x + bx + c ax + bx + c ax x ±a a −x ax + bx + c
Time : 1 hr 15 min.
Full marks : 50
CATEGORY-I For each correct answer one mark will be awarded, whereas, for each wrong answer, 25% of total marks (1/4) will be deducted. If candidates mark more than one answer, negative marking will be done.
p p and with positive 4 3 direction of x-axis and z-axis respectively. Then the acute angle made by the line with y-axis is (a) p/6 (b) p/4 (c) p/3 (d) cos–1(1/3)
1. A straight line makes angles
2. The angle between the lines x +1 y − 2 z + 4 x − 3 2y + 3 z − 2 = = and = = is 3 −2 1 1 5 2 (a) p/3 (b) p/2 (c) cos–1(3/5) (d) cos–1(4/5) 3. The coordinates of the foot of perpendicular drawn from the point A(2, 4, –1) on the line x +5 y +3 z −6 = = are 1 4 −9 (a) (1, –3, 4) (b) (–4, 1, –3) (c) (4, 1, 3) (d) none of these
4. The equation of the line joining the points (2, –1, 4) and (1, 1, –2) is x − 2 y +1 z − 4 = = (a) 1 2 6 x − 2 y +1 z − 4 (b) = = −1 2 −6 x −1 y −1 z + 2 (c) = = 1 2 6 x −1 y +1 z + 2 (d) = = 2 −6 −1 5. A plane meets the coordinate axes at P, Q, R such that the centroid of the triangle PQR is (a, b, c). If x y z the equation of the plane is + + = m , then the a b c value of m is (a) 2 (b) 3 (c) 1 (d) 6 6. The equation of the plane passing through the points (0, 1, 0) and (3, 4, 1) and parallel to the line x +3 y −3 z −2 is = = 2 7 5
By : Sankar Ghosh, S.G.M.C, Kolkata, Ph: 09831244397.
32
mathematics today | February ‘17
(a) 4x – 13y + 15z = 13 (b) 8x – 13y + 15z = 15 (c) 8x – 13y + 15z + 13 = 0 (d) none of these 7. The vector equation of the line passing through the points (2, –3, 1) and (–4, 3, 6) is ^ ^ ^ ^ ^ ^ (a) r = −4 i + 3 j + 6 k + l(2 i − 3 j + k) ^ ^ ^ ^ ^ ^ (b) r = 2 i − 3 j + k + l(−4 i + 3 j + 6 k) ^ ^ ^ ^ ^ ^ (c) r = −6 i + 6 j + 5 k + l(2 i − 6 j − 5 k) ^ ^ ^ ^ ^ ^ (d) r = 2 i − 3 j + k + l(−6 i + 6 j + 5 k) ^ ^ ^ ^ ^ ^ 8. If the line r = 2 i − j + 3 k + l(2 i + j + 2 k) is parallel ^ ^ ^ to the plane r ⋅ (3 i − 2 j + p k ) = 4, then the value of p is (a) 2 (b) –2 (c) 3 (d) –3 9. The vector equation of a plane passing through the point (1, –1, 2) and having 2, 3, 2 as direction number of normal to the plane is ^ ^ ^ (a) r ⋅ (2 i + 3 j + 2 k ) + 4 = 0 ^ ^ ^ ^ ^ ^ ^ ^ ^ (b) r ⋅ (i − j + 2 k ) + (i − j + 2 k ) ⋅ (2 i + 3 j + 2 k ) = 0 ^ ^ ^ (c) r ⋅ (2 i + 3 j + 2 k ) + 5 = 0 ^ ^ ^ (d) r ⋅ (2 i + 3 j + 2 k ) = 3 10. The distance of the point (–3, 2, 4) from the plane ^ ^ ^ r ⋅ (2 i − 3 j + 6 k ) + 7 = 0 is 29 (a) 5 units (b) units 7 18 19 (c) (d) units units 7 7 11. In the mean value theorem f(b) – f(a) = (b – a)f ′(c) (a < c < b), if a = 4, b = 9 and f (x) = x , then the value of c is (a) 8 (b) 5.25 (c) 4 (d) 6.25 12. If the function f (x) = 4x3 + ax2 + bx – 1 satisfies 1 all the conditions of Rolle’s theorem in − ≤ x ≤ 1 4 1 and if f ′ = 0 , then the values of a and b are 2 (a) a = 2, b = –3 (b) a = 1, b = –4 (c) a = –1, b = 4 (d) a = –4, b = –1 13. Let f (x) be a continuous in (–∞, ∞) and f ′(x) exists in (–∞, ∞). If f (3) = –6 and f ′(x) ≥ 6 for all x ∈ [3, 6] then (a) f (6) ≥ 24 (b) f (6) ≥ 12 (c) f (6) ≤ 24 (d) f (6) ≤ 12 14. A spherical balloon is being inflated at the rate of 35 cm3/min. Then the rate of increase of surface area of the balloon when its diameter is 14 cm, is
(a) 7 cm2/min (c) 17.5 cm2/min
(b) 10 cm2/min (d) 28 cm2/min
15. If h(x) = f(x) – (f (x))2 + (f (x))3 and f ′(x) > 0 " x then (a) h(x) is an increasing function of x in some specified interval (b) h(x) is an increasing function " x ∈ R (c) h(x) is a decreasing function " x ∈ R (d) h(x) is a decreasing function of x in some specified interval . 16. Electric current C, measured by a galvanometer, is given by the equation C = k tanq, where k is a constant. Then the percentage error in the current corresponding to an error 0.7 percent in the measurement of q when q = 45° is (a) 1.4 (b) 2.8 (c) 1.1 (d) 2.2 17. The point on the parabola 2y = x2, which is nearest to the point (0, 3) is 1 (a) (± 4, 8) (b) ± 1, 2 9 (d) ± 3, 2 18. If m be the slope of the tangent to the curve ey = 1 + x2, then (a) |m| > 1 (b) |m| < 1 (c) m < 1 (d) |m| ≤ 1 (c) (±2, 2)
19. The normal to the curve x = a(cosq + qsinq), y = a(sinq – q cosq) at any point q is such that (a) it passes through the origin (b) it passes through (a, –a) (c) it is at a constant distance from origin p (d) it makes angle + q with the x-axis. 4 20. values of a for which the function (a + 2)x3 – 3ax2 + 9ax – 1 decreases monotonically throughout for all real values of x, are (a) a < –2 (b) a > –2 (c) –3 < a < 0 (d) a ≤ –3 sin x dx = Ax + B log |sin(x − a)| + c, then the sin(x − a) value of (A, B) is (a) (cosa, sina) (b) (–sina, cosa) (c) (sina, cosa) (d) (–cosa, sina)
21. If ∫
22.
∫e
x
dx is
(a) e x + c 1 (c) e x + c 2
(b) 2( x − 1)e x + c (d) 2( x + 1)e x + c
mathematics today | February ‘17
33
23.
∫
−1 e tan x
dx is 1 + x2
(a) tan–1x + C
24.
1
(b)
−1 (c) e tan x + C
dx 30. If ∫ = f (a)log cos x + cos a
1 + x2 2xe
+C
+C (1 + x 2 )2
(d)
2
∫ f (x)dx = f (x), then ∫ { f (x)} 1 { f (x)}2 + C 2
(b) {f (x)}3 + C
(c)
{ f (x)}3 +C 3
(d) {f (x)}2 + C
(a)
∫2
dx x (x + 1)
1 tan −1( x ) + C 2
(c) tan −1( x ) + C
Every correct answer will yield 2 marks. For incorrect response, 25% of full mark (1/2) would be deducted. If candidates mark more than one answer, negative marking will be done.
x y z x −1 y − 2 z − 3 and = = , = = 1 2 3 3 −1 4 x + k y −1 z − 2 are concurrent. Then, = = 3 2 h (a) h = –2, k = –6 (b) h = 1/2, k = 2 (c) h = 6, k = 2 (d) h = 2, k = 1/2
31. If the lines
is
−1 (b) 2 tan ( x ) + C −1 (d) tan (2 x ) + C
26. The value of the integral
∫
dx
2
x + 4x + 13
is
(a)
x +2 1 + C (b) log(x2 + 4x + 13) + C tan −1 3 3
(c)
1 x +5 log +C 6 x −1
(d)
2
x +2
(x + 4x + 13)2
+C
32. The distance between the plane whose equation is ^ ^ ^ r ⋅ (2 i + j − 3 k) = 5 and the line whose equation is ^ ^ ^ ^ r = i + l (2 i + 5 j + 3 k) , is 3 5 (a) unit (b) units 14 14 (c) 5 units (d) none of these 33. The minimum area of the triangle formed by a x2 y2 tangent to the ellipse + = 1 and the coordinate a 2 b2 axes, is (a) ab
1 + sin x 27. The value of ∫ e x dx = 1 + cos x x +C 2 x x (c) e tan + C 2
x (b) e x sec + C 2
(a) e x sec2
(d) ex tanx + C
n
28. In = ∫ ( log x ) dx, then the value of (In + nIn−1) is (a) (logx)n–1 + C (b) n(logx)n + C n (c) (xlogx) + C (d) x(logx)n + C 29.
∫
e x dx
(e x + 2)(e x + 1) x
is equal to
e +1 +C (a) log x e +2 (c) 34
ex + 1
ex + 2
+C
e +2 +C (b) log x e +1 (d)
(c)
ex + 2 ex + 1
+C
mathematics today | February ‘17
(b)
(a + b)2 2
(d)
a 2 + b2 2
a2 + ab + b2 3
34. The four common tangents to the ellipses x2 y2 x2 y2 + = 1 form + = 1 and 4 9 9 4 (a) a rectangle of area 13 unit2 (b) a parallelogram which is neither a square nor a rectangle (c) a rhombus (d) none of these 35.
x
(c) coseca (d) seca
CATEGORY-II
dx is equal to
(a)
25. The value of
the value of f (a) is (a) sina (b) cosa
tan −1 x
x−a 2 + C , then x+a cos 2 cos
x2 − 1
dx = x 3 ⋅ 2x 4 − 2x 2 + 1 1 4 2 (a) 2 ⋅ 2x − 2x + 1 + C x 1 (b) 3 ⋅ 2x 4 − 2x 2 + 1 + C x
∫
1 . 2x 4 − 2x 2 + 1 + C x 1 . 2x 4 − 2x 2 + 1 + C (d) 2 2x (c)
(c)
36. Let A be a vector parallel to the line of intersection
of the planes P1 and P2 through the origin. The plane ^
^
^
^
P1 is parallel to the vectors 2 j + 3 k and 4 j − 3 k ^
^
while the plane P2 is parallel to the vectors j − k and ^ ^ ^ ^ ^ 3 i + 3 j . The angle between A and 2 i + j − 2 k is (a) p/2 (b) p/4 (c) p/6 (d) 3p/4 37. f (x) is a polynomial of third degree which has a local maxima at x = –1. If f (1) = –1, f (2) = 18 and f ′(x) has a local minimum at x = 0 then (a) f (0) = 5 (b) f (x) has local minimum at x = 1 (c) f (x) is increasing in [1, 2 5] (d) The distance between (–1, 2) and (a, f (a)) is 2 5, where a is a point of local minimum. 38. A tangent to the curve y = f (x) at P(x, y) cuts the x-axis and y-axis at A and B, respectively, such that BP : AP = 3 : 1. If f (1) = 1 then dy (a) the equation of the curve is x + 3 y = 0 dx 1 (b) the curve passes through 2, 8 dy (c) the equation of the curve is x − 3 y = 0 dx (d) the normal at (1, 1) is x + 3y = 4
∫
11 cos x − 16 sin x dx 2 cos x + 5 sin x = −lx + m log l cos x + d sin x + C, then
(a) l = 2 (c) d = l + m 40. If f(x) = ∫ (a) m = 2
(b) m = 3 (d) d = m – l
−1 e 2 tan x (1 + x)2
1+ x
2
mp
3 + 1 e 3 (d) f′( 3) = m SOLUTIONS
CATEGORY-III In this section more than 1 answer can be correct. Candidates will have to mark all the correct answers, for which 2 marks will be awarded. If, candidates marks one correct and one incorrect answer then no marks will be awarded. But if, candidate makes only correct, without making any incorrect, formula below will be used to allot marks. 2×(no. of correct response/total no. of correct options)
39. If
p m f′(1) = me
−1 dx =xe m tan x + C then
(b) f′(0) = 1
1. (c): Let the line makes an acute angle q with the y-axis. Therefore, the direction cosines are p p l = cos , m = cos and n = cos q 4 3 But l2 + m2 + n2 = 1 p p ⇒ cos2 + cos2 + cos2 q = 1 4 3 3 1 1 1 ⇒ cos2 q = 1 − + = 1 − = 2 4 4 4 1 ⇒ cos q = [ q is acute] 2 So, q = p/3 2. (b) : The given equations of the lines are x +1 y − 2 z + 4 x − 3 2y + 3 z − 2 = = and = = 3 1 1 5 2 −2 Let the angle between them be q. Then 5 3 ⋅1 + (−2) ⋅ + 1 ⋅ 2 2 =0 cos q = 2
5 32 + (−2)2 + 12 12 + + 22 2 Hence, q = p/2 3. (b) : The general point on the straight line x +5 y +3 z −6 ... (1) = = = l (say) 1 4 −9 is (l – 5, 4l – 3, –9l + 6) Let the foot of perpendicular L ≡ (l – 5, 4l – 3, –9l + 6) \ The direction ratios of AL are l – 7, 4l – 7, –9l + 7 Since AL is perpendicular to the line (1), therefore 1(l – 7) + 4(4l – 7) – 9(–9l + 7) = 0 ⇒ l = 1 Thus coordinates of foot of perpendicular are (–4, 1, –3) 4. (b) : The equation of the line joining the points (2, –1, 4) and (1, 1, –2) is x − 2 y +1 z − 4 x − 2 y +1 z − 4 = = = i.e., = 2 − 1 −1 − 1 4 + 2 1 −2 6 x − 2 y +1 z − 4 i.e., = = −1 2 −6 5. (b) : Let the vertices of the triangle PQR be P(x, 0, 0), Q(0, y, 0) and R(0, 0, z) x y \ a = ⇒ x = 3a, b = ⇒ y = 3b 3 3 z and c = ⇒ z = 3c. 3 mathematics today | February ‘17
35
Thus, the equation of the plane is x y z x y z + + =1⇒ + + = 3 3a 3b 3c a b c
=
6. (c) : The equation of the plane passing through the point (0, 1, 0) is ax + b(y – 1) + cz = 0 ... (1) Since (1) is also passing through the point (3, 4, 1) \ 3a + 3b + c = 0 ... (2) Again the plane (1) is parallel to the line x +3 y −3 z −2 = = 2 7 5 \ 2a + 7b + 5c = 0 ... (3) Now solving (2) and (3), we get a b c a b c = = ⇒ = = = l (say) 15 − 7 2 − 15 21 − 6 8 −13 15 \ a = 8l, b = –13l, c = 15l Now (1) becomes 8lx –13ly + 15lz = –13l ⇒ 8x – 13y + 15z + 13 = 0 is the required equation of plane 7. (d) : The vector equation of the line passing through the points (2, –3, 1) and (–4, 3, 6) ^ ^ ^ ^ ^ ^ i.e. a = 2 i − 3 j + k and b = −4 i + 3 j + 6 k is r = a + l(b − a) ^ ^ ^ ^ ^ ^ ⇒ r = 2 i − 3 j + k + l(−6 i + 6 j + 5 k) 8. (b) : Here the given equation of the line is ^ ^ ^ ^ ^ ^ ... (1) r = 2 i − j + 3 k + l(2 i + j + 2 k) ^ ^ ^ And equation of the plane is r ⋅ (3 i − 2 j + p k) = 4 ... (2) Since the line (1) is parallel to the plane (2) \ 2(3) + 1(–2) + 2 × p = 0 ⇒ p = –2 9. (d) : The position vector of the point having ^
^
^
coordinates (1, –1, 2) is i − j + 2 k and normal to the ^
^
^
plane is 2 i + 3 j + 2 k Thus, the equation of the plane is ^ ^ ^ ^ ^ ^ ^ ^ ^ r ⋅ (2 i + 3 j + 2 k ) = (i − j + 2 k) ⋅ (2 i + 3 j + 2 k) ^ ^ ^ i.e. r ⋅ (2 i + 3 j + 2 k ) = 3 10. (d) : The position vector of the point having ^
^
^
coordinates (–3, 2, 4) is −3 i + 2 j + 4 k ^ ^ ^ ^ ^ ^ Let a = −3 i + 2 j + 4 k and n = 2 i − 3 j + 6 k Then the distance of the point (–3, 2, 4) from the plane ^ ^ ^ r ⋅ (2 i − 3 j + 6 k) + 7 = 0 is ^ ^ ^ ^ ^ ^ a ⋅ n + 7 |(−3 i + 2 j + 4 k) ⋅ (2 i − 3 j + 6 k) + 7 | = n 22 + (−3)2 + 62 36
mathematics today | February ‘17
−6 − 6 + 24 + 7 4 + 9 + 36
=
19 units 7
11. (d) : Here f (x) = x 1 1 ⇒ f ′(x) = ⇒ f ′(c) = 2 x 2 c Now by mean value theorem, we have f (b) − f (a) f ′(c) = b−a ⇒
1 2 c
=
f (9) − f (4) 3 − 2 1 = = 9−4 5 5
⇒ 2 c = 5 ⇒ 4c = 25 ⇒ c =
25 = 6.25 (4 < 6.25 < 9) 4
12. (b) : Here, the given function is f (x) = 4x3 + ax2 + bx – 1 ⇒ f ′(x) = 12x2 + 2ax + b 1 Given that f ′ = 0 ⇒ 3 + a + b = 0 2 ⇒ a + b = –3 ... (1) Since f (x) satisfies all the conditions of Rolle’s Theorem 1 a b 1 \ f (1) = f − ⇒ 4 + a + b − 1 = − + − − 1 4 16 16 4 a b 65 15 5 65 +b+ = − ⇒ a+ b=− 16 4 4 16 4 16 ⇒ 3a + 4b = –13 Solving (1) and (2) we get a = 1 and b = –4 ⇒ a−
... (2)
13. (b) : Given that f (x) is continuous in (–∞, ∞) and f ′(x) exists in (–∞, ∞) f (6) − f (3) \ f ′(c) = where c ∈(3, 6) 6−3 ⇒
f (6) − f (3) ≥ 6 ⇒ f (6) − (−6) ≥ 18 ⇒ f (6) ≥ 12 3
14. (b) : Let the volume of the spherical balloon be V 4 dV dr \ V = pr 3 ⇒ = 4pr 2 3 dt dt dS dr Again S = 4pr 2 ⇒ = 8pr dt dt dS 2 dS 2 dV 2 Now dt = ⇒ = ⋅ = × 35 = 10 dV r dt r dt 7 dt Thus, the rate of increase of the surface area is 10 cm2/min
15. (b) : h(x) = f (x) − ( f (x))2 + ( f (x))3 h′(x) = f ′(x) − 2 f (x) f ′(x) + 3( f (x))2 f ′(x)
\
2 1 1 1 1 = 3 f ′(x) ( f (x))2 − 2 f (x) ⋅ + + − 3 3 3 9 2 1 2 = 3 f ′(x) f (x) − + 3 3 It is given that f ′(x) > 0 " x Thus h′(x) > 0 for all real values of x. Therefore h(x) is an increasing function for all real values of x. 16. (c) : We have, C = k tanq dC dC ⇒ = k sec2 q ⇒ dq = k sec2 qdq dq dq dC ⇒ dC = k sec2 qdq dC = dq dq
k sec2 qdq sec2 q dC × 100 = × 100 = dq × 100 tan q C k tan q p dq p =2× × 100 = × 0.7 = 1.57 × 0.7 = 1.099 4 q 2 \ Percentage error in C is 1.1 17. (c) : Let the point on the parabola 2y = x2 be x2 x, 2 Now,
x2 Now the distance between the point x, and 2 (0, 3) be L (say) x2 So, L = x + − 3 2 2
2
2
x2 1 dL2 = 2x + 2 − 3 × × 2x dx 2 2 x2 x2 = 2x 1 + − 3 = 2x − 2 2 2 d 2(L2 ) dx 2
= 3x 2 − 4 2
dL = 0 gives x3 – 4x = 0 dx ⇒ x(x2 – 4) = 0 ⇒ x = 0 or x = ±2 d 2(L2 ) \ =8>0 dx 2 x =±2 Now
So L2 is minimum when x = ±2 Therefore the required point is (±2, 2). 18. (d) : The given equation of the curve is e y = 1 + x2 ... (1) Differentiating both sides w.r.t. x, we get dy dy 2x 2x ey = 2x ⇒ = = = m (using (1)) dx dx e y 1 + x 2 ⇒ |m| =
2|x |
... (2)
1+ | x |2
But (1 – |x|)2 ≥ 0 ⇒ 1 + |x|2 – 2|x| ≥ 0 ⇒ 1 + |x|2 ≥ 2|x| Thus, from (2) we get |m| ≤ 1 19. (c) : Given that x = a(cosq + qsinq) dx ⇒ = a(− sin q + sin q + q cos q) = aq cos q dq and y = a(sinq – qcosq) dy ⇒ = a(cos q − cos q + q sin q) = aq sin q dq dx cos q \ = dy sin q Therefore, the equation of the normal to the given curve at q is cos q y − a(sin q − q cos q) = − {x − a(cos q + q sin q)} sin q ⇒ xcosq + ysinq = a(cos2q + qsinqcosq + sin2q – qsinq cosq) ⇒ xcosq + ysinq = a ⇒ xcosq + ysinq – a = 0 ... (1) Now the distance of the normal from the origin is |0 − a | = a(constant) cos2 q + sin2 q Therefore the normal (1) is at a constant distance from the origin. 20. (d) : Given that f (x) = (a + 2)x3 – 3ax2 + 9ax – 1 \ f ′(x) = 3(a + 2)x2 – 6ax + 9a 2 a a2 a = 3(a + 2) x 2 − 2 ⋅ x ⋅ + − + 9a a + 2 a + 2 (a + 2)2 2
a 3a2 = 3(a + 2) x − + 9 a − a+2 (a + 2) 2
a 6a(a + 3) = 3(a + 2) x − + a+2 a+2 Given that the function decreases monotonically throughout for all real values of x. mathematics today | February ‘17
37
2
a 6a(a + 3) <0 \ f ′(x) < 0 ⇒ 3(a + 2) x − + a+2 a+2 Clearly the above condition is satisfied for all real values of x when a ≤ –3 sin x dx 21. (a) : Let I = ∫ sin(x − a) Put x – a = z ⇒ dx = dz sin(z + a) sin z cos a + cos z sin a \ I=∫ dz = ∫ dz sin z sin z = cos a∫ dz + sin a∫ cot zdz = zcosa + sina⋅log(sinz) + c = (x – a)cosa + sina⋅log|sin(x – a)| + c = xcosa + sina⋅log|sin(x – a)| + k (where k = c – acosa) \ (A, B) = (cosa, sina) 22. (b) : Let I = ∫ e x dx Put x = z ⇒
1
2 x
dx = dz ⇒ dx = 2zdz
n 28. (d) : We have, In = ∫ (log x) dx
1 = x(log x)n − ∫ n(log x)n−1 ⋅ x dx x = x(log x)n − n∫ (log x)n−1dx + C
In = x(log x)n − nIn−1 + C ⇒ In + nIn−1 = x(log x)n + C 29. (a) : Let I =
23. (c) :
∫
−1 e tan x
1+ x
2
dx put tan −1x = z ⇒
1 1+ x
2
24. (a) :
∫ f (x) dx = f (x) 2
∫ { f (x)}
Now
\ f (x) = f ′(x)
dx =∫ f (x) f (x)dx =∫ f (x) f ′(x) dx 1 { f (x)}2 + C 2 dx
= ∫ f (x) d { f (x)} = 25. (c) : Let I =
∫2
x (x + 1)
Put x = z ⇒ dx = 2zdz 2z dz \ I =∫ dz = ∫ = tan −1 z + C = tan −1 x + C 2 2z(z + 1) 1+ z 2 dx =∫ 2 x + 4x + 13 x + 4x + 4 + 9 dx 1 x +2 +C =∫ = tan −1 2 2 3 3 (x + 2) + 3
26. (a) : We have,
∫
2
dx
x 1 + sin x dx 27. (c) : We have ∫ e 1 + cos x
x x 2 sin cos 1 2 2 dx = ∫e + x x 2 2 cos2 2 cos 2 2 x
mathematics today | February ‘17
=∫
dx = dz
−1 I = ∫ e zdz = e z + C = e tan x + C
\
∫
e x dx
(e x + 2)(e x + 1)
Put e x = z ⇒ e x dx = dz dz (z + 2) − (z + 1) \ I=∫ =∫ dz (z + 2)(z + 1) (z + 2)(z + 1)
I = ∫ 2ze zdz = 2(ze z − e z ) + c = 2e x ( x − 1) + c
\
38
x x 1 = ∫ e x sec2 + tan dx 2 2 2 x = e x tan + C ∫ e x { f (x) + f ′(x)} dx = e x f (x) + C 2
dz dz −∫ = log | z + 1 | − log | z + 2 | + C z +1 z +2
= log
ex + 1
z +1 + C = log z +2
30. (c) : Let I =
ex + 2
+C
dx
∫ cos x + cos a
2dz x = z ⇒ dx = 2 1 + z2 2dz 2dz \ I=∫ =∫ 2 (1 + z )cos a + 1 − z 2 1− z2 2 (z + 1) cos a + 1 + z 2 Put tan
= 2∫
dz (1 + cos a) − (1 − cos a)z
2
=
dz 2 ∫ (1 − cos a) a 2 2 cot 2 − z
a 2 +C log = ⋅ a a 2a sin 2 cot 1 − z tan 2 2 2 x a 1 + tan tan 2 2 +C = coseca ⋅ log x a 1 − tan tan 2 2 x − a cos 2 = coseca log +C x+a cos 2 \ f(a) = cosec a 1
1
1 + z tan
31. (d) : The coordinates of any point on the line x y z = = = l (say) 1 2 3 are given by x = l, y = 2l and z = 3l Thus the coordinates of a general point on first line are (l, 2l, 3l) The coordinates of any point on the line x −1 y − 2 z − 3 = = = m (say) 3 −1 4 are given by x = 3m + 1, y = –m + 2 and z = 4m + 3 Thus the coordinates of a general point on second line are (3m + 1, –m + 2, 4m + 3) If the lines intersect, then they have a common point. So, for some values of l and m, we must have l = 3m + 1, 2l = –m + 2 and 3l = 4m + 3 or l – 3m = 1, 2l + m = 2, 3l – 4m = 3 Solving two of these equations, we get l = 1 and m = 0. So, the point of intersection of the first two lines is (1, 2, 3). It is also lies on the third line. 1+ k 2 −1 3 − 2 So, = = 3 2 h 1+ k 1 1 1 1 ⇒ = ⇒ k= and = ⇒h=2 3 2 2 2 h 32. (a) : The vector equations of the plane and the straight line are respectively ^ ^ ^ ... (1) r ⋅ (2 i + j − 3 k) = 5 ^ ^ ^ ^ ... (2) and r = i + l(2 i + 5 j + 3 k) Here, b ⋅ n = (2 ^i + 5 ^j + 3 k^) ⋅ (2 ^i + ^j − 3 k^) = 0 So, b is perpendicular to n . Hence, the given line is parallel to the given plane. Let the required distance be d ^
d=
^
^
^
^
2
2
2
^
(i + 0 j + 0 k) ⋅ (2 i + j − 3 k) − 5 2 +1 + 3
=
3 unit 14
33. (a) : The equation of tangent to the ellipse x2
y2 x cos q y sin q + = 1 at (acosq, bsinq) is + = 1. 2 2 a b a b The intercepts of the tangent on the axes are asecq and bcosecq. 1 \ The area of the triangle = × a sec q × bcosecq 2 ab = ≥ ab sin2q
34. (d)
1
x2 − 1
35. (d) : Let I = ∫
x 3 2x 4 − 2x 2 + 1
dx = ∫
1 − 5 x x dx 2 1 2− 2 + 4 x x 3
2 1 1 1 Put 2 − 2 + 4 = z 2 ⇒ 4 3 − 5 dx = 2zdz x x x x 1 zdz 1 1 1 2 1 = ∫ dz = z + C = 2− 2 + 4 +C ∫ 2 z 2 2 2 x x 1 = 2 2x 4 − 2x 2 + 1 + C 2x 36. (a, d) : A vector along the line of intersection of two planes one of them being parallel to the vectors a and b and theother plane being parallel to the vectors c and d , is (a × b ) × (c × d ). ^ ^ ^ ^ ^ ^ ^ ^ \ A = {(2 j + 3 k) × (4 j − 3 k)} × {(j − k) × (3 i + 3 j)} \
I=
^
^
^
^
^
^
= −18 i × (3 i − 3 j − 3 k) = 54(k − j) Let q be an angle between them, then ^
cos q =
^
54(k − j )
^
^
^
.(2 i + j − 2 k) = −
1
54 2 2 p 3p So, angle , 4 4 37. (b, c) : Let f (x) = ax3 + bx2 + cx + d f(1) = –1 ⇒ a + b + c + d = –1 f(2) = 18 ⇒ 8a + 4b + 2c + d = 18 f ′′(0) = 0 ⇒ b = 0 and f ′(1) = 0 ⇒ 3a + 2b + c = 0 19 57 17 Solving, a = , b = 0, c = − , d = 4 4 2 1 \ f (x) = (19x 3 − 57x + 34) 4
=
1 2
f ′(x) = 0 ⇒ 57x 2 − 57 = 0 ⇒ x = 1, −1. f ′′(x) = 57 × 2x, f ′′(−1) < 0 and f ′′(1) > 0. \ f (x) has a local minimum at x = 1. f (x) is increasing ⇒ f ′(x) ≥ 0 ⇒ x2 – 1 ≥ 0 ⇒ x ≤ –1 or x ≥ 1 \ In [1, 2 5], f (x) is increasing. The distance of (–1, 2) from (1, –1) is not 2 5. 38. (a, b) : The equation of tangent to the curve at P(x, y) is dy dy dy Y − y = ( X − x) ⇒ X −Y = x − y dx dx dx mathematics today | February ‘17
39
X Y =1 + dy dy x − y y−x dx dx dy dx dx dy \ A x − y , 0 and B 0, y − x dy dx BP 3 Now, = . AP 1 3 dx 1 dy So, P = x − y , y − x = (x, y) 4 4 dy dx ⇒
1 dy dy =0 y − x = y ⇒ 3 y + x 4 dx dx dx dy dx dy ⇒ 3 + = 0 ⇒ 3∫ + ∫ = 0 x y x y ⇒ 3logx + logy = logc ⇒ x3y = c Now, f (1) = 1 ⇒ (1, 1) is on it. \ c = 1. Hence, x 3 y = 1 1 Also 2, lies on it. 8 The normal to x3y = 1 at (1, 1) has the equation −1 y −1 = (x − 1) dy dx (11 ,) ⇒
−1 (x − 1) = x − 3 y + 2 = 0 −3 11 cos x − 16 sin x dx 39. (a, b, c) : Let I = ∫ 2 cos x + 5 sin x ⇒
40
y −1 =
mathematics today | February ‘17
Let, 11cosx – 16sinx = l(5cosx – 2sinx) + m(2cosx + 5sinx) Now equating the coefficient of cosx and sinx, we get 5l + 2m = 11 .... (1) and –2l + 5m = –16 ⇒ 2l – 5m = 16 .... (2) Solving (1) and (2), we get l = 3 and m = –2 3(5 cos x − 2 sin x) − 2(2 cos x + 5 sin x) \ I=∫ dx 2 cos x + 5 sin x (5 cos x − 2 sin x) = 3∫ dx − 2∫ dx 2 cos x + 5 sin x = 3log|2cosx + 5sinx| – 2x + C = –2x + 3log|2cosx + 5sinx| + C \ l = 2, m = 3 and d = 5 40. (a, b, c, d) : We have, f(x) = ∫
−1 e 2 tan x (1 + x)2
1 + x2
dx
−1 −1 e2 tan x (1 + x 2 + 2x) 2xe 2 tan x 2 tan −1 x =∫ dx = ∫ e dx + ∫ dx 1 + x2 1 + x2
= xe
2 tan −1 x −1
− ∫ 2e
tan −1 x
×
= xe 2 tan x + C \ m=2 −1 Now f′(x) = xe 2 tan x ×
1
1 + x2 2
1+ x
2
xdx + ∫
−1 2xe 2 tan x
1 + x2
dx + C
−1 + e 2 tan x
\ f′(0) = 1, f′(1) = 2e p/2 = me p/m 2p
mp
3 3 and f′( 3) = + 1 e 3 = + 1 e 3 2 m
10 BEST
PROBLE
MS
Math archives, as the title itself suggests, is a collection of various challenging problems related to the topics of IIT-Jee Syllabus. This section is basically aimed at providing an extra insight and knowledge to the candidates preparing for IIT-Jee. In every issue of MT, challenging problems are offered with detailed solution. The readers’ comments and suggestions regarding the problems and solutions offered are always welcome.
−x 3 + x 2 sin6 – xsin4· sin8 – 5sin–1 3 (a2 – 8a + 17) then f ′(sin8) is (a) < 0 (b) > 0 (c) 12 (d) –12 1.
If f ( x ) =
{ }
1 If 2 < x < 3, and if x 2 = , where {⋅} represents x 1 fractional part function, then the value of x − is x (a) –1 (b) 0 (c) 1 (d) 2 2.
3. If f(x) = a + bcos–1x , (b > 0) has same domain and range then the value of a + b is 2 2 p p (a) 1 − (b) 1 − (c) −1 − 1 (d) p p 2 2 4. If x2 + pxy + 4y2 + x – 2y + q = 0 represent the pair of parallel lines then p + q may be (a) –5 (b) 1 (c) 2 (d) 5 5. Consider the circle x2 + y2 = 8. If from point P(2, 2), two chords PA and PB drawn of length 1 unit then the equation of AB is (a) 4x + 4y + 15 = 0 (b) 4x + 4y – 15 =0 (c) x + y + 15 = 0 (d) x + y – 15 = 0 6.
x3 − 1 is x −1 3 (b) , ∞ 4
Range of the function f ( x ) =
(a) 0, ∞ ) 3 (c) , ∞ − {3} 4
(d) R
7. Let A and B be two sets containing 3 and 4 elements respectively. The number of subsets of A × B having 2 or more elements is (a) 4083 (b) 4096 (c) 4046 (d) 4076 ^ ^ ^ ^ ^ 8. If the vectors AB = 2 i − 2 k and AC = 4 i + j+ 2 k are sides of a DABC, then the length of the median through A is 33 4
(a) 9.
(b)
33 2
(c)
37 4
(d)
37 2
5
∫1 (x − 1) (x − 2) (x − 3) (x − 4) (x − 5) dx =
(a) 3
(b) –3
(c) 0
(d) 1
10. The number of all possible positive triplets(x, y, z) such that (2y – x)sinq + (x – z)cos2q – 2xcos2q = 0 for all q is (a) infinite (b) one (c) zero (d) greater than 10 but less than 100 soLUtioNs 3
−x + x 2 sin6 – xsin4 · sin8 – 5sin–1 3 (a2 – 8a + 17) 2 f ′(x) = – x + 2x · sin6 – sin4 · sin8 f ′(sin8) = –(sin8)2 + 2sin8 · sin6 – sin4 · sin8 = –sin28 + 2sin8 · sin6 – sin4 · sin8 = –sin8[sin8 + sin4 – 2sin6] = –sin8[2sin6 · cos2 – 2sin6] = –2sin6 · sin8[cos2 – 1] = 2sin8 · sin6(1 – cos2) > 0 1. (b) : f ( x ) =
By : Prof. Shyam Bhushan, Director, Narayana IIT Academy, Jamshedpur. Mob. : 09334870021
mathematics today | February ‘17
41
A
1 2. (c) : Since, 2 < x < 3 and {x 2} = x 1 ⇒ x 2 − 2 = ⇒ x3 – 2x – 1 = 0 x ⇒ x3 + x2 – x2 – x – x – 1 = 0 ⇒ x2(x + 1) – x(x + 1) – 1(x + 1) = 0 ⇒ (x + 1)(x2 – x – 1) = 0 x ≠ –1, x2 – x – 1 = 0 ⇒ x= ⇒
8. (c) :
B
1± 5 1− 5 1+ 5 ⇒ x= But x ≠ 2 2 2
1 2 5 −1 5 −1 = × = x 2 5 +1 5 −1
1 5 +1 5 −1 5 +1− 5 +1 = − = =1 x 2 2 2 3. (d) : Domain : x ∈ [–1, 1] Q 0 ≤ cos–1x ≤ p ⇒ 0 ≤ bcos–1x ≤ bp ⇒ a ≤ a + bcos–1x ≤ a + bp From (i) and (ii), a = –1 2 ⇒ a + bp = 1 ⇒ bp = 2 ⇒ b = p \ x−
...(i) ...(ii)
4. (a) : For pair of lines to be parallel, h2 – ab = 0 ⇒ p = –4 Now (x – 2y)2 + (x – 2y) + q = 0, lines to be parallel root is real and distinct 1 ⇒ 1 − 4q > 0 ⇒ q < 4 ⇒ p+q=–5 5. (b) : Required equation of AB is S – S′ = 0 ((x – 2)2 + (y – 2)2 – 1) – (x2 + y2 – 8) = 0 ⇒ 4x + 4y – 15 = 0
(
)
2 x 3 − 1 ( x − 1) x + x + 1 = x −1 ( x − 1) 2 ⇒ f(x) = x + x + 1, x ≠ 1
6. (b) : f ( x ) =
f(x) = x2 + x + 1
3
3
3/4
x=1
7. (a) : n (A × B) = 12 No. of subset each contain 2 or more elements is = (12C2 + 12C3 + ... + 12C12)
= (12C0 + 12C1 + 12C2 + 12C3 + ... + 12C12) – (12C0 + 12C1) = 212 – (1 + 12) = 4096 – 13 = 4083 42
mathematics today | February ‘17
C
M Since, AB + BC + CA = 0 AC − AB ⇒ BC = AC − AB ⇒ BM = 2 Again, AB + BM + MA = 0 AC − AB \ AM = AB + BM = AB + 2 ^ ^ ^ AB + AC 6 i + j ^ j = = = 3 i+ 2 2 2 1 37 AM = 9 + = 4 4 9. (c) : Put x – 3 = t ⇒ dx = dt ⇒
\
5
∫ (x − 1)(x − 2)(x − 3)(x − 4)(x − 5)dx 1
=
2
∫ (t + 2)(t + 1)t ⋅ (t − 1) ⋅ (t − 2) dt
−2
= 0 (since function is odd). 10. (c) : 2ysinq – xsinq + (x – z)(1 – sin2q) – 2x (1 – 2sin2q) = 0 ⇒ 2ysinq – xsinq + x – xsin2q – z + zsin2q – 2x + 4xsin2q = 0 2 ⇒ (3x + z)sin q + (2y –x)sinq – x – z = 0 ⇒ (3x + z)sin2q + (2y – x)sinq – (x + z) = 0 This equation is zero for all values of q. Thus, 3x + z = 0, 2y – x = 0, x + z = 0 −z x So, x = , y = , x = −z 3 2 −z −z −z So the triplet formed is −z, , z and , , z 3 6 2 Thus no positive triplet is formed.
ANSWER KEY
MPP-8 CLASS XI 1.
(a)
2.
(a)
6.
(d)
7.
(b, c) 8.
10. (b) 15. (d) 20. (4)
3.
(a)
4.
(a, b) 9 .
(a)
5.
(d)
(a, b, c, d)
11. (a, b, c) 12. (a, c) 13. (a, b, c)14. (b) 16. (c) 17. (5) 18. (4) 19. (0)
ELLIPSE This article is a collection of shortcut methods, important formulas and MCQs along with their detailed solutions which provides an extra edge to the readers who are preparing for various competitive exams like Jee(Main & advanced) and other PeTs.
Definition anD StanDarD equation of ellipSe An ellipse is the locus of a point in a plane which moves in such a way that the ratio of its distance from a fixed point (called focus) in the same plane to its distance from a fixed straight line (called directrix) is always constant which is less than unity. The constant ratio is generally denoted by e (0 < e < 1) and is known as the eccentricity of the ellipse. The general equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents an ellipse, if D = abc + 2fgh – af 2 – bg2 – ch2 ≠ 0 and h2 < ab. equation of the ellipSe The equation of the ellipse, whose focus is the point (h, k) and directrix is lx + my + n = 0 and having eccentricity e, is (lh + mk + n)2 (x − h)2 + ( y − k)2 = e 2 ⋅ (l 2 + m2 )
z
Equation of the ellipse whose focus is (ae, 0) and directrix is x = a/e and having eccentricity e is given by x
2
a
2
+
y 2
2 2
a (1 − e )
= 1 or
x
2
a
2
+
y b
2
2
=1
X
Directrix
(0, b)B P(x1, y1)
M
Y
x2
Ellipse y2 b
2
a2
= 1−
(PN )2 b2
=
+
x2 a
+
2
y2 b2 =
= 1 can also be defined as
a2 − x 2 a
a2
⇒
2
A ′N ⋅ AN
y2 b
2
=
(a + x )(a − x ) a2
(PN )2 b2 (BC )2 = = AN ⋅ A ′N a2 ( AC )2
or Y
B p2
M
C A S S A (–ae, 0) (ae, 0) (a, 0) (–a, 0) B(0, –b) x = –a/e
z
or Y
y2
...(1) =1 a 2 b2 The foci S and S′ are (ae, 0) and (–ae, 0). The equation of its directrices are x = a/e and x = –a/e. Let P(x1, y1) be any point on (1), Now, SP = ePM = e (a/e – x1) = a – ex1 and S′P = ePM′ = e(a/e + x1) = a + ex1 \ SP + S′P = (a – ex1) + (a + ex1) = 2a = AA′ = Length of major axis The sum of focal distances of any point on the ellipse is equal to the length of major axis. note : (a) Two ellipses are said to be similar, if they have the same value of eccentricity. (b) Distance of every focus from the extremity of minor axis is equal to a, as b2 + a2e2 = a2.
⇒
where b2 = a2 · (1 – e2) Directrix
z
x2
The ellipse is
x = a/e
X
X
A (–a, 0)
P p1 N
C
A (a, 0)
X
B Y
Sanjay Singh Mathematics Classes, Chandigarh, Ph : 9888228231, 9216338231 mathematics today | February‘17
43
equation of ellipse referred to two perpendicular lines
z
(a) Ellipse
x2
a2
+
y2
b2
= 1 can also be defined as
Let Q be a point on auxiliary circle x2 + y2 = a2 such that QP produced is perpendicular to the x-axis. Then P and Q are the corresponding points on the ellipse and the auxiliary circle respectively. Y
(L.P. from any point on ellipseto the major axis)2
Q
2
P
(Length of semi minor axis) +
(L.P. from any point on ellipseto the minor axis)2 2
(Length of semi major axis)
X
=1
x2 + y2 = a2
(where L.P. means length of perpendicular) ⇒
p12
p2 + 2 =1 b2 a2
Y
(b) If L1 : a1x + b1y + c1 = 0 and L2 : b1x – a1y + c2 = 0, then
the equation
a x + b1 y + c1 1 a 2 + b 2 1 1
2
b x − a1 y + c2 1 a 2 + b 2 1 1
2
+ =1 a2 b2 represents an ellipse in the plane such that (i) The cent re of t he el l ip s e is t he p oi nt of intersection of the lines L1 = 0 and L2 = 0. (ii) The major axis lies along L2 = 0 and the minor axis along L1 = 0, if a > b. If a < b, then the major axis is along L1 = 0 and minor axis is along L2 = 0.
p2
P(x, y) p1
b
L1
a
L2
(iii) If a > b, then the lengths of the major and minor axes are 2a and 2b respectively and if a < b, then the lengths of major and minor axes are 2b and 2a respectively. parametric equation of the ellipse The circle described on the major axis of an ellipse as diameter is called the auxiliary circle of the ellipse. z
Let the equation of ellipse be \
x2
+ = 1 (a > b). a 2 b2 Equation of its auxiliary circle is x 2 + y 2 = a 2
mathematics today | February‘17
NS
X A (a, 0)
x2 + y 2 = 1 a2 b2
Let ∠QCA = q (0 ≤ q < 2p) i.e., the eccentric angle of P on an ellipse is the angle which the radius (or radius vector) through the corresponding point on the auxiliary circle makes with the major axis. \ Q ≡ (a cosq, a sinq) and P ≡ (a cosq , b sinq) The equation x = a cosq and y = b sinq taken together 2 2 are called parametric equation of ellipse x + y = 1 a 2 b2 where q is real parameter and P ≡ (a cosq , b sinq ) is any point on the ellipse also known as P(q) or ‘q’ point on the ellipse. note: (i) Eccentric angles of the extremities of latus
x2 y2 rectum of the ellipse + = 1 are given by 2 2 a b b tan −1 ± . ae x2 y2 (ii) Area of the ellipse + = 1 is pab sq. units. a 2 b2 position of points w.r.t. an ellipse x2
y2
= 1 and P ≡ (h, k) be any a 2 b2 point then P will lie outside, on or inside the ellipse Let the ellipse be x2
a
2
+
+
y2
2 2 = 1 according as h + k >, =, < 1 b a 2 b2 2
Y
P(outside)
y2
( AA′ is diameter of the circle) 44
C
A (–a, 0)
X
A
B P(inside)
C
B P(on) Y
A
X
important points The equation of the ellipse whose axes are parallel to the coordinate axes and whose centre is at the origin, is x2
a2
+
y2
b2
= 1 with the following properties :
S. no.
x2
ellipse
(i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix)
Basic fundamentals Coordinates of the centre Coordinates of the vertices Coordinates of the foci Length of the major axis Length of the minor axis Equation of the major axis Equation of the minor axis Equations of the directrices Eccentricity
(x) (xi) (xii) (xiii)
Ends of the latus rectum Length of the latus rectum Parametric coordinates Auxilliary circle
a2
x
2
a
2
+
y b
2
2
=1
If the chord joining two points whose eccentric angles are a and b, cut the major axis of the ellipse a2 tan
z
+
y2 b2
= 1 (a > b) at the point (ae, 0), then
a b e −1 tan = . z z e +1
If a and b are the eccentric angles of extremities of a focal chord of the ellipse
x
2
a
2
+
y
2
b2
a2
+
= 1 (a > b) , then
a −b cos 2 a b e −1 e +1 ⇒ tan ⋅ tan = or . e= 2 2 e +1 e −1 a +b cos 2
z
y2 b2
= 1, a < b
(0, 0) (0, ±b) (0, ±be) 2b 2a x=0 y=0 y = ± b/e
b2 e = 1− a2
x a −b a +b y a +b cos + sin = cos . 2 b 2 2 a
x2
b2
x2
= 1, a > b
a2 e = 1− a2
(± ae, ± b2/a) 2b2/a (a cosq, b sinq) x2 + y2 = a2
eccentric angles are a and b on the ellipse
z
y2
(0, 0) (±a, 0) (±ae, 0) 2a 2b y=0 x=0 x = ±a/e
equation of Chord z Equation of the chord joining the points whose is given by
+
(± a2/b, ± be) 2a2/b (a cosq, b sinq) x2 + y2 = b2
Locus of mid-points of focal chords of an ellipse x2
a
2
+
y2
b
2
= 1 (a > b) is
x2
a
2
+
y2
b
2
=±
ex . a
equation of tangents in Different forms point form : Equation of tangent to the ellipse xx yy x2 y2 + = 1 at the point (x1, y1) is 1 + 1 = 1. 2 2 2 a b2 a b parametric form : Equation of tangent to the x2
a2
the ellipse
+
y2
= 1 at the point (a cosq, b sinq) b2 y is x cos q + sin q = 1. a b Slope form : Equation of tangent of slope m to ellipse
x2
a2
+
y2
b2
= 1 are given by
y = mx ± (a2m2 + b2 ) and the coordinates of the points of contact are a2 m b2 ,± . (a2 m2 + b2 ) (a2 m2 + b2 ) mathematics today | February‘17
45
results related to tangents (i) The equations of the tangents to the ellipse at points P(a cosq1, b sinq1) and Q(a cosq2, b sinq2) y y x x are cos q1 + sin q1 = 1 and cos q2 + sin q2 = 1 a b a b and these two intersect at the point q1 + q2 q1 + q2 a cos 2 b sin 2 . , q1 − q2 q1 − q2 cos 2 cos 2 (ii) The line joining two points on an ellipse, the difference of whose eccentric angle is constant, touches an other ellipse of same eccentricity. (iii) The locus of the point of intersection of tangents to an ellipse at the points whose eccentric angles differs by constant is an ellipse of the same eccentricity. If the eccentric angles differs by a 2 2 right angle then the locus is x + y = 2. a 2 b2 (iv) The locus of the point of intersection of tangents
to the ellipse
x2
y2
= 1 at the points the sum a 2 b2 of whose eccentric angles is constant is a straight line passing through the centre of the ellipse. (v) If the tangent at P on an ellipse meets the directrix in F, then PF will subtend a right angle at the corresponding focus i.e., ∠PSF = p/2. (vi) If SM and S′M′ are perpendiculars from the foci upon the tangent at any point of the ellipse, then SM ·S′M′ = b2 and the point M & M′ lie on the auxiliary circle (where a > b). (vii) The tangents drawn from any point on the director circle of a given ellipse to the ellipse are always at right angle. +
equations of normals in Different forms poi nt for m : E quat ion of nor ma l to t he y2
+ = 1 at t h e p oi nt ( x 1 , y 1 ) i s a 2 b2 b2 y − = a 2 − b2 . y1
e l l ip s e a2 x x1
x2
parametric form : Equation of normal to x2
y2
+ = 1 at (a cosq, b sinq) is a 2 b2 ax sec q – by cosecq = a2 – b2. the ellipse
46
mathematics today | February‘17
Slope form : Equation of normal of slope m to the 2 2 m(a2 − b2 ) ellipse x + y = 1 are given y = mx (a2 + b2m2 ) a 2 b2 a2 b2 m ,± at the points ± a2 + b2m2 a2 + b2m2 results related to normals (i) If the line lx + my + n = 0 be a normal to the ellipse x2
+
y2
= 1 then
a2
+
b2
=
(a2 − b2 )2
l 2 m2 a 2 b2 n2 (ii) Locus of mid-points of normal chords of an ellipse x2
2
y2
x 2 y 2 a 6 b6 = 1 is + + = (a2 − b2 )2 a 2 b2 a 2 b2 x 2 y 2 (iii) Four normals can be drawn from a point to an ellipse. intersection of a Circle and an ellipse (i) The circle x2 + y2 + 2gx + 2fy + c = 0 and ellipse x2 y2 + = 1 can intersect each other at maximum a 2 b2 of four points. (ii) The maximum number of common chord of a circle and an ellipse is six. (iii) The maximum number of common tangent of a circle and an ellipse is four. Director Circle The locus of the point of intersection of the tangents to +
an ellipse
x2
y2
= 1 which are perpendicular to each a 2 b2 other is called director circle and its equation is given by x2 + y2 = a2 + b2. Sub-tangent and Sub-normal +
Length of sub-tangent NT = Length of sub-normal GN = Y
a2 − x1 x1 b2 a
T X
C
Y
2
|x1| = (1 – e2) |x1|
P(x1, y1) G
N
T
X
reflection property of an ellipse If an incoming light ray passes through the focus (S) strike the concave side of the ellipse then it will get reflected towards other focus (S′) and ∠SPS′ = ∠SQS′ Y
Q X
S
A
Y C
B
Tang ent P
N S Normal B
A
X
+
y
2
= 1 (|b| > |a|) , provided a b2 (a) c2 = a2 + b2m2 (b) c2 = b2 + a2m2 2 2 2 2 (c) c = (a + b )m (d) c2 = (b2 – a2)m2 2
2. Tangents are drawn to the ellipse x2 + 2y2 = 4 from any arbitrary point on the line x + y = 4, the corresponding chord of contact will always pass through a fixed point, whose coordinates are 1 1 (a) 1, (b) , 1 2 2 1 1 (c) 1, − (d) , − 1 2 2 3. Equation of the ellipse whose axes are along the coordinate axes and whose length of latus rectum and eccentricity are equal and equal to 1/2 each , is (a) 6x2 + 12y2 = 1 (b) 12x2 + 6y2 = 1 2 2 (c) 3x + 12y = 1 (d) 9x2 + 12y2 = 1 The line y = x - 1 touches the ellipse 3x2 + 4y2 = 12, at 1 1 (a) , − (b) (3, 2) 2 2 (c) (–1, –2) (d) None of these 4.
5. One foot of normal of the ellipse 4x2 + 9y2 = 36, that is parallel to the line 2x + y = 3, is 9 8 (a) , 5 5 9 8 (c) − , − 5 5
(a)
x2
9 8 (b) − , 5 5 (d) None of these
x2
y2
= 1 at a 2 b2 point ‘P’ meets the coordinate axes at points A and B respectively. Locus of mid-point of segment AB is a2 a2
(c)
The tangent and normal at any point of an ellipse bisect the external and internal angles between the focal radii to the point. problems 1. The line y = mx + c, will be a tangent to the ellipse x
Tangent drawn to the ellipse
Light ray
Re ected ray Y
2
6.
7.
x2
+ +
y2 b2 b2 y2
=2
(b)
=4
(d)
a2
x2 x2 a2
+ +
b2
y2 y2 b2
Normals drawn to the ellipse
+
=2 =4 x2
y2
= 1 at a 2 b2 point ‘P’ meets the coordinate axes at points A and B respectively. Locus of mid point of segment AB is (a) 4x2a2 + 4y2b2 = (a2 – b2)2 (b) 4x2b2 + 4y2a2 = (a2 – b2) (c) 16x2a2 + 16y2b2 = (a2 – b2)2 (d) 16x2b2 + 16y2a2 = (a2 – b2) +
8. If the line y = mx + c is a tangent to the ellipse x2 y2 + = 1 then corresponding point of contact is a 2 b2 2 a2 m 2 b2 (a) a m , b (b) − ,− c c c c a2 m b2 (c) ,− c c 9. x
a 2 m b2 (d) − , c c
If the line y = mx + c is a normal to the ellipse 2
a2
+
y2
b2
= 1, then corresponding foot of normal is
a2 c b2 c (a) , m(b2 − a2 ) (b2 − a2 ) a2 c b2 c (b) , m(a2 − b2 ) (a2 − b2 ) a2 m b2 c (c) , c(b2 − a2 ) (b2 − a2 ) a2 m b2 c (d) , c(a2 − b2 ) (a2 − b2 ) 10. Locus of the mid-point of chords of the ellipse x2
+
y2
= 1 that are parallel to the line y = 2x + c, is a 2 b2 (a) 2b2y – a2x = 0 (b) 2a2y – b2 x = 0 2 2 (c) 2b y + a x = 0 (d) 2a2y + b2 x = 0 mathematics today | February‘17
51
11. The locus of the moving point P(x, y) satisfying (x − 1)2 + y 2 + (x + 1)2 + ( y − 12 )2 = a will be an ellipse if (a) a < 4 (b) a > 2
(c) a > 4
(d) a < 2
y2 x2 12. The equation + = 1 will represent an 6 − − 2 a a ellipse if (a) a ∈ (1, 3) (b) a ∈ (1, 6) (c) a ∈ (–∞, 2) ∪ (6, ∞) (d) a ∈ (2, 6) ~ {4} 13. Foci of the ellipse 16x2 + 25y2 = 400 are (a) (± 2, 0) (b) (±3, 0) 48 (d) ± , 0 5 14. Equation of an ellipse whose focus is S ≡ (1, 0), corresponding directrix being the line x + y = 2 and 1 eccentricity being is 2 (a) 3x2 + 3y2 – 2xy + 4x – 4y = 0 (b) 3x2 + 3y2 + 2xy – 4x + 4y = 0 (c) 3x2 + 3y2 + 2xy + 4x – 4y = 0 (d) 3x2 + 3y2 – 2xy – 4x + 4y = 0 15. The equation 3(x + y – 5)2 + 2(x – y + 7)2 = 6 represents an ellipse, whose centre is (a) (–1, 6) (b) (6, –1) (c) (1, –6) (d) (–6, 1) x2 2
+
y2 2
= 1, centered at
a b point ‘O’ and having AB and CD as it’s major and minor axes respectively. If S1 be one of the focus of the ellipse, radius of incircle of triangle OCS1 be 1 unit and OS1 = 6 units, then area of the ellipse is equal to 65 (a) 16 p sq. units (b) p sq. units 4 (c) 65 p sq. units (d) 65 p sq. units 2 x2 y2 17. Normal drawn to the ellipse + = 1 at the 42 32 point P(q) meets the ellipse again at point Q(2q), then value of cos q can be 1 16 (a) − (b) − (c) − 19 (d) − 21 2 23 23 23 18. P is any variable point on the ellipse
x2
y2
+ =1 a 2 b2 having the points S1 and S2 as its foci. Then maximum 52
mathematics today | February‘17
2 y2 19. P is any variable point on the ellipse x + =1 a 2 b2 having the points S1 and S2 as its foci. Then Locus of
incentre of triangle PS1S2 will be (a) a straight line (b) a circle (c) a parabola (d) an ellipse 20. P1 and P2 are the foot of altitudes drawn from the
24 (c) ± , 0 5
16. Consider an ellipse
area of the triangle PS1S2 is equal to (a) b2e sq. units (b) a2e sq. units (c) ab sq. units (d) abe sq. units
foci S1 and S2 respectively of the ellipse
x2
y2
=1 a 2 b2 on one of its variable tangent. Maximum value of (S1P1) (S2P2) is equal to (a) b2
(b) b
(c) a2
+
(d) a
21. The equation of the ellipse centered at (1, 2) having the point (6, 2) as one of its focus and passing through the point (4, 6) is 2 3( y − 2)2 (a) (x − 1) + =1 36 64
(b)
(x − 1)2 ( y − 2)2 + =1 18 32
(c)
(x − 1)2 7( y − 2)2 + =1 72 128
2 2 (d) (x − 1) + ( y − 2) = 1 45 20 22. The angle between the tangents drawn to the ellipse 3x2 + 2y2 = 5, from the point P(1, 2) is equal to
(a) tan −1 (24 5 )
(b) tan −1 (12 5 )
24 (c) tan −1 5
12 (d) tan −1 5
23. The line 5x – 3y = 8 2 is a normal to the ellipse
x2 y2 + = 1. If ‘q’ be the eccentric angle of the foot of 25 9 this normal then ‘q’ is equal to p 3
(a)
p 6
(b)
(c)
p 4
(d) None of these
24. The distance of a point P (lying in the first quadrant) on the ellipse x2 + 3y2 = 6 from the center of the ellipse is 2 units. Eccentric angle of the point ‘P’ is (a) p (b) p 3 6 p (c) (d) None of these 4 x 2 11 y 2 25. The tangent drawn to the ellipse + = 1 at 16 256 the point P(q), touches the circle (x – 1)2 + y2 = 16. ‘q’ is equal to p (a) p (b) 3 6 p (c) (d) None of these 4 26. There are exactly two points on the ellipse x2
30. Length of the focal chord of the ellipse
27. For all admissible values of the parameter ‘a’ the straight line 2ax + y 1 − a2 = 1 will touch an ellipse whose eccentricity is equal to 1 3 2 (a) (b) (c) 1 (d) 3 2 2 3 28. The normal to the ellipse 4x2 + 5y2 = 20 at the point ‘P’ touches the parabola y2 = 4x, the eccentric angle of the point ‘P’ is 1 p 1 p (a) (b) + tan −1 + cos −1 5 2 5 2 1 (c) p − tan −1 5
1 (d) p − cos −1 5
29. The tangent drawn to the ellipse
x2
y2
= 1 , at a 2 b2 the point ‘P’ meets the circle x2 + y2 = a2, (a > b) at the points A and B. The line segment AB subtends an angle p/2 at the center of the ellipse. If ‘e’ be the eccentricity of the ellipse and ‘q’ be the eccentric angle of point ‘P’ then (a) e2(1 + sin2q) = 1 (b) e2(1 + cos2q) = 1 2 2 (c) e (2 + sin q) = 1 (d) e2(2 + cos2q) = 1 +
2b2a
(a)
a2 sin2 q + b2 cos2 q 2a2b
(c)
a2 sin2 q + b2 cos2 q
(b) (d)
+
y2
= 1, a 2 b2 that is inclined at an angle ‘q’ with the x-axis, is equal to 2b2a
a2 cos2 q + b2 sin2 q 2a2b a2 cos2 q + b2 sin2 q
31. The line y = mx + c, will be a normal to the ellipse , x2
a2 (a)
y2
+ = 1 whose distance from the center of the a 2 b2 a2 + 2b2 ellipse are equal and equal to . Eccentricity 2 of this ellipse is equal to 1 1 3 2 (a) (b) (c) (d) 2 3 3 2
x2
+
y2
b2 a2 m2
= 1, if + b2 =
2 (c) a +
b2 m2
=
2 (b2 − a2 )2 (b2 − a2 )2 (b) c + b2 = m2 c2 c2
(b2 − a2 )2 c2
b2 (b2 − a2 )2 (d) c 2 + = m2 a2
32. Locus of the point of intersection of tangents
2 2 drawn to the given ellipse x + y = 1, at points P(q1) a 2 b2 2p and Q(q2), such that q1 – q2 = , is 3
(a) (c)
a2 x2 x2 a2
+ +
b2 y2 y2 b2
=4
(b)
=4
(d)
33. Consider the ellipse
a2 x2 x2 a2 x2
+ +
b2 y2 y2 b2
=2 =2
y2
= 1, having it’s a 2 b2 eccentricity equal to e. P is any variable point on it and P1, P2 are the foot of perpendiculars drawn from P to the x and y-axis respectively. The line P1P2 will always be a normal to an ellipse whose eccentricity is equal to (a) e2
(b)
e
(c)
+
2e (d) e 1+ e
34. The normal drawn to the ellipse
x2
+
y2
=1 a 2 b2 (a > b) at the extremity of the latus rectum passes through the extremity of the minor axis. Eccentricity of this ellipse is equal to mathematics today | February‘17
53
(a)
5 −1 2
(b)
5 −1 2
(c)
3 −1 2
(d)
3 −1 2
35. The equation of common x2 + 4y2 = 8 and y2 = 4x are (a) y – 2x – 4 = 0, y + 2x + 4 (b) y – 2x – 2 = 0, y + 2x + 2 (c) 2y – x – 4 = 0, 2y + x + 4 (d) None of these
tangents of the curves =0 =0 =0
36. If L is the length of perpendicular drawn from the 2 2 origin to any normal of the ellipse x + y = 1, then 25 16 maximum value of L is (a) 5 (b) 4 (c) 1 (d) None of these
37. The maximum distance of the centre of the ellipse x2 y2 + = 1 from the chord of contact of mutually 9 4 perpendicular tangents of the ellipse is 3 6 36 9 (a) (b) (c) (d) 13 13 13 13 38. Tangents PA and PB are drawn to the ellipse x2 y2 + = 1 from the point P(0, 5). Area of triangle 16 9 PAB is equal to 256 16 sq. units (a) (b) sq. units 25 5
1024 32 (d) sq. units sq. units 25 5 39. The straight line x – 2y + 4 = 0 is one of the common x2 y2 tangents of the parabola y2 = 4x and + = 1. The 4 b2 equation of another common tangent of these curves is (a) x + 2y + 4 = 0 (b) x + 2y – 4 = 0 (c) x + 2y + 2 = 0 (d) x + 2y – 2 = 0 (c)
x2 y2 + = 1 at the 36 9 point ‘P’ meets the y-axis at A and normal drawn to the ellipse at point ‘P’ meets the x-axis at B. If area of 27 triangle OAB is sq. units, then eccentric angle of 4 point ‘P’ is 40. Tangent drawn to the ellipse
54
mathematics today | February‘17
p (b) p 3 6 (c) p (d) None of these 4 41. The tangent and normal drawn to the ellipse x2 + 4y2 = 4 at the point P, meet the x-axis at A and B respectively. If AB = 2 then cosq is equal to, (‘q’ being the eccentric angle of point P) (a)
8 2 5 (a) 1 (b) (c) (d) 3 3 3 3 42. The chord joining the points P1(a cosq1, b sinq1) and P2(a cosq2, b sinq2) meets the x-axis at A. If OA = c(‘O’ being the origin), then the value of q q tan 1 ⋅ tan 2 is equal to 2 2 c−a c −b c−a (a) (b) (c) c − b (d) c +b c+a c+a c +b 2 2 y x 43. A tangent to the ellipse + = 1, having 18 32 4 slope − cuts the x and y-axis at the points A and B 3 respectively. If O is the origin then area of triangle OAB is equal to (a) 12 sq. units (b) 24 sq. units (c) 36 sq. units (d) None of these 44. Which of the following is a common tangent to the ellipses
x2
a 2 + b2
+
y2
b2
= 1 and
x2
a2
+
y2
a 2 + b2
= 1?
(a) by = ax − a 4 + a2b2 + b 4 (b) ay = bx − a 4 + a2b2 + b 4 (c) by = ax − a 4 − a2b2 + b 4 (d) ay = bx − a 4 − a2b2 + b 4 45. Maximum length of the chord of the ellipse x2 y2 + = 1, such that eccentric angles of its extremities a 2 b2 p differ by is (a > b) 2 (a) a 2 (b) b 2 (c) ab 2 (d) None of these 46. Consider an ellipse having its axes along the coordinate axes and passing through the point (4, –1).
If the line x + 4y – 10 = 0 is one of its tangent, then area of the ellipse is equal to (a) 20 p sq. units (b) 30 p sq. units (c) 15 p sq. units (d) 10 p sq. units 47. S1 and S2 are the foci of an ellipse. ‘B’ be one of the extremity of its minor axis. If triangle S1S2B is right angled then eccentricity of the ellipse is equal to 1 3 (a) (b) 2 2 3 (d) None of these 2 48. If the chord of contact of tangents drawn from a x2 y2 point P on the ellipse + = 1 touches the circles a 2 b2 x2 + y2 = c2, then locus of P is (c)
(a) (c)
x2 a2 x2 a4
+ +
y2 b2 y2 b4
= =
a2
(b)
c2 1
(d)
c2
x2
a2 x2
+ +
y2
b2 y2
= =
b4 c4
a4
a 2 b2 c 4 49. The point on the ellipse 4x2 + 9y2 = 1 such that tangent drawn to the ellipse at this point is perpendicular to the line 8x – 9y = 0, is 27 16 27 16 ,− , (a) (b) − 2 985 3 985 2 985 2 985 27 16 , (c) 2 985 3 985
(d) None of these
x2 y2 + = 1, from 16 9 the point ‘P’, meets the coordinate axes at concyclic points, then locus of point ‘P ’ is (a) x2 + y2 = 7 (b) x2 – y2 = 7 2 2 (c) x + y = 25 (d) x2 – y2 = 25 50. Tangents drawn to the ellipse
solutions 1. (b) : If the point of contact is P(a cosq, b sinq), y x then equation of tangent cos q + sin q = 1 a b and given line y = mx + c will be identical. Comparing the coefficients, we get cos q sin q 1 = =− am −b c am b ⇒ cos q = − , sin q = ⇒ c c
a2m2 + b2 = c2
2. (a) : Let the aribitrary point be (a, 4 – a), then equation of corresponding chord of contact is, x · a + 2y (4 – a) = 4 ⇒ a(x – 2y) + 8y – 4 = 0 It is indeed a family of concurrent lines. The point of concurrency being the intersection point of the lines x – 2y = 0 and 8y – 4 = 0 1 \ Required point is 1, 2 3. (d) : e = 1−
2b2 1 b2 1 = ⇒ = a 2 a 4 b2 a
2
=
...(i)
(Given)
1 2
...(ii)
1 2 1 From (i) and (ii), we get, a = , b = 3 12 x2 y2 + =1 \ Equation of ellipse is 1 1 9 12 i.e., 9x2 + 12y2 = 1 2 2 4. (d) : Given ellipse is, x + y = 1 4 3 and line is y = x – 1 Let the point of contact be (2 cosq, 3 sinq).
\
Equation of tangent becomes
y x cos q + sin q = 1 2 3
cos q sin q = =1 2 − 3 ⇒ cosq = 2, sinq = – 3 which is not possible. Thus, the given line can’t be a tangent to the given ellipse. ⇒
2 2 5. (b) : Given ellipse is, x + y = 1 9 4 Let the foot of normal be (3 cosq, 2 sinq) Equation of normal at this point is, 2y 3x − =5 cos q sin q 3 Slope = tanq = – 2 2 4 ⇒ tanq = – 3 4 3 or sin q = , cos q = − 5 5 9 8 Thus foot of normal be − , . 5 5 mathematics today | February‘17
55
6. (c) : Let P ≡ (a cosq, b sinq) y x \ Equation of tangent is cos q + sin q = 1 a b ⇒ A ≡ (a secq, 0), B ≡ (0, b cosecq) If the mid-point of segment AB be Q(h, k), then 2h = a secq, 2k = b cosecq ⇒
a2
b2
+
=1 4h 2 4 k 2 a 2 b2 + =4 \ Locus of Q is x2 y2 7. (a) : Let P ≡ (a cosq, b sinq) by ax \ Equation of normal is − = a 2 − b2 cos q sin q a 2 − b2 a 2 − b2 ⇒ A≡ cos q, 0 and B ≡ 0, sin q b a If (h, k) be the mid-point of segment AB, then 2
2
2
2
a −b a −b cos q, 2k = sin q a b ⇒ 4h2a2 + 4k2b2 = (a2 – b2)2 Thus, locus of the point is 4a2x2 + 4b2y2 = (a2 – b2)2 2h =
8. (d) : Let the point of contact be P(a cosq, b sinq). y x Then equation of tangent is cos q + sin q = 1 a b cos q sin q 1 Thus on comparing, = =− am −b c ⇒ cos q = −
am b , sin q = c c
a 2 m b2 Thus the point of contact is − , c c 9. (a) : Let the foot of normal be P(a cosq, b sinq) by ax Equation of normal at P is − = a 2 − b2 cos q sin q m cos q sin q c \ = = 2 a b (b − a2 ) ⇒ cos q =
ac 2
2
m(b − a )
, sin q =
bc 2
(b − a2 )
a2 c b2 c Thus the foot of normal is , m(b2 − a2 ) (b2 − a2 ) 10. (d) : Let the mid-point of chord be P(h, k). Equation of this chord will be T = S1 56
mathematics today | February‘17
i.e.,
xh a2
+
yk b2
=
h2 a2
+
k2
b2 h2 k 2 2+ 2 b a h k b2 ⇒ =− =− ⇒ k =− ⋅h c 2a2 b2 2a2 b2 Thus the required locus is y = − x or 2a2y + b2x = 0 2a2 11. (c) : Let A ≡ (1, 0), B ≡ (–1, 12 ) \ Given expression is, PA + PB = a Here, AB = 4 + 12 = 4 Thus P(x, y) will lie on an ellipse having foci as A and B, provided a > 4 12. (d) : Clearly, 6 – a > 0, a – 2 > 0 and 6 – a ≠ a – 2 ⇒ a ∈ (2, 6) ~ {4} 13. (b) : Equation of the ellipse is
x2 y2 + =1 25 16
16 9 3 = ⇒ e= 25 25 5 3 3 Thus, foci are 5 ⋅ , 0 and −5 ⋅ , 0 5 5 i.e. (± 3, 0) ⇒
e2 = 1 −
14. (d) : If P(x, y) be any point on the ellipse, then PS = e · PM, where PM is perpendicular distance of P from directrix. 1 x + y −2 ⇒ (x − 1)2 + y 2 = 2 2
2
Thus required equation is 3x2 + 3y2 – 2xy – 4x + 4y = 0 15. (a) : Given equation can be rewritten as 2
2
x + y −5 x − y +7 2 2 + =1 1 3/2 It is clearly an ellipse whose major and minor axes are along the lines x – y + 7 = 0 and x + y – 5 = 0 Thus, centre of this ellipse is the intersection point of these lines i.e., the point (–1, 6) 16. (b) : OS1 = ae = 6, OC = b (say), CS1 = a 1 DOCS = ⋅ (OS1 )(OC ) = 3b 1 2 Semi perimeter of triangle OCS1 1 1 = (OS1 + OC + CS1) = (6 + a + b) 2 2
Now, inradius of triangle OCS1 3b = = 1 (given) 1 (6 + a + b) 2 ⇒ 5b = 6 + a Also b2 = a2 (1 – e2) = a2 – 36 Thus, 25b2 = 36 + a2 + 12a ⇒ 25(a2 – 36) = 36 + a2 + 12a 13 5 ⇒ 2a2 – a – 78 = 0 ⇒ a = , b = 2 2 65p Hence, area of ellipse = pab = sq. units. 4 17. (a) : Equation of normal at P(q) is, 3y 4x − =7 cos q sin q Since it passes through the point Q(2q), thus 16 ⋅ cos 2q 9 ⋅ sin 2q − =7 cos q sin q 16(2 cos2 q − 1) − 18 cos q = 7 cos q ⇒ 23 cos2q – 7 cosq – 16 = 0 16 ⇒ cosq = – [ as cosq ≠ 1] 23 18. (d) : In this case, base S1S2 is fixed and PS1 + PS2 is fixed. Hence area will be maximum when PS1 = PS2 1 \ Maximum area = (2ae)b = abe sq. units. 2 19. (d) : Here, S1S2 = 2ae, PS1 + PS2 = 2a If the normal to ellipse at P, meets the x-axis at Q, then incentre of DPS1S2 will lie on segment PQ and PS + PS2 1 will divide it in the ratio 1 i.e., S S e 1 2 Let P ≡ (a cosq, b sinq) ⇒
Equation of normal at ‘P’ is by ax − = a 2 − b2 = a 2 e 2 cos q sin q ⇒ Q ≡ (ae2 cosq , 0) Let R(h, k) be the incenter, then h=
a cos q ⋅ e + ae 2 cos q b sin q ⋅ e + 0 ⋅1 , k= 1 1 1+ 1+ e e
1 1 ⇒ h 1 + = (ae + ae 2 ) cos q and be ⋅ sin q = k 1 + e e Thus locus of R is
1 x 1 + e 2
2
1 y 1 + e 2
+
a2e 2 (1 + e)2
2
b2 e 2
=1
which is clearly an ellipse. 20. (a) : S1 = (ae, 0), S2 = (–ae, 0) Let the tangent be xb cosq + ay sinq – ab = 0 S1P1 = ⇒ S1P1 =
| abe cos q − ab | b2 cos2 q + a2 sin2 q ab(1 − e cos q) b2 cos2 q + a2 sin2 q ab(1 + e cos q)
Similarly, S2P2 = ⇒ S1P1·S2P2 = =
2
b cos2 q + a2 sin2 q a2b2 (1 − e 2 cos2 q)
b2 cos2 q + a2 sin2 q
a2b2 (1 − e 2 cos2 q) 2
2
2
2
a + (b − a )cos q
=
a2b2 (1 − e 2 cos2 q) 2
2 2
2
a − a e cos q
= b2
which is clearly a constant Thus, (S1P1) (S2P2) = b2 21. (d) : Let the equation of ellipse having centre (1, 2) be
(x − 1)2
( y − 2)2
=1 a2 b2 We have, ae = 5 ⇒ a2e2 = 25 Also,
9 a
2
+
16 b2
+
=1
Now, b2 = a2(1 – e2) = a2 – 25 9 16 ⇒ + = 1 ⇒ a4 – 50a2 + 225 = 0 2 2 a a − 25 ⇒ a2 = 5, 45 But a2 = 5 is clearly rejected \ a2 = 45 ⇒ b2 = 20 (x − 1)2 ( y − 2)2 Thus, required equation is + =1 45 20 22. (d) : Tangent to the ellipse is 5m2 5 + 3 2 5m2 5 If it passes through P(1, 2), then (2 – m)2 = + 3 2 ⇒ 4m2 + 24m – 9 = 0 If it’s roots are m1 and m2 then y = mx ±
mathematics today | February‘17
57
m1 + m2 = – 6, m1m2 = –
9 4
⇒ | m1 − m2 | = 36 + 9 = 3 5 \
⇒
m1 − m2 3 5 12 = = 9 1 + m1m2 5 1− 4 12 q = tan −1 5
⇒ A2 ⋅
23. (c) : Equation of normal at point ‘q’ is 3y 5x − = 16 cos q sin q On comparing with 5x – 3y = 8 2, we get p 1 ⇒ q= ⇒ cos q = sin q = 4 2
⇒
25. (b) : Equation of tangent at point ‘q’ is, x 11 cos q + y sin q = 1 4 16 It will touch the given circle, if cos q −1 4 =4 cos2 q 11sin2 q + 16 256 2 ⇒ 4 cos q + 8 cosq – 5 = 0 1 ⇒ cos q = ⇒ q = p , 5p 2 3 3 26. (c) : The given distance is clearly the length of semi major axis a2 + 2b2 Thus, =a 2 ⇒ 2b2 = a2 ⇒ 2a2 (1 – e2) = a2 1 1 ⇒ e2 = ⇒ e = 2 2
mathematics today | February‘17
1 − a2
x2 A
2
+
y2 B
2
4a 2
1− a
2
+ B2 =
1 1 − a2
a 2 ( 4 A2 − B 2 ) + B 2 2
=
1 2
⇒ A2 =
1 4
1− a 1− a If ‘e’ be the eccentricity of the ellipse, then A2 = B2(1 – e2) 1 3 ⇒ 1 − e2 = ⇒ e= 4 2 28. (d) : Any normal to the ellipse is 2y 5x − =1 cos q sin q
24. (c) : Let the point ‘P’ be ( 6 cos q, 2 sin q) A.T.Q. 6 cos2q + 2 sin2q = 4 1 ⇒ 6 – 4 sin2q = 4 ⇒ sin2q = 2 p ⇒ q = ( q lies in first quadrant) 4
58
1 − a2
1
x+
2 2 2 Comparing it with y = mx ± A m + B , we get 2a m=− 1 − a2 1 A2m2 + B2 = 1 − a2
tan q =
27. (a) : Let the equation of ellipse be
2a
We have, y = −
sin q i.e., y = x . 5 tanq – 2 2 2 Since (i) is tangent to y = 4x 1 \ y = mx + m ⇒ m= ⇒
...(i)
5 1 sin q tan q, = − 2 m 2
1 5 sin2 q , and cos q = 5 ⋅ = −1 ⇒ cos q = − 5 4 cos q
1 1 ⇒ q = p − cos −1 or p + cos −1 5 5 29. (a) : Let line AB be bx cosq + ay sinq – ab = 0 Now, combined equation of lines OA and OB is x2
+
y2
=
y a cos q + b sin q
x a2
2
a2 a i.e., x2 · sin2q + y2 1 − sin2 q − sin 2q · xy = 0 b2 b Since the lines OA and OB are mutually perpendicular, hence =1
sin2 q + 1 −
a2
b
2
sin2 q = 0
a2 1 ⇒ sin q − 1 = 1 ⇒ sin2 q − 1 = 1 2 2 1 − e b 2
⇒ e2 · sin2q = 1 – e2 ⇒ e2 (1 + sin2q ) = 1 30. (a) : Let the extremities of the focal chord be P1(q1) and P2(q2). Equation of chord P1P2 is x q +q y q +q q −q cos 1 2 + sin 1 2 = cos 1 2 2 b 2 2 a It shou ld p ass t hroug h (ae, 0) and its slop e should be tan q q +q q −q ⇒ e cos 1 2 = cos 1 2 2 2 b q +q and − cot 1 2 = tan q 2 a a tan q q +q ⇒ cot 1 2 = − 2 b Now, P1P2 = P1S1 + P2S1 = a – ea cosq1 + a– ea cosq2 = 2a – ea(cosq1 + cosq2) q −q q +q = 2a – 2ae cos 1 2 cos 1 2 2 2 q +q = 2a – 2ae2 cos2 1 2 2 2 2 = 2a – 2ae2 · a tan q a2 tan2 q + b2 3 2
= =
2
a2 sin2 q + b2 cos2 q 2a sin2 q(1 − e 2 ) + 2ab2 cos2 q 3
a2 sin2 q + b2 cos2 q
2ab2 sin2 q + 2ab2 cos2 q a2 sin2 q + b2 cos2 q
=
2ab2 a2 sin2 q + b2 cos2 q
31. (a) : Let the foot of normal is (a cosq, b sinq), by ax then − = a 2 − b2 cos q sin q and the given lines will be identical, m cos q sin q c Thus, = = 2 a b (b − a2 ) ac bc ⇒ cos q = , sin q = 2 2 2 m(b − a ) (b − a2 ) ⇒
a2 m2
+ b2 =
(b2 − a2 )2 c2
Thus,
h2 a2
+
k2 b2
=4
Hence, locus is,
x2 a2
+
y2 b2
=4
33. (d) : Let P ≡ (a cosq, b sinq) ⇒ P1 ≡ (a cosq, 0), P2 ≡ (0, b sinq) y x + =1 Thus, equation of line P1P2 is a cos q b sin q y /b x /a ⇒ − =1 cos(−q) sin(−q) which is clearly a normal to the ellipse of the form x2
2
+
y2
2
A
= 1 where,
2
2
=
B λ λ and = 2 2 a b A −B
A −B A B If a > b, then B > A Let the eccentricity of the second ellipse be e1 ⇒ 1 − e12 =
2a e sin q
= 2a −
32. (c) : If the point of intersection be R(h, k), then q +q q +q a cos 1 2 b sin 1 2 2 2 , k= h= q −q q −q cos 1 2 cos 1 2 2 2 q +q q +q b sin 1 2 a cos 1 2 2 2 , k= ⇒ h= 1 1 2 2
A2 B
2
=
b2 a
2
= 1 − e 2 ⇒ e1 = e
b2 34. (a) : Equation of normal at ae, is a a(x − ae) = (ay – b2) e It should pass through (0, – b) a(0 − ae) = – a2 – b2 e b2 b ⇒ a2 = ab + b2 ⇒ 2 + − 1 = 0 a a ⇒
⇒
b −1 + 5 b2 3 − 5 = ⇒ = = 1 − e2 2 a 2 2 a
⇒ e2 =
5 −1 ⇒ e= 2
5 −1 2
35. (c) : Tangent for y2 = 4x is y = mx +
1 m
mathematics today | February‘17
59
x2 y2 + = 1 is y = mx ± 8m2 + 2 8 2 For common tangents we must have 1 1 = ± 8m2 + 2 ⇒ m = ± m 2 Thus, common tangents are x x y = + 2 and y = − − 2 2 2 i.e. 2y – x – 4 = 0 and 2y + x + 4 = 0 36. (c) : Any normal to the ellipse is 4y 5x − = 9 i.e., 5 secq x – 4 cosecq y – 9 = 0 cos q sin q 9 \ L= 2 25 sec q + 16 cosec2q 9 = 2 25(1 + tan q) + 16(1 + cot 2 q) 9 = 2 25 tan q + 16 cot 2 q + 41 25 tan2 q + 16 cot 2 q Now, ≥ 20 2 ⇒ 25 tan2q + 16 cot2q + 41 ≥ 81 9 ⇒ ≤1 ⇒ L ≤ 1 2 25 tan q + 16 cot 2 q + 41 \ Maximum value of L is 1. 37. (a) : Clearly, the tangents have been drawn by taking any point on the director circle of the ellipse. Any point on the director circle can be taken as ( 13 cos q, 13 sin q). Equation of corresponding chord of contact is 13 13 .cos q.x + sin q y = 1 9 4 and for
\
Distance from the origin =
1
13 2 13 cos q + sin2 q 81 16 36 36 9 = = = 208 13 1053 − 845 cos2 q
38. (b) : Equation of AB is T = 0, 5y 9 i.e. =1 ⇒ y = 9 5 9 Putting y = in the equation of ellipse, we get 5 81 16 32 x2 + = 1 ⇒ x = ± . Thus, AB = 16 25 ⋅ 9 5 5
60
mathematics today | February‘17
9 16 = 5 5 1 32 16 256 Thus, area of D = ⋅ ⋅ = sq. units PAB 2 5 5 25 39. (a) : Other common tangent will clearly be the reflection of x – 2y + 4 = 0 in the x-axis. Thus other common tangent will be x + 2y + 4 = 0 40. (c) : Let the eccentric angle of point ‘P’ be q, then equation of tangent and normal at point are respectively 3y y 6x x − = 27 cos q + sin q = 1 and cos q sin q 6 3 Distance of P from AB = 5 –
27 3 Thus, A ≡ 0, and B ≡ cos q, 0 6 sin q 1 3 27 27 Area of DOAB = ⋅ (given) ⋅ ⋅ cos q = 2 sin q 6 4 p 5p ⇒ cotq = 1 ⇒ q = , 4 4 41. (c) : Let P ≡ (2 cosq, sinq) x Equation of tangent at ‘P’ is cosq + y sinq = 1 2 ⇒ A ≡ (2 secq, 0)
y 2x − =3 Equation of normal at ‘P’ is cos q sin q 3 ⇒ B ≡ cos q, 0 2 3 Now, AB = | 2 secq – cosq | = 2 (given) 2 ⇒ |4 – 3 cos2q| = 4 |cosq| ⇒ 3 cos2q ± 4 cosq – 4 = 0 2 2 ⇒ cosq = , − 3 3 42. (a) : Equation of chord is x q +q y q +q q −q cos 1 2 + sin 1 2 = cos 1 2 2 b 2 2 a q1 − q2 a cos 2 , 0 ⇒ A≡ q1 + q2 cos 2 Since OA = c, therefore, q −q q +q a cos 1 2 = c.cos 1 2 2 2 q1 q2 q q ⇒ cos .cos (a − c) = sin 1 .sin 2 (−c − a) 2 2 2 2 q q c−a ⇒ tan 1 . tan 2 = 2 2 c+a
mathematics today | February‘17
61
43. (b) : Equation of tangent at point y x cos q + sin q = 1 ( 18 cos q, 32 sin q) , is 18 32 32 4 Its slope = – . cotq = – (given) 3 18 ⇒ cotq = 1 Also, A ≡ ( 18 sec q, 0), B ≡ (0, 32 cosec q ) 1 ⇒ DOAB = 3 2 sec q ⋅ 4 2cosec q 2
Thus, OS1 = OB = OS2 ⇒ ae = b ⇒ a2e2 = a2 – a2 e2 1 1 ⇒ e= ⇒ e2 = 2 2
= 12 ⋅ 2 ⋅ 2 = 24 sq. units
=c b 4 h2 + a 4 k 2 Hence, locus of point ‘P ’ is
44. (a) : Any tangent to
x2
a 2 + b2
+
y2 b2
= 1 is,
y = mx ± (a2 + b2 )m2 + b2 It will also be a tangent to second ellipse if a2m2 + (a2 + b2) = (a2 + b2) m2 + b2 a ⇒ b2m2 = a2 ⇒ m = ± b Thus common tangents are y=±
2
a a x ± (a2 + b2 ) ⋅ + b2 b b2
i.e., by = ± ax ± a 4 + b 4 + a2b2 45. (a) : Let the extremities of the chord be P1 ≡ (a cosq, b sinq) and P2 ≡ (–a sinq, b cosq) Now, P1P22 = a2 (cosq + sinq)2 + b2(sinq – cosq)2 = a2 + b2 + (a2 – b2)2sinqcosq ≤ a2 + b2 + a2 – b2 = 2a2 ⇒ P 1P 2 ≤ a 2 46. (d) : Let the ellipse be
x
2
y
=1 a b2 If the line y = mx + c be its tangent, then c2 = a2m2 + b2 Tangent is given to be x + 4y – 10 = 0 2
+
2
m 1 c 1 5 25 a2 =− = ⇒m= − ,c = \ = + b2 1 4 −10 4 2 4 16 ⇒ a2 + 16b2 = 100 We also have P(4, –1) to be a point on the ellipse, 16 1 \ + =1 a 2 b2 ⇒ a2 + 16b2 = a2b2 ⇒ a2b2 = 100 ⇒ ab = 10 \ Area of ellipse = pab = 10p sq. units 47. (b) : If ‘O’ be the center of ellipse then for triangle S1S2B to be right angled, we must have p ∠OBS1 = ∠OBS2 = . 4 ⇒
62
mathematics today | February‘17
48. (c) : Let P ≡ (h, k)
xh
Equation of chord of contact is
a
+
2
i.e., xb2h + ya2k – a2b2 = 0 It will touch the circle x2 + y2 = c2 , if
yk b2
=1
a 2 b2
b4 x2 + a4y2 =
a 4b 4 c
2
i.e.,
x2 a
4
+
y2 b
4
=
1 c2
9 . 8 Slope of tangent at any point P(q) on the ellipse 49. (c) : Slope of tangent must be – x2
a
2
+
y2
= 1 is − b cot q a b 2
9 2 16 Thus, − = − cot q ⇒ tan q = 8 3 27 ⇒ cos q = − or cos q = −
27 985 27
, sin q = , sin q = −
985 Thus, required point is 27 16 , or 2 985 3 985
16 985 16 985
27 16 ,− − 2 985 3 985
50. (b) : Let P ≡ (h, k) Any tangent to the ellipse is y = mx ± 16m2 + 9 If it passes through ‘P’, then (k – mh)2 = 16m2 + 9 ⇒ m2(h2 – 16) – 2mhk + (k2 – 9) = 0 If its roots are m1 and m2, then m1m2 =
k2 − 9
h2 − 16 Since, the drawn tangents meet the coordinate axes at concyclic points, thus, m1m2 = 1 ⇒ k2 – 9 = h2 – 16 Hence required locus is, x2 – y2 = 7.
on
Conic Sections
*ALOK KUMAR, B.Tech, IIT Kanpur
important facts and formulae z
z
z
z
z
z
z
z
z
z
z
z
The area of the triangle inscribed in the parabola 1 y2 = 4ax is |( y − y )( y − y3)( y3 − y1)|, where 8a 1 2 2 y1, y2, y3 are the ordinates of the vertices. The length of the side of an equilateral triangle inscribed in the parabola y2 = 4ax is 8a 3. (one angular point is at the vertex). y2 = 4a(x + a) is the equation of the parabola whose focus is the origin and the axis is x-axis. y2 = 4a(x – a) is the equation of parabola whose axis is x-axis and y-axis is directrix. x2 = 4a(y + a) is the equation of parabola whose focus is the origin and the axis is y-axis. x2 = 4a(y – a) is the equation of parabola whose axis is y-axis and the x-axis is directrix. The equation of the parabola whose vertex and focus are on x-axis at a distance a and a′ respectively from the origin is y2 = 4(a′ – a)(x – a). The equation of the parabola whose axis is parallel to x-axis is x = Ay2 + By + C and y = Ax2 + Bx + C is a parabola with its axis parallel to y-axis. If the straight line lx + my + n = 0 touches the parabola y2 = 4ax, then ln = am2. If the line xcosa + ysina = p touches the parabola y2 = 4ax then pcosa + asin2a = 0 and point of contact is (atan2a, –2atana). x y If the line + = 1 touches the parabola l m y2 = 4a(x + b), then m2(l + b) + al2 = 0. If the two parabolas y2 = 4x and x2 = 4y intersect at point P, whose absiccae is not zero, then the tangent to each curve at P, make complementary angle with the x-axis.
z
z
z
z
z
Tangents at the extremities of any focal chord of a parabola meet at right angle on the directrix. Area of the triangle formed by three points on a parabola is twice the area of the triangle formed by the tangents at these points. If the tangents at the points P and Q on a parabola meet in T, then ST is the geometric mean between SP and SQ, i.e., ST2 = SP·SQ Tangent at one extremity of the focal chord of a parabola is parallel to the normal at the other extremity. The angle of intersection of two parabolas y2 = 4ax and x2 = 4by is given by tan −1
z
z
z
z
z
2(a2/3 + b2/3)
.
The equation of the common tangent to the parabolas y2
z
3a1/3b1/3
= 4ax and
x2
= 4by is
1 1 2 2 3 3 a x + b y + a 3b 3
=0.
The line lx + my + n = 0 is a normal to the parabola y2 = 4ax, if al(l2 + 2m2) + m2n = 0. If the normals at points (at12, 2at1) and (at22, 2at2) on the parabola y2 = 4ax meet on the parabola, then t1t2 = 2. If the normal at a point P(at2, 2at) to the parabola y2 = 4ax subtends a right angle at the vertex of the parabola then t2 = 2. If the normal to a parabola y2 = 4ax, makes an angle f with the axis, then it will cut the curve again at an 1 angle tan −1 tan f . 2 If the normal at two points P and Q of a parabola y2 = 4ax intersect at a third point R on the curve. Then the product of the ordinate of P and Q is 8a2.
* Alok Kumar is a winner of INDIAN NATIONAL MATHEMATICS OLYMPIAD (INMO-91). He trains IIT and Olympiad aspirants.
mathematics today | FEBRUARY ‘17
63
z
z
z
The chord of contact joining the point of contact of two perpendicular tangents always passes through focus. If tangents are drawn (at12 , 2at1) Q from the point (x1, y1) to the parabola y2 = 4ax, (x1, y1)P then the length of their (at22 , 2at2)R chord of contact is 1 ( y12 − 4ax1)( y12 + 4a2 ) |a | The area of the triangle formed by the tangents drawn from (x1, y1) to y2 = 4ax and their chord of ( y12 − 4ax1)3/2 . 2a If one extremity of a focal chord is (at12, 2at1), then a −2a the other extremity (at22, 2at2) becomes , t12 t1 by virtue of relation t1t2 = –1 and length of chord
z
x2
y2 + = 1 , if a2cos2a + b2sin2a = p2 and that a 2 b2 2 2 point of contact is a cos a , b sin a . p p z
y2 + = 1 , then the common tangent is inclined a 2 b2 r 2 − b2 to the major axis at an angle tan −1 2 2 . a −r z
z
z
on the parabola y2 = 4ax is a(t1 − t2 ) (t1 + t2 )2 + 4 The length of intercept made by line y = mx + c between 4 the parabola y2 = 4ax is a(1 + m2 )(a − mc) . m2 Locus of mid-points of all chords which is inscribed a right angle on the vertices of parabola is parabola. The focal chord of parabola y2 = 4ax making an angle a with the x-axis is of length 4acosec2a and perpendicular on it from the vertex is asina. The length of a focal chord of a parabola varies inversely as the square of its distance from the vertex. If l1 and l2 are the length of segments of a focal chord 4l l of a parabola, then its latus-rectum is 1 2 . l1 + l2 The semi-latus rectum of the parabola y2 = 4ax is
z
the harmonic mean between the segments of any focal chord of the parabola. The straight line lx + my + n = 0 touches the ellipse
z
z
z
z
z
x2
a2 64
y2
+ 2 = 1 , if b
a2l2
+
b2m2
=
n2.
mathematics today | FEBRUARY‘17
y2 + = 1 is (x2 + y2)2 = a2 x2 + b2y2 or r2 = a 2 b2 a2cos2q + b2sin2q. (In terms of polar coordinates) The locus of the mid points of the portion of the
y2 + = 1 intercepted a 2 b2 between the axes is a2y2 + b2x2 = 4x2y2. tangents to the ellipse
2
The length of the chord joining two points 't1' and 't2'
The locus of the foot of the perpendicular drawn from centre upon any tangent to the ellipse x2
1 = a t + . t z
A circle of radius r is concentric with the ellipse x2
contact is z
The line xcosa + ysina = p touches the ellipse
x2
x2
y2 + = 1, a 2 b2
z
If y = mx + c is the normal to the ellipse
z
m2(a2 − b2 )2 then condition of normality is c 2 = 2 2 2 . (a + b m ) The straight line lx + my + n = 0 is a normal to the
2 2 2 2 2 y2 if a + b = (a − b ) . + = 1 , a 2 b2 l 2 m2 n2 Four normals can be drawn from a point to an ellipse. Y Normal If S be the focus P(x1, y1) and G be the point where the normal at X X S C G S T P meets the axis of an el l ip s e, t he n Y SG = e. SP, and the tangent and normal at P bisect the external and internal angles between the focal distances of P. Any point P of an ellipse is joined to the extremities of the major axis then the portion of a directrix intercepted by them subtends a right angle at the corresponding focus. The equations to the normals at the end of the latus rectum and that each passes through an end of the minor axis, if e4 + e2 – 1 = 0.
ellipse z
z
z
z
x2
z
The area of the triangle formed by the three points, on 2
y + 2 = 1 , whose eccentric angles are q, a b q−f f − y y −q f and y is 2absin sin sin 2 2 2 the ellipse
z
x
2
z
2
z
If the point of intersection of the ellipses x2
y2 x2 y2 + = 1 and + = 1 be at the extremities a 2 b2 a 2 b2 of the conjugate diameters of the former, then a2
b2 + =2. a 2 b2
z
z
z
z
z
y2 − = 1 , then a2l2 – b2m2 = n2. a 2 b2 If t h e s t r ai g ht l i n e x c o s a + y s i n a = p hyperbola
z
x2
tou che s t he hy p e r b ol as a2 cos2 a − b2 sin2 a = p2 . z
x2
normal to the hyperbola a2
z
hyperbola from any point. If a, b, g are the eccentric angles of three points on
y2 − = 1 , the normals at which a 2 b2 are concurrent, then sin(a + b) + sin(b + g) + sin(g + a) = 0
q q 1− e y2 . − = 1 , then tan 1 tan 2 = 2 2 2 2 1+ e a b
If the polars of (x1, y1) and (x2, y2) with respect
then z
z
x2
y2 − = 1 are at right angles, a 2 b2
x1x2 a 4 + =0. y1 y2 b 4
The parallelogram formed by the tangents at the extremities of conjugate diameters of a hyperbola has its vertices lying on the asymptotes and is of constant area. The product of length of perpendiculars drawn from any point on the hyperbola the asymptotes is
z
y2 − = 1 , then a 2 b2
z
x2
to the hyperbola
x2
b2 (a2 + b2 )2 − = l 2 m2 n2 In general, four normals can be drawn to a
the hyperbola
z
z
z
x2
z
.
If the chord joining two points (asecq1, btanq1) (asecq2, btanq2) passes through the focus of the hyperbola
y2
If the line lx + my + n = 0 will be
x2 y2 The length of chord cut off by hyperbola 2 − 2 = 1 a b from the line y = mx + c is (b2 − a2m2 )
− 2 = 1 , t he n a b 2
y2 − = 1 from (h, k) a 2 b2 lie on a2y(x – h) + b2x(y – k) = 0.
2ab [c 2 − (a2m2 − b2 )](1 + m2 )
z
The sum of the squares of the reciprocal of two perpendicular diameters of an ellipse is constant. In an ellipse, the major axis bisects all chords parallel to the minor axis and vice-versa, therefore major and minor axes of an ellipse are conjugate diameters of the ellipse but they do not satisfy the condition m1 · m2 = –b2/a2 and are the only perpendicular conjugate diameters. The foci of a hyperbola and its conjugate are concyclic. Two tangents can be drawn from an outside point to a hyperbola. If the straight line lx + my + n = 0 touches the
x2
The feet of the normals to
a2b2
a 2 + b2
x2
y2 − = 1 to a 2 b2
.
c If the normal at ct , on the curve xy = c2 meets t 1 the curve again in 't', then t ′ = − 3 . t A triangle has its vertices on a rectangular hyperbola; then the orthocentre of the triangle also lies on the same hyperbola. All conics passing through the intersection of two rectangular hyperbolas are themselves rectangular hyperbolas. An infinite number of triangles can be inscribed in the rectangular hyperbola xy = c2 whose all sides touch the parabola y2 = 4ax. mathematics today | FEBRUARY ‘17
65
proBlems Single Correct Answer Type
1. The equation of the common tangent of the parabolas x2 = 108y and y2 = 32x, is (a) 2x + 3y = 36 (b) 2x + 3y + 36 = 0 (c) 3x + 2y = 36 (d) 3x + 2y + 36 = 0 2. The line xcosa + ysina = p will touch the parabola y2 = 4a(x + a), if (a) pcosa + a = 0 (b) pcosa – a = 0 (c) acosa + p = 0 (d) acosa – p = 0 3. The angle of intersection between the curves y2 = 4x and x2 = 32y at point (16, 8), is 4 3 (a) tan −1 (b) tan −1 5 5 p (c) p (d) 2 4. If y1, y2 are the ordinates of two points P and Q on the parabola and y3 is the ordinate of the point of intersection of tangents at P and Q, then (a) y1, y2, y3 are in A.P. (b) y1, y3, y2 are in A.P. (c) y1, y2, y3 are in G.P. (d) y1, y3, y2 are in G.P. 5. The equation of the common tangent touching the circle (x – 3)2 + y2 = 9 and the parabola y2 = 4x above the x-axis, is (a) (b) 3 y = 3x + 1 3 y = −(x + 3) (c)
3y = x + 3
(d)
3 y = −(3x + 1)
6. The angle between the tangents drawn from the points (1,4) to the parabola y2 = 4x is p p p p (a) (b) (c) (d) 2 3 6 4 7. Let a circle tangent to the directrix of a parabola y2 =2ax has its centre coinciding with the focus of the parabola. Then the point of intersection of the parabola and circle is (a) (a, –a) (b) (a/2, a/2) (c) (a/2, ± a) (d) (± a, a/2) 8.
An ellipse passes through the point (–3, 1) and its
2 The equation of the ellipse is . 5 (a) 3x 2 + 5 y 2 = 32 (b) 3x 2 + 5 y 2 = 25 (c) 3x 2 + y 2 = 4 (d) 3x 2 + y 2 = 9 eccentricity is
9.
The locus of the point of intersection of 2
2
perpendicular tangents to the ellipse x + y = 1 , is a 2 b2 66
mathematics today | FEBRUARY‘17
(a) x2 + y2 = a2 – b2 (c) x2 + y2 = a2 + b2
(b) x2 – y2 = a2 – b2 (d) x2 – y2 = a2 + b2
2 2 10. If the eccentricity of the two ellipse x + y = 1 169 25 2 2 and x + y = 1 are equal, then the value of a/b is a 2 b2 (a) 5/13 (b) 6/13 (c) 13/5 (d) 13/6
11. The eccentricity of the curve represented by the equation x2 + 2y2 – 2x + 3y + 2 = 0 is (a) 0 (b) 1/2 (c) 1 / 2 (d) 2 12. The equation of the hyperbola whose conjugate axis is 5 and the distance between the foci is 13, is (a) 25x2 – 144y2 = 900 (b) 144x2 – 25y2 = 900 (c) 144x2 + 25y2 = 900 (d) 25x2 + 144y2 = 900 13. The vertices of a hyperbola are at (0, 0) and (10, 0) and one of its foci is at (18, 0). The equation of the hyperbola is (a)
x2 y2 − =1 25 144
2 2 (b) (x − 5) − y = 1 25 144
2 2 x 2 ( y − 5)2 (d) (x − 5) − ( y − 5) = 1 − =1 25 144 25 144 14. The equation of the hyperbola whose foci are (6, 4) and (–4, 4) and eccentricity 2 is given by (a) 12x 2 − 4 y 2 − 24x + 32 y − 127 = 0
(c)
(b) (c)
12x 2 + 4 y 2 + 24x − 32 y − 127 = 0 12x 2 − 4 y 2 − 24x − 32 y + 127 = 0
(d) 12x 2 − 4 y 2 + 24x + 32 y + 127 = 0 15. The latus rectum of the hyperbola 9x2 – 16y2 + 72x – 32y – 16 = 0 is 32 9 32 9 (a) (b) − (c) (d) − 3 2 3 2 16. If m1 and m2 are the slopes of the tangents to the x2 y2 − = 1 which pass through the point 25 16 (6, 2), then hyperbola
24 11 48 (c) m1 + m2 = 11 (a) m1 + m2 =
(b) (d)
10 11 11 m1m2 = 20 m1m2 =
2 2 17. Let E be the ellipse x + y = 1 and C be the circle 9 4 x2 + y2 = 9. Let P and Q be the points (1, 2) and (2, 1)
respectively. Then (a) Q lies inside C but outside E (b) Q lies outside both C and E (c) P lies inside both C and E (d) P lies inside C but outside E
Comprehension Type
18. The value of m, for which the line y = mx + 2 2 is a normal to the conic x − y = 1, is 16 9 2 3 (a) (b) − (c) − (d) 1 3 3 2
25 3 , 3
19. If q is the acute angle of intersection at a real point of intersection of the circle x2 + y2 = 5 and the parabola y2 = 4x then tanq is equal to 1 (a) 1 (b) (c) 3 (d) 3 3 20. The equation of the hyperbola in the standard form (with transverse axis along the x-axis) having the length of the latus rectum = 9 units and eccentricity = 5/4 is (a)
2
2
x y − =1 16 18
(c)
x2 y2 − =1 64 36
(e)
x2 y2 − =1 16 9
(b) (d)
2
2
x y − =1 36 27 x2 y2 − =1 36 64
Paragraph for Q. No. 24 to 26 Consider the circles x2 + y2 = a2 and x2 + y2 = b2 where b > a > 0. Let A(–a, 0), B(a, 0). A parabola passes through the points A, B and its directrix is a tangent to x2 + y2 = b2. If the locus of focus of the parabola is a conic then 24. The eccentricity of the conic is (a) 2a/b (b) b/a (c) a/b
(d) 1
25. The foci of the conic are (a) (±2a, 0) (b) (0, ±a) (c) (±a, 2a) (d) (±a, 0) 26. Area of triangle formed by a latusrectum and the lines joining the end points of the latusrectum and the centre of the conic is a 2 2 (a) (b) 2ab (b − a ) b (c) ab/2 (d) 4ab/3 Paragraph for Q. No. 27 to 29 y2 + = 1. S and S′ are a 2 b2 foci and e is the eccentricity of ellipse. ∠PSS′ = a and ∠PS′S = b P is any point of an ellipse
27. tan
x2
a b tan is equal to 2 2
2e 1− e 1− e (c) 1+ e
1+ e 1− e 2e (d) 1+ e
21. The equations of the common tangents of the curves x2 + 4y2 = 8 and y2 = 4x are (a) x + 2y + 4 = 0 (b) x – 2y + 4 = 0 (c) 2x + y = 4 (d) 2x – y + 4 = 0
(a)
22. A straight line touches the rectangular hyperbola 9x2 – 9y2 = 8 and the parabola y2 = 32x. An equation of the line is (a) 9x + 3y – 8 = 0 (b) 9x – 3y + 8 = 0 (c) 9x + 3y + 8 = 0 (d) 9x – 3y – 8 = 0
28. Locus of incentre of triangle PSS′ is (a) an ellipse (b) hyperbola (c) parabola (c) circle
23. A hyperbola having the transverse axis of length 1 unit is confocal with the ellipse 3x2 + 4y2 = 12, then 2 x2 y2 1 (a) Equation of the hyperbola is − = 15 1 16 (b) Eccentricity of the hyperbola is 4 (c) Distance between the directrices of the hyperbola 1 is units 8 15 (d) Length of latus rectum of the hyperbola is 2 units
(b)
29. Eccentricity of conic, which is locus of incentre of triangle PSS′ (a)
e 1+ e
(b)
2e 1+ e
(c)
2e 1− e
(d)
e 1− e
Paragraph for Q. No. 30 to 32 A point P moves such that sum of the slopes of the normals drawn from it to the hyperbola xy = 16 is equal to the sum of ordinates of feet of normals. The locus of P is a curve C. mathematics today | FEBRUARY ‘17
67
30. The equation of the curve C is (a) x2 = 4y (b) x2 = 16y 2 (c) x = 12y (d) y2 = 8x
36. The equation of the curve on reflection of the ellipse
31. If the tangent to the curve C cuts the coordinate axis in A and B, then the locus of the middle point of AB is (a) x2 = 4y (b) x2 = 2y 2 (c) x + 2y = 0 (d) x2 + 4y = 0 32. Area of the equilateral triangle inscribed in a curve C having one vertex is the vertex of curve C. (a) 772 3 sq. units (b) 776 3 sq. units (c) 760 3 sq. units
(d) 768 3 sq. units
33. Match the following Consider the parabola y2 = 12x Column II
A. Tangent and normal at the extremities of the latus rectum intersect the x-axis at T & G respectively. The coordinates of middle point of T & G are
P.
B.
Variable chords of the parabola passing through a fixed point K on the axis, such that sum of the reciprocals of two parts of the chord through K, is a constant. Coordinates of K are
Q.
(3, 0)
C. AB and CD are the chords of a parabola which intersect at a point E on the axis. The radical axis of the two circles described on AB and CD as diameter always passes through the point
R.
(12, 0)
(0, 0)
2
circle
concentric
to
an
ellipse
cuts the ellipse at ‘P’ such that area of triangle P F1 F2 is 30 sq.units. If F1F2 = 13K where K ∈ Z then K = mathematics today | FEBRUARY‘17
x2
π y2 − = 1 is . Then the eccentricity of conjugate 2 2 3 a b hyperbola is x2
– 108y = 0
x2 T ≡ xx1 − 2a( y + y1) = 0 ⇒ xx1 − 54 y + 1 = 0 108 2 S2 ≡ y – 32x = 0 y2 T ≡ yy2 − 2a(x + x2 ) = 0 ⇒ yy2 − 16 x + 2 = 0 32 x1 54 −x12 54 = = 2 = r ⇒ x1 = 16r and y2 = r 16 y2 y2 2 −(16r) 9 \ =r ⇒ r =− 2 4 (54 / r) \
x1 = −36, y2 = −24, y1 =
4x 4y 17 + = 1 λ < passes through foci F1 and F2 289 λ2 2
68
x2 y2 + = 1 is 3K. Then K is equal to 9 5 38. The equation of Asymptotes of xy + 2x + 4y + 6 = 0 is xy + 2x + 4y + C = 0, then C = ___
1. (b) : S1 ≡
34. A line passing through (21, 30) and normal to the curve y = 2 x . If m is slope of the normal then m+6= 2
37. The area of the quadrilateral formed by the tangents at the end point of latus rectum to the ellipse
soLUtioNs
Integer Answer Type
35. A
is 16x2 + 9y2 + ax – 36y + b = 0 then the value of a + b – 125 =
39. If e is the eccentricity of the hyperbola (5x – 10)2 + 25e (5y + 15)2 = (12x – 5y + 1)2 then is equal to 13 40. If the angle between the asymptotes of hyperbola
Matrix – Match Type
Column I
(x − 4)2 ( y − 3)2 + = 1 about the line x – y – 2 = 0 16 9
(−36)2 (−24)2 = 12, x2 = = 18 108 32
\ Equation of common tangent is −36 ( y − 12) = (x + 36) ⇒ 2x + 3 y + 36 = 0 54 2. (a) : xcosa + ysina – p = 0 2ax – yy1 + 2a(x1 + 2a) = 0 −p cos a sin a From (i) and (ii), = = − y1 2a(x1 + 2a) 2a
...(i) ...(ii)
⇒ y1 = –2a tana and x1 = –pseca – 2a \ y2 = 4a(x + a) ⇒ 4a2 tan2a = –4a(pseca + a) ⇒ pcosa + a = 0
3. (a) : Using formula q = tan −1 \
3a1/3b1/3
2(a2/3 + b2/3)
q = tan −1
, where a = 1 and b = 8
6 3 = tan −1 5 2(1 + 4)
4. (b) : L e t t h e c o o r d i n a t e s o f P a n d Q b e (at12 , 2at1) and (at22 , 2at2 ) respectively. Then y1 = 2at1 and y2 = 2at2. The coordinates of the point of intersection of the tangents at P and Q are {at1t2, a(t1 + t2)} \ y3 = a(t1 + t2) y +y ⇒ y3 = 1 2 ⇒ y1, y3 and y2 are in A.P. 2 1 5. (c) : Any tangent to y2 = 4x is y = mx + . It touches m 1 3m + m the circle, if 3 = 1 + m2 1 ⇒ 9(1 + m ) = 3m + m 1 ⇒ 2 =3 m 2
\ m=±
2
1
Common tangent is, y =
1 3
x+ 3 ⇒
3y = x + 3
6. (b) 7. (c) : Given parabola is y2 = 2ax \ Focus (a/2, 0) and directrix is given by x = –a/2, as circle touches the directrix. \ Radius of circle = distance from the point (a/2, 0) y to the line x = –a/2 2
a a = + =a 2 2
(–a/2, 0) 2
(a/2, 0)
Equation of circle be x − a + y 2 = a2 2 Also y2 = 2ax a 3a Solving (i) and (ii) we get, x = , − 2 2 \
x2 y2 8. (a) : Let the equation of ellipse be 2 + 2 = 1 a b Q It passes through (– 3, 1) 9 1 a2 So, 2 + 2 = 1 ⇒ 9 + 2 = a2 a b b Given eccentricity is 2 / 5
...(i)
2 b2 b2 3 So, = 1 − 2 ⇒ 2 = 5 5 a a
...(ii)
32 32 From equation (i) and (ii), a2 = , b2 = 3 5 Hence, required equation of ellipse is 3x2 + 5y2 = 32 9. (c) : Let point be (h, k). Their pair of tangent will be x 2 y 2 h2 k 2 hx yk 2 2 + 2 − 1 2 + 2 − 1 = 2 + 2 − 1 b b a b a a
3 For the common tangent to be above the x-axis, 1 m= 3 \
Putting these values in y2 = 2ax, we get, a y = ±a at x = 2 3a and y = a −3 at x = − 2 [Which is imaginary value of y] \ Required points are (a/2, ±a)
x
...(i) ...(ii)
Pair of tangents will be perpendicular, if coefficient of x2 + coefficient of y2 = 0 ⇒
k2
h2 1 1 + = 2 + 2 ⇒ h 2 + k 2 = a 2 + b2 2 2 2 2 ab ab a b
Replace (h, k) by (x, y) ⇒ x2 + y2 = a2 + b2 10. (c) : In the first case, e = 1 − (25 / 169) In the second case, e′ = 1 − (b2 / a2) According to the given condition, 1 − b2 / a2 = 1 − (25 / 169) ⇒ b / a = 5 / 13 (Q a > 0, b > 0) ⇒ a / b = 13 / 5 11. (c) : Equation x2 + 2y2 – 2x + 3y + 2 = 0 can be written as 2
3 y+ 3 1 (x − 1) (x − 1) 4 = 1, +y+ = ⇒ + 2 4 16 (1 / 8) (1 / 16) 2
2
2
1 1 which is an ellipse with a2 = and b2 = 8 16 1 1 1 1 1 \ = (1 − e 2 ) ⇒ = (1 − e 2 ) ⇒ e = 16 8 16 8 2 mathematics today | FEBRUARY ‘17
69
12. (a) : Conjugate axis is 5 and distance between foci = 13 ⇒ 2b = 5 and 2ae = 13 Now, also we know for hyperbola 25 (13)2 2 = 2 (e − 1) 4 4e 25 169 169 169 13 ⇒ = − 2 or e 2 = ⇒ e= 4 4 4e 144 12 5 ⇒ a = 6, b = 2
b2 = a2(e 2 − 1) ⇒
\
Required equation is 25x2 – 144y2 = 900
13. (b) : 2a = 10, ⇒ a = 5 8 13 ae − a = 8 or e = 1 + = 5 5 \ b=5
132 2
−1 = 5 ×
12 = 12 5
5 and centre of hyperbola ≡(5, 0) (x − 5)2 ( y − 0)2 \ Required equation of hyperbola is − =1 25 144
14. (a) : Foci are (6,4) and (–4,4), e = 2 \
6− 4 4+ 4 , Centre is = (1, 4) 2 2
5 5 ⇒ 6 = 1 + ae ⇒ ae = 5 ⇒ a = and b = ( 3) 2 2
2 2 Hence, the required equation is (x − 1) − ( y − 4) = 1 (25 / 4) (75 / 4) or 12x2 – 4y2 – 24x + 32y – 127 = 0
15. (a) : Given equation of hyperbola is 9x2 – 16y2 + 72x – 32y – 16 = 0 ⇒ 9(x2 + 8x) – 16(y2 + 2y) – 16 = 0 ⇒ 9(x + 4)2 – 16(y + 1)2 = 144 (x + 4)2 ( y + 1)2 − =1 16 9 2b2 9 9 =2× = Therefore, latus rectum = 4 2 a ⇒
16. (a) : The line through (6,2) is y – 2 = m(x – 6) ⇒ y = mx + 2 – 6m Now from condition of tangency, (2 – 6m)2 = 25m2 – 16 ⇒ 36m2 + 4 – 24m – 25m2 + 16 = 0 ⇒ 11m2 – 24m + 20 = 0 If m1, m2 are its roots then 24 20 m1 + m2 = and m1m2 = 11 11 70
mathematics today | FEBRUARY‘17
2 2 17. (d) : The given ellipse is x + y = 1. The value 9 4 x2 y2 of the expression + − 1 is positive for x = 1, 9 4 y = 2 and negative for x = 2, y = 1. Therefore P lies outside E and Q lies inside E. The value of the expression x2 + y2 – 9 is negative for both the points P and Q. Therefore P and Q both lie inside C. Hence P lies inside C but outside E. 18. (b) : Any normal to the hyperbola is ax by ...(i) + = a 2 + b2 sec q tan q But it is given by lx + m′y – n = 0 ...(ii) Comparing (i) and (ii), we get a n b n sec q = 2 2 and tan q = l a +b m′ a2 + b2
b2 (a2 + b2 )2 − = ...(iii) l 2 m′2 n2 25 3 Since a2 = 16, b2 = 9, l = m, m′ = –1 and n = 3 2 . \ On substituting in (iii), we get m = ± 3 19. (c) : Solving equations x2 + y2 = 5 and y2 = 4x we get, x2 + 4x – 5 = 0 i.e., x = 1, – 5 For x = 1; y2 = 4 ⇒ y = ± 2 For x = –5 ; y2 = –20 (imaginary values) \ Points are (1, 2)(1, –2) m1 for x2 + y2 = 5 at (1, 2) is dy x 1 =− =− dx y (1, 2) 2
Hence eliminating q, we get
a2
Similarly, m2 for y2 = 4x at (1, 2) is 1 1 − −1 m1 − m2 \ tan q = = 2 =3 1 1 + m1m2 1− 2 2 20. (c) : Q 2b = 9 ⇒ 2b2 = 9a2 a2 9 4 Now b2 = a2(e 2 − 1) = a2 ⇒ a = b 16 3
From (i) and (ii), b = 6, a = 8 Hence, equation of hyperbola
x2 y2 − =1 64 36
...(i) ...(ii) 5 Q e = 4
21. (a, b) :
x2 y2 + = 1, y 2 = 4x 8 2
Any tangent to parabola is y = mx +
1 m
If this line is tangent to ellipse then 1
m2
= 8m2 + 2 ⇒ 8m4 + 2m2 − 1 = 0
−2 ± 4 + 32 −2 ± 6 = 16 16 1 1 ⇒ m2 = ⇒ m = ± 4 2 x x y = + 2 or y = − − 2 2 2 x – 2y + 4 = 0 or x + 2y + 4 = 0 \
m2 =
22. (b, c) : y2 = 32x
8 Let equation of tangent be y = mx + m 64 8 2 8 = m − 9 m2 9 m = ± 3, y = ± 3x ± 8/3
2 2 23. (b, c, d) : Ellipse is x + y = 1 4 3 3 1 Here, = 1 − e 2 ⇒ e = 4 2 Foci are (± 1, 0) Now the hyperbola is having same focus i.e. (± 1, 0) Let e1 be the eccentricity of hyperbola 2ae1 = 2 1 But 2a = So, e1 = 4 2 1 15 b2 = a2 e 2 − 1 = (16 − 1) = 16 16 So, the equation of the hyperbola is
(
2
2
1
)
2
2
1 x y x y − =1⇒ − = 1 15 1 15 16 16 16 Its distance between the directrices 2a 1 1 = = = units e1 2 × 4 8 2b2 2 × 15 × 4 15 = = units \ Length of latusrectum = 16 × 1 2 a (24 - 26) : 24. (c) 25. (d) 26. (a) x 2 + y 2 = a 2 ; x 2 + y 2 = b 2 ; b > a > 0, A = (–a, 0); B = (a, 0)
Let (h, k) be a point on the locus. Any tangent to circle x2 + y2 = b2 is xcosq + ysinq = b \ Equation of parabola is (x − h)2 + ( y − k)2 =| x cos q + y sin q − b | i.e., (x – h)2 + (y – k)2 = (xcosq + ysinq – b)2 The points (±a, 0) satisfy this equation \ (a – h)2 + k2 = (acosq – b)2 (a + h)2 + k2 = (acosq + b)2 Subtracting (1) from (2), we get h = bcosq 2 ax \ Required locus is (a + x)2 + y 2 = + b b 2 2 x y i.e., + 2 2 = 1 which is an ellipse. 2 b b −a 27. (c) : or
PS PS′ 2ae = = sin b sin a sin(π − (a + b)
2a 2ae = sin a + sin b sin(a + b)
a +b a −b 1 2 sin 2 .cos 2 ⇒ = e 2 sin a + b cos a + b 2 2 1− e a b \ = tan tan 1+ e 2 2
y β
S′ O
...(1) ...(2)
P α S
b 28 (a) : y − 0 = tan (x + ae) 2 a y − 0 = − tan (x − ae) 2 1− e 2 2 2 or, y 2 = − [x − a e ] 1 + e
x
... (i) ... (ii)
2 y2 1− e 2 2 1 − e 2 2 or, x + =1 ⇒ + = x y a e 1 + e 1+ e a2e 2 1 − e a2e 2 1 + e
which is clearly an ellipse.
1− e 2e = 1+ e 1+ e 4 30. (b) : Any point on the hyperbola xy = 16 is 4t , t of the normal passes through P(h, k), then k – 4/t = t2(h – 4t) ⇒ 4t4 – t3h + tk – 4 = 0 h \ ∑ t1 = and St1t2 = 0 4 k ∑ t1t2t3 = − 4 and t1t2t3t 4 = −1 1 1 1 1 k \ + + + = ⇒ y1 + y2 + y3 + y 4 = k t1 t2 t3 t 4 4 29. (b) : e ′ = 1 −
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71
h2 Now, = =k 16 ⇒ Locus of (h, k) is x2 = 16y. 31. (c) : x2 = 16y Equation of tangent of P is t12 + t22 + t32 + t 42
2
16( y + t ) 2 4tx = 8y + 8t2 tx = 2y + 2t2 A = (2t, 0), B = (0, –t2) M(h, k) is the middle point of AB x ⋅ 4t =
(4t, t2) P A B M
t2 ⇒ 2k = −h2 2 Locus of M(h, k) is x 2 + 2 y = 0.
h = t, k = −
4t 4 32. (d) : tan 30° = 21 = t1 t1 1 4 ⇒ = ⇒ t1 = 4 3 3 t1
(–4t,
t2
1
)A
(0, 3) B
O (0, 0)
3 Area of ∆OAB = × 32 3 × 32 3 = 768 3 sq.units 4 33. A–Q, B–Q, C–P (A) Equation of tangent at (3, 6) : y = x + 3 \ T( – 3, 0) Equation of normal at (3, 6) : y = –x + 9 \ G( 9, 0) Hence middle point is (3, 0) (B) Point is obviously focus (3, 0) (C) Let A(t1) and B(t2), C(t3) and D(t4) If AB and CD intersect at a point E on the axis, then by solving the equations of AB and CD we get the relation t1t2 = t3t4 Now equations of the circles with AB and CD as diameters are (x − at12 )(x − at22 ) + ( y − 2at1)( y − 2at2 ) = 0 (x − at32 )(x − at 42 ) + ( y − 2at3)( y − 2at 4 ) = 0 If we solve these two circles, then the equation of their radical axis is of the form y = mx. So it passes through the origin. 34. (1) : Equation of the normal is y = mx – 2m – m3 If it pass through (21, 30) we have 30 = 21m – 2m – m3 ⇒ m3 – 19m + 30 = 0 Then m = –5, 2, 3 But if m = 2 or 3 then the point where the normal mathematics today | FEBRUARY‘17
Y
B (+4t, t2 ) 1 30°
AB = 8t1 = 32 3
72
meets the curve will be (am2 , –2am) where the curve does not exist. Therefore m = –5 \ m+6=1 35. (1) : Since F1 and F2 are the ends of the diameter 1 1 Area of ∆PF1F2 = ( F1P )( F2P ) = x (17 − x ) = 30 2 2 ⇒ x = 5 or 12 ⇒ F1F2 = 13 \ K = 1 36. (7) : Let P(4, 0) and Q(0, 3) are two points on given ellipse E1 P1 and Q1 are images of P,Q w.r.t x – y – 2 = 0 \ P1(2, 2) Q1(5, –2) lies on E2 \ a = –160, b = 292 ⇒ a + b – 125 = 7 37. (9) : Eccentricity of given ellipse e = 2/3 2x y Equation of tangent at L is + = 1 . It meets 9 3
(–2, 0)
5 L (2, 3 )
(0, 0) (2, 0)
9 ( 2 , 0)
X
9 x-axis at A , 0 and y axis at B(0, 3) 2 1 9 \ Area = 4 ⋅ ⋅ 3 = 27 \ K = 9. 2 2 38. (8) : xy + 2x + 4y + C = 0 represents pair of lines ⇒C=8 39. (5) : Equation can be rewritten as 13 12x − 5 y + 1 13 (x − 2)2 + ( y + 3)2 = So, e = . 5 13 5 25e \ =5 13 b π 40. (2) : 2 tan −1 = a 3 b 1 , e2 = 1 + 1 = 4 = a 3 3 3 Let eccentricity of conjugate hyperbola be e′. 1 3 1 1 1 1 \ + 2 = 1 ⇒ 2 + = 1 ⇒ 2 = ⇒ e′ = 2 2 4 4 e′ e′ e′ e
8
T
Class XI
his specially designed column enables students to self analyse their extent of understanding of specified chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness.
STATISTICS AND PROBABILITY Total Marks : 80
Time Taken : 60 Min.
Only One Option Correct Type 1. A five digit number be selected at random. The probability that the digits in the odd places are odd and the even places are even (repetition is not allowed) 5 5 P2 × 5 P3 P2 × 5 P3 (a) (b) 104 × 9 105 5
5
C2 × 5C3
C2 × 5C3
×2 (d) 9 × 104 104 × 9 2. The mean of the data is 30 and data are given as (c)
C.I. 0-10 10-20 20-30 30-40 40-50 Frequency 8 10 f1 15 f2 If total of frequencies is 70, then the missing numbers are (a) 14, 23 (b) 25, 21 (c) 24, 20 (d) none of these 4 3. The probability that A speaks truth is , B speaks 5 truth is 3 . The probability they contradict each 4 other is 4 7 3 (a) (b) 1 (c) (d) 5 20 20 5 4. In a series there are 2n observations half of them equal to ‘a’ and half equal to – a. If the standard deviation of the observations is 2, then |a| equals 2 1 (d) 2 n n 5. If ten rupee coins which are 10 in number and five rupee coins which are 5 in number are to be placed in a line, then the probability that the extreme coins are five rupee coins is (a) 2
(b)
(c)
1 1 (b) 10! 15! 5!10 ! (c) (d) none of these 15! 6. In an experiment with 15 observations on x, the following results were available : Σx2 = 2830 and Σx = 170. One observation i.e., + 20 was found to be wrong and was replaced by the correct value 30, then the corrected variance is (a) 188.66 (b) 177.33 (c) 8.33 (d) 78.00 One or More Than One Option(s) Correct Type (a)
7. The probabilities that a student pass in Mathematics, Physics and Chemistry are a, b and g respectively. Of these subjects, student has a 75% chance of passing in atleast one, a 50% chance of passing in atleast two and a 40% chance of passing in exactly two subjects. Which of the following relations are true? (a) a + b +g = 19/20 (b) a + b + g = 27/20 (c) abg = 1/10 (d) abg = 1/4 8. The variable x takes two values x1 and x2 with frequencies f1 and f2, respectively. If s denotes the standard deviation of x, then f1x12 + f 2 x22 f1x1 + f 2 x2 − f1 + f 2 f1 + f 2 f1 f 2 (x − x )2 (b) s 2 = ( f1 + f 2 )2 1 2
2
2 (a) s =
(x1 − x2 )2 (d) None of these f1 + f 2 9. If A and B are two events such that P(A) = 1/2 and P(B) = 2/3, then (a) P(A∪B) ≥ 2/3 (b) P(A∩B′)≤1/3 (c) 1/6 ≤P(A∩B) ≤ 1/2 (d) 1/6 ≤P(A′∩B)≤1/2 (c) s 2 =
mathematics today | February‘17
73
10. For two data sets, each of size 5, the variance are given to be 4 and 5 and the corresponding means are given to be 2 and 4, respectively. The variance of the combined data set is 13 11 5 (a) (b) (c) 6 (d) 2 2 2 11. The letters of the word PROBABILITY are written down at random in a row. Let E1 denotes the event that two I’s are together and E2 denotes the event that two B’s are together, then 2 2 (a) P(E1) = P(E2) = (b) P(E1∩E2) = 11 55 18 (c) P(E1∪E2) = (d) none of these 55 12. There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test Marks 0 1 2 3 4 5 2 2 Frequency x–2 x x (x + 1) 2x x + 1 where, x is a positive integer. Then, (a) Mean = 2.8 (b) Variance = 1.12 (c) Standard Deviation = 1.12 (d) None of these
Matrix Match Type 16. Match the following: Column I Column II P. Two balls are drawn from an urn containing 8 2 white, 3 red and 4 black balls one by one 1. m =
Q. A bag contains 2 white and 4 black balls while 11 another bag contains 6 white and 4 black balls. 2. m = 12 A bag is selected at random and a ball is drawn. If l be the probability that the ball drawn is of white colour and m be the probability that the ball drawn is black colour, then R. Bag A contains 4 red and 5 black balls and
5 bag B contains 3 red and 7 black balls. One 3. l = 18 ball is drawn from bag A and two from bag B. If l be the probability that out of 3 balls drawn two are black and one is red and m be the probability that out of three balls drawn two are red and one is black, then 4. m =
13. For two events A and B, P(A∩B) is (a) not less than P(A) + P(B) – 1 (b) not greater than P(A) + P(B) (c) equal to P(A) + P(B) – P(A∪B) (d) equal to P(A) + P(B) + P(A∪B) Comprehension Type
(a) (b) (c) (d)
A JEE aspirant estimates that she will be successful with an 80% chance if she studies 10 hours per day, with a 60% chance if she studies 7 hours per day and with a 40% chance if she studies 4 hours per day. She further believes that she will study 10 hours, 7 hours and 4 hours per day with probabilities 0.1, 0.2 and 0.7, respectively: 14. Given that she is successful, the probability that she studied for 4 hours, is 6 7 8 9 (a) (b) (c) (d) 12 12 12 12 15. Given that she does not achieve success, the probability she studied for 4 hours, is 18 21 19 20 (a) (b) (c) (d) 26 26 26 26
15
without replacement. If the probability that both the balls are of same colour is l and at least one ball is red is m, then
P 2 3 3 1
11 45
Q R 1 3 4 2 1 4 2 3 Integer Answer Type
17. If four squares are chosen at random on a chess board. If the probability that they lie on a diagonal xyz line is 64 , then the value of x + y – z must be C4 18. If the mean deviation about the median of the numbers a, 2a, ......., 50a is 50, then |a| equals 19. Two integers x and y are chosen (without random) at random from the set {x : 0 ≤ x ≤ 10,} x is an integer}. If the probability for |x – y| ≤ 5 is p, then the value of 121 p – 91 must be 20. If the variance of first 50 even natural numbers is abc. Then find the value of a + b – c. Keys are published in this issue. Search now! J
Check your score! If your score is No. of questions attempted …… No. of questions correct …… Marks scored in percentage ……
74
> 90%
excellent work ! You are well prepared to take the challenge of final exam.
90-75%
Good work !
You can score good in the final exam.
74-60%
satisfactory !
You need to score more next time.
< 60%
mathematics today | February ‘17
not satisfactory! Revise thoroughly and strengthen your concepts.
mathematics today | February‘17
75
Exam on 20th March 2017
PRACTICE PAPER 2017 Time Allowed : 3 hours
Maximum Marks : 100 GENERAL INSTRUCTIONS
(i) (ii) (iii) (iv) (v) (vi)
All questions are compulsory. This question paper contains 29 questions. Questions 1-4 in Section-A are very short-answer type questions carrying 1 mark each. Questions 5-12 in Section-B are short-answer type questions carrying 2 marks each. Questions 13-23 in Section-C are long-answer-I type questions carrying 4 marks each. Questions 24-29 in Section-D are long-answer-II type questions carrying 6 marks each.
section-a
1. Write the vector equation of a line passing through the point (1, –1, 2) and parallel to the line x − 3 y −1 z +1 . = = 1 2 −2 2. Evaluate :
∫
x3 − x2 + x − 1 dx x −1
3. If the binary operation * on the set of integers Z is defined by a * b = a + 3b2, then find the value of 2 * 4. x sin q cos q 4. If − sin q − x 1 = 8 , then find the value of x. cos q 1 x section-B
5. Prove that : 1 1 1 1 p tan −1 + tan −1 + tan −1 + tan −1 = . 3 5 7 8 4
2 0 1 6. If A = 2 1 3 , find A2 – 5A + 4I. 1 −1 0 76
mathematics today | february ‘17
7. Show that f : N → N, defined by n + 1 , if n is odd f (n) = 2 n , if n is even 2 is a many-one onto function. 8. For what value of k, the function defined by k(x 2 + 2), if x ≤ 0 f (x ) = if x > 0 3x + 1, is continuous at x = 0 ? 9. Form the differential equation of the family of curves y = a cos (x + b), where a and b are arbitrary constants. 10. If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximate error in calculating its surface area. 11. Let * be a binary operation on the set of rational numbers given as a * b = (2a – b)2, a, b ∈ Q. Find 3 * 5 and 5 * 3. Is 3 * 5 = 5 * 3 ? 12. If xy + y2 = tan x + y, find
dy . dx
section-c
13. Using integration, find the area of the region bounded by the curves y = x2 and y = x. 14. Solve the differential equation
dy + x 2 y 2 = 0. dx 15. In a parliament election, a political party hired a public relations firm to promote its candidates in three ways-telephone, house calls and letters. The cost per contact (in paise) is given in matrix A as 140 Telephone A = 200 House call 150 Letters The number of contacts of each type made in two cities X and Y is given in matrix B as Telephone House call Letters cosec x log y
500 5000 City X 1000 B= 3000 1000 10000 City Y Find the total amount spent by the party in the two cities. What should one consider before casting his/ her vote-party’s promotional activity or their social activities?
OR 1 2 2 If A = 2 1 2 , then verify that A2 – 4A – 5I = O. 2 2 1
Hence find A–1. 16. If a ,b and c are three vectors such that each one is perpendicular to the vector obtained by sum of the other two and a = 3, b = 4 and c = 5, then prove that a + b + c = 5 2 . 17. An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be at least 4 successes. 18. If y = 3cos(logx) + 4sin(logx), then show that d2 y dy x2 2 + x + y = 0 dx dx OR For what value of l, the function defined by l(x 2 + 2), if x ≤ 0 f (x ) = is continuous at if x > 0 4 x + 6, x = 0? Hence check the differentiability of f(x) at x = 0.
19. Find the equation of the plane passing through the intersection of the planes x + 3y + 6 = 0 and 3x – y – 4z = 0 and whose perpendicular distance from the origin is unity. 20. Evaluate :
cos x
∫ (1 − sin x)(2 − sin x) dx
OR Using properties of definite integrals prove that p
x tan x p2 dx = ∫ sec x cosec x 4 0
21. Prove that :
9 p 9 −1 1 9 −1 2 2 − sin = sin . 3 4 3 8 4
22. Find the intervals in which the following function is (a) increasing (b) decreasing : f(x) = x4 – 8x3 + 22x2 – 24x + 21 23. Using properties of determinants, show that
x+y x x 5x + 4 y 4 x 2 x = x 3 . 10 x + 8 y 8 x 3x
section-d
24. Find the equation of the plane through the line of intersection of the planes 2x + y – z = 3 and 5x – 3y + 4z + 9 = 0 x −1 y − 3 z − 5 and parallel to the line = = . 2 4 5 Also, find the distance of the plane from origin. OR Find the equation of the plane passing through the point A(1, 2, 1) and perpendicular to the line joining the points P(1, 4, 2) and Q(2, 3, 5). Also, find the distance of this plane from the line x +3 y −5 z −7 = = . 2 −1 −1 25. Find the particular solution of the differential dy equation x + y – x + xy cot x = 0; x ≠ 0, dx p given that when x = , y = 0. 2 1 −1 0 2 2 −4 26. If A = 2 3 4 and B = −4 2 −4 are two 0 1 2 2 −1 5 square matrices, find AB and hence solve the system of equations x – y = 3, 2x + 3y + 4z = 17 and y + 2z = 7. mathematics today | february ‘17
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OR Two schools P and Q want to award their selected students on the values of discipline, politeness and punctuality. The school P wants to award ` x each, ` y each and ` z each for the three respective values to its 3, 2 and 1 students with a total award money of ` 1,000. School Q wants to spend ` 1,500 to award its 4, 1 and 3 students on the respective values (by giving the same award money for the three values as before). If the total amount of awards for one prize on each value is ` 600, using matrices, find the award money for each value. 27. A dealer in rural area wishes to purchase a number of sewing machines. He has only ` 5760 to invest and has space for at most 20 items. An electronic sewing machine costs him ` 360 and a manually operated sewing machine costs ` 240. He can sell an electronic sewing machine at a profit of ` 22 and a manually operated sewing machine at a profit of ` 18. Assuming that, he can sell all the items that he can buy, how should he invest his money in order to maximize his profit. Make it as a linear programming problem and solve it graphically. Keeping the rural background in mind justify the ‘values’ to be promoted for the selection of manually operated machine. 28. If the length of three sides of a trapezium other than base is 10 cm each, then find the area of the trapezium when it is maximum. 29. From a well shuffled pack of 52 cards, 3 cards are drawn one-by-one with replacement. Find the probability distribution of number of queens. OR Suppose the probability for A to win a game against B is 0.4. If A has an option of playing a ‘best of 3 games’ or a ‘best of 5 games’ match against B, which option should A choose so that the probability of his winning the match a higher? (No game ends in draw). solutions
1. Vector equation of the line passing through (1, –1, 2) x − 3 y −1 z +1 and parallel to the line is = = 1 2 −2 r = (i − j + 2k ) + l (i + 2 j − 2k ) 78
mathematics today | february ‘17
2. Let I = ∫
x3 − x2 + x − 1 dx x −1
x 2 (x − 1) + 1(x − 1) (x 2 + 1)(x − 1) =∫ dx x −1 x −1 1 = ∫ (x 2 + 1)dx = x 3 + x + C 3
=∫
3. Here, a * b = a + 3b2 \ 2 * 4 = 2 + 3(4)2 = 2 + 3 × 16 = 50 x sin q cos q 4. Given, − sin q − x 1 =8 cos q 1 x ⇒ ⇒ ⇒ ⇒ ⇒ ⇒
x(–x2 – 1) –sin q(–xsin q – cos q) + cos q (–sin q + xcos q) = 8 –x3 – x + xsin2q + sin q cos q – sin q cos q + xcos2q = 8 3 2 2 –x – x + x(sin q + cos q) = 8 –x3 – x + x = 8 ⇒ x3 + 8 = 0 (x + 2) (x2 – 2x + 4) = 0 ⇒ x + 2 = 0 [Q x2 – 2x + 4 > 0 "x] x = –2
1 1 1 −1 1 + tan −1 + tan −1 + tan −1 5. L.H.S. = tan 3 5 7 8
1 1 1 1 + + −1 7 8 3 5 + tan = tan 1 1 1 1 1− × 1− × 7 8 3 5 15 8 −1 15 −1 56 = tan + tan 14 55 56 15 4 3 = tan −1 + tan −1 11 7 65 4 3 + −1 7 11 −1 77 = tan = tan 65 1 − 4 × 3 77 7 11 p = tan −1 1 = = R.H.S. 4 −1
6. We have, A2 – 5A + 4I 2 0 1 2 0 1 1 0 0 2 0 1 = 2 1 3 2 1 3 −5 2 1 3 + 4 0 1 0 1 −1 0 1 −1 0 0 0 1 1 −1 0
5 −1 2 = 9 −2 5 − 0 −1 −2
dS = 8pr dr dS \ DS = Dr = 8pr Dr dr = 8p × 9 × 0.03 = 2.16 p cm2
10 0 5 4 0 0 10 5 15 + 0 4 0 5 −5 0 0 0 4
⇒
−1 −1 −3 = −1 −3 −10 −5 4 2
This is the approximate error in calculating surface area.
(1 + 1) 2 2 7. We have, f (1) = = = 1 and f (2) = = 1 2 2 2 Thus, f(1) = f(2) while 1 ≠ 2 \ f is many-one. In order to show that f is onto, consider an arbitrary element n ∈N. If n is odd, then (2n – 1) is odd (2n − 1 + 1) 2n \ f (2n − 1) = = =n 2 2 If n is even, then 2n is even 2n \ f (2n) = = n 2 Thus, for each n ∈ N (whether even or odd) there exists its pre-image in N. \ f is onto. Hence, f is many-one onto function.
11. We have, a * b = (2a – b)2 \ 3 * 5 = (2 × 3 – 5)2 = (6 – 5)2 = 1 5 * 3 = (2 × 5 – 3)2 = (10 – 3)2 = 49 Thus, 3 * 5 ≠ 5 * 3
12. Given, xy + y2 = tan x + y Differentiating w.r.t. x, we get dy dy dy x + y + 2y = sec2 x + dx dx dx 2 dy sec x − y dy ⇒ (x + 2 y − 1) = sec2 x − y ⇒ = dx x + 2 y − 1 dx 13. The given curves are y = x2 ... (i) and y = x ...(ii)
8. We have, lim f (x ) = lim k(h2 + 2) = 2k x →0−
h →0
The point of intersection of (i) and (ii) are A (1, 1) and O(0, 0). \ Required area = Area of shaded region
lim f (x ) = lim (3h + 1) = 1 and f(0) = 2k
x →0+
h →0
As f(x) is continuous at x = 0 \
x → 0−
x → 0+
⇒ 2k = 1 ⇒ k =
1 2
9. Here, y = a cos (x + b) Differentiating (i) w.r.t. x, we get dy = −a sin ( x + b ) dx Again differentiating w.r.t. x, we get d2 y dx 2
⇒
1
lim f ( x ) = lim f ( x ) = f (0)
d2 y dx 2
...(i)
= −a cos ( x + b ) = −y ⇒
d2 y dx 2
+ y = 0.
10. Let r be the radius of sphere and Dr be the error in measuring the radius. Then, r = 9 cm, Dr = 0.03 cm Now, surface area (S) of the sphere is given by 4pr2
1
3 2 1 1 1 = ∫ (x − x )dx = x − x = − = sq.unit. 2 3 0 2 3 6 0 dy 14. We have, cosec x log y + x 2 y 2 = 0 dx log y x2 dy + dx = 0 cosec x y2 2
Integrating both sides, we get log y 2 ∫ y 2 dy + ∫ x sin x dx = 0 1 Put log y = t ⇒ dy = dt and y = et y
Solution Sender of Maths Musing set-169 1. N. Jayanthi Hyderabad 2. ravinder Gajula Karimnagar mathematics today | february ‘17
79
⇒
∫ t. e
⇒ t.
⇒
−t
dt + ∫ x 2 sin x dx = 0
e −t e −t − ∫ 1. dt + x 2 (− cos x ) − ∫ 2x ( − cos x ) dx = C −1 −1
–t e–t –e–t –x2 cos x + 2x sin x −2 ∫ 1 ⋅ sin x dx = C
⇒ −
1 + log y − x 2 cos x + 2 x sin x + 2 cos x = C y
is the required solution. 15. The total amount spent by the party in two cities X and Y is represented in the matrix equation by matrix C as, C = BA 140 X 1000 500 5000 ⇒ = 200 Y 3000 1000 10000 150 ⇒
\
X 1000 × 140 + 500 × 200 + 5000 × 150 = Y 3000 × 140 + 1000 × 200 + 10000 × 150 990000 = 2120000 X = ` 990000 and Y = ` 2120000
Thus, amount spent by the party in city X and Y is ` 990000 and ` 2120000 respectively. One should consider about the social activity before casting his/her vote. OR 1 2 2 Given, A = 2 1 2 2 2 1 1 2 2 1 Now, A2 = 2 1 2 2 2 2 1 2 2 \ A – 4A – 5I 1 9 8 8 = 8 9 8 − 4 2 2 8 8 9
2 2 9 8 8 1 2 = 8 9 8 2 1 8 8 9
1 2 2 1 2 − 5 0 2 1 0 9 8 8 4 8 8 5 0 = 8 9 8 − 8 4 8 − 0 5 8 8 9 8 8 4 0 0 0 0 0 = 0 0 0 = O 0 0 0
80
mathematics today | february ‘17
0 0 1 0 0 1 0 0 5
Now, A2 – 4 A – 5 I = O Pre-multiplying by A–1 both sides, we get (A–1 A)A – 4 (A–1A) – 5(A–1 I) = O ⇒ IA – 4I = 5A–1 −3 2 2 1 −1 1 ⇒ A = (A − 4 I ) = 2 −3 2 5 2 2 2 −3 16. Given, a = 3, b = 4, c = 5 ...(i) ( ) ( ) and a ⋅ b + c = 0 , b ⋅ (c + a ) = 0 , c ⋅ a + b = 0 \ a ⋅b + a ⋅c + b ⋅c + b ⋅a + c ⋅a + c ⋅b = 0 ...(ii) ⇒ 2(a ⋅ b + b ⋅ c + c ⋅ a ) = 0 2 Now, a + b + c 2 2 2 = a + b + c + 2 ( a ⋅ b ) + 2 (b ⋅ c ) + 2 ( c ⋅ a ) = 32 + 42+ 52 + 0 = 50 [Using (i) and (ii)] \ a +b + c = 5 2. 2 2 = 17. Here, p = P (success of the experiment) = 2 +1 3 2 1 \ q = 1− p = 1− = 3 3 Also n = 6 Let X denote the number of successes. \ Required probability P(X ≥ 4) = P(X = 4) + P(X = 5) + P(X = 6) 2
4
1
5
0
1 2 1 2 1 2 = 6C4 + 6C5 + 6C6 3 3 3 3 3 3 16 32 64 = 15 × 6 + 6 × 6 + 6 = 240 + 192 + 64 = 496 3 3 3 729 36 18. We have, y = 3 cos (log x) + 4 sin (log x) Differentiating w.r.t. x, we get 1 1 dy = −3 sin(log x ) × + 4 cos(log x ) × x x dx dy ⇒ x = −3 sin(log x ) + 4 cos(log x ) dx Again differentiating w.r.t. x, we get x ⇒ x2
⇒ x2
d2 y dx
2
d2 y dx 2
d2 y dx 2
+
dy 1 1 = − 3 cos(log x ) × − 4 sin(log x ) × dx x x
+x
dy = −[3 cos(log x ) + 4 sin(log x )] dx
+x
2 dy dy 2d y + x + y =0 =− y ⇒x 2 dx dx dx
6
OR
dt −dt + = − log 1 − t + log 2 − t + C 1−t ∫ 2 −t 2−t 2 − sin x = log + C = log +C 1−t 1 − sin x
I=∫
\
l(x 2 + 2), if x ≤ 0 Here, f (x ) = if x > 0 4 x + 6,
At x = 0, f(0) = l(02 + 2) = 2l L.H. L. = lim f (x ) = lim l[(−h)2 + 2] = 2l x →0
−
OR p
h →0
Let I = ∫
R.H.L. = lim f (x ) = lim[4(h) + 6] = 6 x →0
+
h →0
For f to be continuous at x = 0, 2l = 6 ⇒ l = 3. Hence, the function becomes 3(x 2 + 2), if x ≤ 0 f (x ) = if x > 0 4 x + 6, \
f ′(0− ) = lim
h →0
f (0 − h) − f (0) 0−h
f (0 + h) − f (0) 0+h
h →0
⇒ \
–
I=∫ 0
0
0
= lim
0
p
4h + 6 − 6
h →0
h
0 p
=4
f ′(0 ) ≠ f ′(0 ) f is not differentiable at x = 0.
= p∫ ⇒ I=
26 = 26l2 ⇒ l2 = 1 ⇒ l = ±1 Equation of the required plane is r ⋅ (4i + 2 j − 4k ) + 6 = 0 or r ⋅ (−2i + 4 j + 4k ) + 6 = 0
⇒ \
21. L.H.S. =
0 2
cos x dx (1 − sin x )(2 − sin x ) Put sinx = t ⇒ cosxdx = dt dt \ I=∫ (1 − t )(2 − t ) 1 A B We write, = + (1 − t )(2 − t ) 1 − t 2 − t
...(1)
...(2)
p
tan x dx = p ∫ sin2 x dx sec x cos ec x 0
p
p sin 2 x p2 1 − cos 2 x = dx = x − 2 2 2 0 2
p 4
9 p 9 −1 1 9 p 1 − sin = − sin −1 3 8 4 4 2 3 9 1 = cos −1 3 4
p −1 −1 Qsin x + cos x = 2
Q cos −1 x = sin −1 1 − x 2 for 0 ≤ x ≤ 1 8 9 −1 2 2 = sin = R.H.S. 3 9 4
9 1 = sin −1 1 − 4 9 9 = sin −1 4
20. Let I = ∫
p
( p − x )tan(p − x ) ( p − x )tan x dx = ∫ dx sec( p − x )cosec (p − x ) sec x cos ec x
+
1 = A (2 – t) + B(1 – t) Putting t = 1 in (1), we get 1 = A(2 – 1) ⇒ A = 1 Putting t = 2 in (1), we get 1 = B(1 – 2) ⇒ B = –1
a
2I = p∫
19. Equation of plane passing through intersection of given planes is x + 3y + 6 + l(3x – y – 4z) = 0 ⇒ r ⋅[(1 + 3 l) i + (3 − l)j − 4 lk ] + 6 = 0 Q Distance of plane from origin is unity. 6 \ =1 2 (1 + 3l) + (3 − l)2 + 16 l2
⇒
p
...(1)
∫ f (x)dx = ∫ f (a − x)dx, we get
Adding (1) and (2), we get
3(h2 + 2) − 6 = lim (−3h) = 0 = lim −h h →0 h →0 and f ′(0+ ) = lim
Using
0 a
x tan x dx sec x cosec x
22. The given function is f(x) = x4 – 8x3 + 22x2 – 24x + 21 ⇒ f ′(x) = 4x3 – 24x2 + 44x – 24 = 4(x3 – 6x2 + 11x – 6) = 4(x – 1)(x2 – 5x + 6) = 4(x – 1)(x – 2)(x – 3) Thus f ′(x) = 0 ⇒ x = 1, 2, 3. Hence, possible disjoint intervals are (–∞, 1), (1, 2), (2, 3) and (3, ∞). In the interval (–∞, 1), f ′ (x) < 0 In the interval (1, 2), f ′(x) > 0 In the interval (2, 3), f ′(x) < 0 In the interval (3, ∞), f ′(x) > 0 \ f is increasing in (1, 2) ∪ (3, ∞) and f is decreasing in (–∞, 1) ∪ (2, 3). mathematics today | february ‘17
81
x+y x x 23. L.H.S. = 5x + 4 y 4 x 2 x . 10 x + 8 y 8 x 3x
Since each element in the first column of determinant is the sum of two elements, therefore, determinant can be expressed as the sum of two determinants given by x x x y x x 5x 4 x 2 x + 4 y 4 x 2 x 10 x 8 x 3x 8 y 8 x 3x Taking x common from R1, R2, R3 in first determinant and x common from C2, C3, y common from C1 in second determinant, we get 1 1 1 1 1 1 1 1 1 3 2 3 x 5 4 2 + yx 4 4 2 = x 5 4 2 + yx 2 ⋅ 0 10 8 3 8 8 3 10 8 3 (Q C1 and C2 are identical in the second determinant) Applying C1 → C1 – C3 and C2 → C2 – C3, we get 0 0 1 3 3 x 3 3 2 2 = x ⋅1⋅(15 – 14) = x = R.H.S. 24. The equation of the plane passing through the line of intersection of the planes 2x + y – z = 3 and 5x – 3y + 4z + 9 = 0 is (2x + y – z – 3) + l(5x – 3y + 4z + 9) = 0 ⇒ x(2 + 5l) + y(1 – 3l) + z(4l – 1) + 9l – 3 = 0 ...(i) Since, plane (i) is parallel to the line x −1 y − 3 z − 5 = = 2 4 5 \ 2(2 + 5l) + 4(1 – 3l) + 5(4l – 1) = 0 1 ⇒ 18l + 3 = 0 ⇒ l = − 6 Putting the value of l in (i), we obtain 7x + 9y – 10z – 27 = 0 This is the equation of the required plane. Now, distance of the plane 7x + 9y – 10z – 27 = 0 from the origin is, −27 27 unit = 72 + 92 + 102 230 OR The line joining the given points P(1, 4, 2) and Q(2, 3, 5) has direction ratios <1 –2, 4 –3, 2 – 5> i.e., <– 1, 1, –3> The plane through (1, 2, 1) and perpendicular to the line PQ is –1(x – 1) + 1(y – 2) – 3(z – 1) = 0 mathematics today | february ‘17
x – y + 3z – 2 = 0
x +3 y −5 z −7 Now, direction ratios of line = = 2 −1 −1 are 2, –1, –1. Now, 2(1) + (–1) (–1) + (3) (–1) = 2 + 1 – 3 = 0 \ Line is parallel to the plane. Since, (–3, 5, 7) lies on the given line. \ Distance of the point (–3, 5, 7) from plane is −3 − 5 + 3(7) − 2 11 d= ⇒ d= = 11 units. 1+1+ 9 11 dy 25. We have, x + y − x + xy cot x = 0,(x ≠ 0) dx dy ⇒ x + (1 + x cot x ) ⋅ y = x dx dy 1 + x cot x .y = 1 ⇒ + dx x dy This is linear D.E. of the form + Py = Q dx 1 + x cot x 1 where P = = + cot x , Q = 1 x x Pdx log (x sin x) ∫ Now, I.F. = e =e = x sin x \
7 5 3
82
⇒
...(i)
The solution of (i) is
y ⋅ x sin x = ∫ 1⋅ x sin x dx + C
= x(− cos x ) + ∫ 1 ⋅ cos x dx + C
...(ii) p Putting x = , y = 0 in (ii), we get 2 p p p 0 = − cos + sin + C ⇒ C = −1 2 2 2 \ xy sinx = sinx – xcos x – 1 is the required particular solution. ⇒ x y sin x = − x cos x + sin x + C
1 −1 0 2 2 −4 26. Here, A = 2 3 4 ; B = −4 2 −4 0 1 2 2 −1 5 1 −1 0 2 2 −4 6 0 \ AB = 2 3 4 ⋅ −4 2 −4 = 0 6 0 1 2 2 −1 5 0 0 ⇒
0 0 = 6I 6
1 1 A ⋅ B = I ⇒ A is invertible and A−1 = B 6 6
Now, the given system of equations is x–y=3 2x + 3y + 4z = 17 y + 2z = 7 The system of equations can be written as AX = P
mathematics today | february ‘17
83
1 −1 0 x 3 where, A = 2 3 4 ; X = y ; P = 17 0 1 2 z 7
Since A–1 exists, so system of equations has a unique solution given by X = A–1 P x 2 2 −4 3 1 1 ⇒ y = BP = −4 2 −4 17 6 6 z 2 −1 5 7 12 2 1 = −6 = −1 6 24 4 ⇒ x = 2, y = – 1, z = 4.
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mathematics today | february ‘17
3 2 1 A = 4 1 3 = −5 ≠ 0 1 1 1
A is invertible and system of equations has a unique solution given by X = A–1 B Now, A11 = –2, A12 = –1, A13 = 3, A21 = –1, A22 = 2, A23 = –1, A31 = 5, A32 = –5, A33 = –5 \
OR
According to question, we have, 3x + 2y + z = 1000
4x + y + 3z = 1500 ...(ii) x + y + z = 600 ...(iii) The given system of equations can be written as AX = B 3 2 1 x 1000 where, A = 4 1 3 , X = y and B = 1500 1 1 1 z 6000
...(i)
\
−2 −1 5 adj A = −1 2 −5 3 −1 −5
\ A −1
−2 −1 5 adj( A) −1 = = −1 2 −5 | A| 5 3 − 1 −5
Now, X = A–1B x −2 −1 5 1000 −500 −1 −1 ⇒ y = −1000 −1 2 −5 1500 = 5 5 z 3 −1 −5 600 −1500 x 100 ⇒ y = 200 z 300 ⇒ x = 100, y = 200, z = 300
Hence, the money awarded for discipline, politeness and punctuality are ` 100, ` 200 and ` 300 respectively. 27. Suppose, number of electronic sewing machines purchased = x and number of manually operated sewing machines purchased = y Mathematical formulation of given problem is : Maximize Z = 22x + 18y Subject to constraints : x + y ≤ 20 ...(i) 360x + 240y ≤ 5760 or 3x + 2y ≤ 48 ... (ii) x ≥ 0, y ≥ 0 Solving equations x + y = 20 and 3x + 2y = 48, we get x = 8, y = 12 Let P ≡ (8, 12) Now, we plot the graph of system of inequalities.
The shaded portion OCPB represents the feasible region which is bounded.
The corner points of feasible region are : O(0, 0), C(16, 0), P(8, 12) and B(0, 20). Now, we calculate Z at each corner point. Corner points Value of Z = 22x + 18y O(0, 0) 22(0) + 18(0) = 0 C(16, 0) 22(16) + 18(0) = 352 P(8, 12) 22(8) + 18(12) = 392 (Maximum) B(0, 20) 22(0) + 18(20) = 360 \ Maximum value of Z is 392 which occurs at the point P(8, 12). Hence, the maximum profit is ` 392 when 8 electronic machines and 12 manually operated sewing machines are purchased. Manually operated machine should be promoted, to save energy and increase employment for rural people. 28. Let ABCD be the given trapezium. Then AD = DC = CB = 10 cm In DAPD and DBQC, DP = CQ = h AD = BC = 10 cm ∠DPA = ∠CQB = 90° \ DAPD @ DBQC (by R.H.S. congruency) ⇒ AP = QB = x cm (say) \ AB = AP + PQ + QB = x + 10 + x = (2x + 10) cm Also from DAPD, AP2 + PD2 = AD2 ⇒ x2 + h2 = 102 ⇒ h = 100 − x 2
...(i)
Now, area S of this trapezium is given by 1 1 S = ( AB + DC ) ⋅ h = (2 x + 10 + 10)h 2 2 = (x + 10) ⋅ 100 − x 2
...(ii)
[Using (i)]
Differentiating (i) w.r.t. x, we get 1 dS = 1 ⋅ 100 − x 2 + (x + 10) ⋅ ⋅ (−2 x ) dx 2 100 − x 2 100 − x 2 − x 2 − 10 x −2(x 2 + 5x − 50) = = 100 − x 2 100 − x 2 −2(x + 10)(x − 5) = 100 − x 2 dS For max. or min. value of S, =0 dx ⇒ (x + 10)(x – 5) = 0 ⇒ x = 5 (Reject x = – 10 as x
85
A will win the best of 3 games match if he wins in 2 or 3 games E2 = the event that A wins ‘a best of 5 games’ match. A will win a best of 5 games match if he wins in 3 or 4 or 5 games. Now P(E1) = P(X = 2 or X = 3) = P(X = 2) + P(X = 3) = 3C2 p2q + 3C3p3= 3(0.4)2(0.6) + (0.4)3 = (0.4)2[1.8 + 0.4] = (0.4)2(2.2) = 0.352 P(E2) = P(X = 3 or X = 4 or X = 5) = P(X = 3) + P(X = 4) + P(X = 5) =5C3 p3q2 + 5C4 p4q + 5C5 p5 = (10p3q2 + 5p4q + p5) = p3(10q2 + 5pq + p2) = (0.4)3 × [10 × (0.6)2 + 5 × (0.4) × (0.6) + (0.4)2] = (0.064) × (3.6 + 1.2 + 0.16) = 0.064 × 4.96 = 0.317 Since P(E1) > P(E2), hence A should choose the first option.
dS <0 For this value of x, dx \ S is maximum at x = 5.
From (ii), the max. value of S = (5 + 10) ⋅ 100 − 52 = 15 75 = 75 3 sq.cm.
29. Let Ei(i = 1, 2, 3) be the event of drawing a queen in the ith draw. Let X denote the discrete random variable “Number of Queens” in 3 draws one by one with replacement. Here X = 0, 1, 2, 3 4 Now P(Ei ) = ; P(Ei ) = 48 (i = 1, 2, 3) 52 52 \ P(X = 0) = P(E1E2 E3 ) = P(E1)P(E2 )P(E3 ) 3
3
48 12 1728 = = = 52 13 2197 P(X = 1) = P(E1E2 E3 or E1E2 E3 or E1E2 E3 ) = P(E1E2 E3 ) + P(E1E2 E3) + P(E1E2 E3) = P(E1)P(E2 )P(E3 ) + P(E1)P(E2 )P(E3 ) = 3×
+ P(E1)P(E2 )P(E3 )
2
2
4 48 1 12 432 × = 3× × = 52 52 13 13 2197
P(X = 2) = P(E1E2 E3 or E1E2 E3 or E1E2 E3)
= P(E1E2 E3) + P (E1E2 E3 ) + P ( E1E2 E3 )
= P(E1)P(E2 )P(E3 ) + P (E1)P(E2 )P(E3 ) + P(E1)P(E2 )P(E3 ) 2
2
36 1 12 4 48 = 3× × = 3× × = 13 13 2197 52 52 P(X = 3) =P(E1E2E3) = P(E1)P(E2)P(E3) 3
\
1 4 = = 52 2197 The required probability distribution of X is X
0
1
2
3
P(X)
1728 2197
432 2197
36 2197
1 2197
OR Let E = the event that A wins a game against B. Let occurrence of the event E be called a success and X denote the number of successes. Let E1 = the event that A wins ‘a best of 3 games’ match. 86
mathematics today | february ‘17
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8
T
Class XII
his specially designed column enables students to self analyse their extent of understanding of specified chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness.
VECTORS & THREE DIMENSIONAL GEOMETRY Total Marks : 80
Time Taken : 60 Min.
Only One Option Correct Type 1. If the volume of parallelopiped with a × b , b × c , c × a as co-terminous edges is 9 cu. units, then the volume of the parallelopiped with (a × b ) × (b × c ), (b × c ) × (c × a ), (c × a ) × (a × b ) as co-terminous edges is (a) 9 (b) 729 (c) 81 (d) 27 2. If the plane 2ax – 3ay + 4az + 6 = 0 passes through the midpoint of the line joining the centres of the spheres, x2 + y2 + z2 + 6x – 8y – 2z =13 and x2 + y2 + z2 – 10x + 4y – 2z = 8, then a equals (a) 1 (b) – 1 (c) 2 (d) – 2 3. If p × q = r and q × r = p, then (a) r = 1, p = q (b) p = 1, q = 1 (c) r = 2p, q = 2 (d) q = 1, p = r x − 2 y −1 z + 2 lie in the plane = = 3 −5 2 x + 3y – az + b = 0. Then (a, b) equals (a) (–6, 7) (b) (5, –15) (c) (–5, 5) (d) (6, –17) 5. Let a = a1 i + a2 j + a3 k , b = b1i + b2 j + b3 k and c = c1 i + c2 j + c3 k be three non-zero vectors such that c is a unit vector perpendicular to both the vectors a and b and if the angles between a and 4. Let the line
2
a1 a2 a3 p b is , then b1 b2 b3 is equal to 6 c1 c2 c3 (a) 1 1 (b) (a12 + a22 + a32 ) (b12 + b22 + b32 ) 4
3 2 2 2 2 2 2 2 2 2 (a + a2 + a3 ) (b1 + b2 + b3 ) (c1 + c2 + c3 ) 4 1 (d) none of these y −1 z − 3 6. If the angle between the line x = and = 2 l 5 , then l the plane x + 2 y + 3z = 4 is cos −1 14 equals (a) 2/5 (b) 5/3 (c) 2/3 (d) 3/2 One or More Than One Option(s) Correct Type (c)
7. If a vector r satisfies the equation r × (i + 2 j + k ) = i − k , then r may be equal to (a) i + 3j + k (b) 3i + 7 j + 3k (c) j + t (i + 2 j + k ), where t is any scalar (d) i + (t + 3)j + k , where t is any scalar
8. If OABC is a tetrahedron such that OA2 + BC2 = OB2 + CA2 = OC2 + AB2, then (a) OA ⊥ BC (b) OB ⊥ CA (c) OC ⊥ AB (d) AB ⊥ BC 9. If r is equally inclined to the co-ordinate axes and magnitude of r = 6 then r equals 2(i + j + k ) (a) 3(k + i + j) (b) 3 6(i + j + k ) (c) (d) − 2 3(i + j + k ) 3 10. If the unit vectors a and b are inclined at an angle 2q and | a − b |< 1, then if 0 ≤ q ≤ p, q lies in the interval (a) [0, p/6) (b) (5p/6, p] (c) [p/6, p/2] (d) [p/2, 5p/6] mathematics today | february‘17
87
11. Let the unit vectors A and B be perpendicular and the q to both at an angle unitvector C be inclined A and B . If C = a A + b B + g( A × B ) then (a) a = b (b) g2 = 1 – 2a2 1 + cos 2q (c) g2 = – cos 2q (d) b2 = 2 12. The equation of a sphere which passes through (1, 0, 0), (0, 1, 0) and (0, 0, 1) and whose centre lies on the curve 4xy = 1 is (a) x2 + y2 + z2 – x – y – z = 0 (b) x2 + y2 + z2 + x + y + z – 2 = 0 (c) x2 + y2 + z2 + x + y + z = 0 (d) x2 + y2 + z2 – x – y – z – 2 = 0 x −1 y −1 z + 3 = = 13. The points in which the line 1 −1 1 cuts the surface x2 + y2 + z2 – 20 = 0 are (a) (0, 2, 4) (b) (0, 2, – 4) (c) (4, 2, 0) (d) (0, – 2, – 4) Comprehension Type If a, b, c are three given non-coplanar vectors and r be any arbitrary vector in space, where r ⋅a b⋅a c⋅a a⋅ a r ⋅ a c ⋅ a ∆1 = r ⋅ b b ⋅ b c ⋅ b , ∆ 2 = a ⋅ b r ⋅ b c ⋅ b , a⋅c r ⋅c c ⋅c r ⋅c b⋅c c⋅c a⋅ a b⋅ a r ⋅ a a⋅ a b⋅ a c ⋅ a ∆ 3 = a ⋅ b b ⋅ b r ⋅ b and ∆ = a ⋅ b b ⋅ b c ⋅ b , then a⋅ c b⋅ c r ⋅ c a⋅c b⋅c c ⋅c 14. If vector r is expressible as, r = xa + yb + zc, then a⋅b a⋅a (a) a = (b × c ) + (c × a) [a b c] [a b c] c⋅a + (a × b) [a b c] (b) a = a ⋅ a(b ×c ) + a ⋅ b(c × a) + c ⋅ a(a × b) (c) a = [ a b c ](b × c) + [ a b c ](c × a ) + [ a b c ](a × b) (d) None of these. a b c 15. The value for a ⋅ p b ⋅ p c ⋅ p , is a⋅q b⋅q c ⋅q
(a) ( p × q)[ a × b b × c c × a] (b) 2( p × q)[ a × b b × c c × a] (c) 4( p × q)[ a × b b × c c × a] (d) ( p × q) [ a × b b × c c × a]. Matrix Match Type
16. Match the following: Column I
Column II
P. A line is perpendicular to x + 2y +
1.
5 3
Q. A plane passes through (1, –2, 1)
2.
2 2
Point (a, b, g) lies on the plane x + y + z = 2. Let a = ai + b j + g k , k × ( k × a ) = 0 , then g =
3.
2
2z = 0 and passes through (0, 1, 0). The perpendicular distance of this line from the origin is and is perpendicular to two planes 2x – 2y + z = 0 and x – y + 2z = 4. The distance of the plane from the point (1, 2, 2) is
R.
(a) (b) (c) (d)
P 2 3 3 1
Q R 1 3 4 2 2 4 2 3 Integer Answer Type
17. The line passes through the points (5, 1, a) and 17 13 (3, b, 1) crosses the yz-plane at the point 0, , − . 2 2 then a – b = 18. If [a × b b × c c × a] = l[abc ]2 , then l is equal to 19. The plane x + 2y – z = 4 cuts the sphere x2 + y2 + z2 – x + z – 2 = 0 in a circle of radius 20. Let u = i + j , v = i − j , w = i + 2 j + 3k . If n is a unit vector such that u . n = 0 and v . n = 0 then w . n equals Keys are published in this issue. Search now! J
Check your score! If your score is > 90%
excellent work ! You are well prepared to take the challenge of final exam.
No. of questions attempted
……
90-75%
Good work !
You can score good in the final exam.
No. of questions correct
……
74-60%
satisfactory !
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Marks scored in percentage ……
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88
mathematics today | february‘17
not satisfactory! Revise thoroughly and strengthen your concepts.
E 4
60° F 60° soLUtioN set-169
120°
1. (c) : z1, z2, z3 are roots ⇒ Sz1 = –p, Sz1z2 = q
60° 60° A 1 B
= 32 ⋅ 112 ⋅ 612 The number of all pairs is the number of divisors of 32 ⋅ 112 ⋅ 612, which is 33 = 27. 27 − 1 = 13 Number of pairs (a, b), a < b is 2 3. (a) : x2 – ix – 1 = 0. Solving, 5π π x = cis and x = cis 6 6 π π 2013 x = cis (2013) and x2013 = cis (10065) 6 6 2013 2013 –2013 ⇒ x = –i, x +x = –i + i = 0 4. (c) : =
100
100
0 9
0
∫ { x }dx = ∫
xdx −
9
(r +1)2
∑ ∫
r =0 r2
rdx
2000 2000 2 − ∑ (2r 2 + r ) = − 615 = 51 3 3 3 r =1
5. (b) : The parabola is open upwards with vertex (2, – 1). The focus is (2, –1 + 1) = (2, 0). The reflected ray passes through the focus (2, 0). 6. (a, d) : The equation represents a pair of planes. 3 1 a 2 2 3 ⇒ 1 1 = 0 ⇒ 4ab – 4a – 9b + 5 = 0 …(i) 2 1 1 b 2 For perpendicular planes a + 1 + b = 0 …(ii) Eliminating b from (i) and (ii), we get 7 4a2 – a – 14 = 0 ⇒ a = 2, − 4 7. (c) : ABEF and BCDE are trapeziums. BE = 2 + 2 · 4 cos60° = 6 AF = BE – 2cos60° = 6 – 1 = 5 Perimeter = 1 + 4 + 2 + 4 + 1 + 5 = 17
2
120° C
p2 = Sz21 + 2q For equilateral triangle, Sz21 = Sz1z2 \ p2 = q + 2q = 3q 1 1 1 2. (b) : + = a b 2013 ⇒ (a – 2013)(b – 2013) = 20132
D
4
8. (d) : Area = Area of BCDE + Area of ABEF 1 1 = (6 + 2)4 sin 60° + (6 + 5)sin 60° 2 2 11 43 =8 3+ 3= 3 4 4 9. (4) : N is the coefficient of x30 in (1 + x + x2 + … + x10)3 (1 + x + x2 + … + x20) (1 – x11)3 (1 – x21)(1 – x)–4 4 5 = (1 – 3x11 + 3x22 – x21 …) 1 + x + x 2 + ... 2 1 33 22 11 12 which is − 3 + 3 − 30 19 8 9 33 22 11 12 = − 3 + 3 − = 1111, 3 3 3 3 with digit sum 4. 10. (a)-(s, t); (b)-(p), (c)-(r), (d)-(r) f ( x ) − f (0) 1 = lim x a −1 sin = 0 x x x →0 x →0
(a) f ′(0) = lim
if a > 1 1 1 x ≠ 0 ⇒ f ′(x ) = ax a −1 sin − x a − 2 cos x x lim f ′( x ) = 0 if a > 2 \ a = 3, 4. x →0
(b) y(1) = 2 ⇒ 2 = a + b + c, y(0) = 0, y′(0) = 1 ⇒ c = 0, b = 1, a = 1 \ y = x2 + x, y(–1) = 0 (c) f(x) = sinx(1 + a), where a = \
f ( x ) = 2 sin x ,
π /2
∫
π /2
∫
cos tdt = 1
0
f ( x )dx = 2.
0
x2 y2 − y 2 = 1 and − x 2 = 1 have 2 2 common tangents with slopes ±1. The four tangents are x + y ± 1 = 0, x – y ± 1 = 0. They form a square of area 2 × 2 = 2
(d) The hyperbolas
mathematics today | February‘17
89
90
mathematics today | February‘17
76
19
Vol. XXXV
No. 2
February 2017
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contents 8
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Maths Musing Problem Set - 170 23
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mathematics today | February‘17
7
M
aths Musing was started in January 2003 issue of Mathematics Today with the suggestion of Shri Mahabir Singh. The aim of Maths Musing is to augment the chances of bright students seeking admission into IITs with additional study material. During the last 10 years there have been several changes in JEE pattern. To suit these changes Maths Musing also adopted the new pattern by changing the style of problems. Some of the Maths Musing problems have been adapted in JEE benefitting thousand of our readers. It is heartening that we receive solutions of Maths Musing problems from all over India. Maths Musing has been receiving tremendous response from candidates preparing for JEE and teachers coaching them. We do hope that students will continue to use Maths Musing to boost up their ranks in JEE Main and Advanced.
jee main
1. If the tangent at the point (a, b) on the curve x3 + y3 = a3 + b3 meets the curve again at the point (p, q), then (a) ap + bq + ab = 0 (b) bp + aq + ab = 0 (c) ap + bq – pq = 0 (d) bp + aq – pq = 0 an +1 3 + 1 1 − 3 an , n ∈ N. 2. Let = bn +1 3 − 1 1 + 3 bn If a1 = b1 = 1, then a22 = (a) 231 (b) 232 (c) 233 (d) 333 3. In a triangle ABC, if tan A : tan B : tan C = 2 : 3 : 4, ∆ then 2 = R 9 6 18 6 36 2 18 3 (a) (b) (c) (d) 35 35 35 35 3 −i 4. If z = , then (i101 + z101)103 = 2 (a) z (b) z (c) iz (d) iz 5. Let m be the number of 5-element subsets that can be chosen from the set of the first 15 natural numbers so that at least two of the five numbers are consecutive. The sum of the digits of m is (a) 11 (b) 12 (c) 13 (d) 14 jee advanced 4 6 6 6. Let A = 1 3 2 . A value of l, such that −1 −5 −2 AX = lX, where X is a non-zero column vector, is (a) 0 (b) 1 (c) 2 (d) 3 comprehension The parametric equations of a parabola are x = u cos a⋅t, 1 y = u sin a⋅t − gt2, t is a parameter. 2 7. Its latus rectum is 2u2 2 2u2 sin a cos2 a (a) (b) g g 8
mathematics today | FeBruary‘17
2 u2 cos2 a (c) u sin2 a (d) 2g 2g 8. Its focus is u2 u2 (a) sin 2a, cos 2a 2g 2g
u2
−u 2
cos 2a (b) sin 2a, 2g 2g
u2 −u 2 2 cos2 a (c) sin a, 2g 2g u2
−u 2
2 sin2 a (d) cos a, 2g 2g integer match 9. I is the incentre of the triangle ABC. AI, BI, CI when produced meet the opposite sides at D, E, F respectively. Then the value of AI BI CI AI BI CI ⋅ ⋅ − + + is ID IE IF ID IE IF matrix match 10. List-I contains S and List-II gives last digit of S. List-I List-II
P. S = Q. S = R. S = S. S = (a) (b) (c) (d)
P 4 4 2 1
11
∑ (2n − 1)2
1.
0
∑ (2n − 1)3
2.
1
∑ (2n − 1)2 (−1)n
3.
5
∑ (2n − 1)3(−1)n−1
4.
8
n =1 10
n =1 18 n =1 15
n =1
Q 2 2 1 2
R 1 3 4 3
S 3 1 3 4
See Solution Set of Maths Musing 169 on page no. 89
Exam Dates OFFLINE : 2nd April ONLINE : 8th & 9th April
1. Th ree numbers are chosen at random from 3 20 21 2 (a) (b) (c) (d) 1, 2, ..., 30. What is the probability that they will 5 25 25 5 form a G.P.? 7. If two independent events A and B are such 10 12 19 20 3 8 (a) (b) (c) (d) that P ( A BC ) and P ( AC B) and 4060 4060 4060 4060 25 25 1 2. If n balls are distributed into m boxes, so that each P ( A) , the value of P(A) + P(B) = 2 ball is equally likely to fall in any box, the probability 8 4 that a specified box will contain r balls is (a) (b) n nr n nr 5 5 Pr (m 1) Cr (m 1) (a) (b) 7 n n Cm Cm (c) (d) None of these 5 n nr n nr Cr (m 1) Cr (m 1) 8. If two events A and B are such that P(AC) = 0.3, (c) (d) n m nm P(B) = 0.4, P(ABC) = 0.5, then P(B|A BC) is 3. Th e probability that at least one of the eventsA and 4 1 1 1 (a) (b) (c) (d) B occur is 0.4. If A and B be occur simultaneously 5 2 4 5 with probability 0.1, then probability of AC or BC is 9. Th e independent probabilities thatA, B, C can equal to 1 1 1 (a) 1.2 (b) 1.4 (c) 1.5 (d) 1.8 solve the problem are , , respectively. Th e 2 3 4 4. In an attempt to land an unmanned rocket on the probability that just two of them only solve the moon, the probability of a successful landing is 0.4 problem is 2 3 and the probability of monitoring system giving the (a) (b) 3 correct information of landing is 0.9 in either case. Th e 4 1 probability of a successful landing, it being known that (c) (d) None of these the monitoring system indicated it correctly 4 10. A bag A contains 2 white and 3 red balls and bag B 6 9 4 2 (a) (b) (c) (d) contains 4 white and 5 red balls. One ball is drawn 7 10 25 5 at random from one of the bags and is found to be 5. Each of N + 1 identical urns marked 0, 1, 2, ..., N th red. Th e probability that it was drawn from bagB is contains N balls so that the i urn contains i black and N – i white balls (0 i N). An urn is chosen 25 8 13 25 (a) (b) (c) (d) at random and n balls are drawn from it one after 51 17 27 52 another with replacement. If all the n balls turn out 11. Th e set of equations to be black, the probability that the next ball will x − y + (cos)z = 0 also black [assume that N is large] is 3x + y + 2z = 0 n n n 1 n 1 (cos)x + y + 2z = 0 (a) (b) (c) (d) n2 n 1 n2 n 1 0 < 2, has non−trivial solutions (a) for no value of and 6. Th e probability that a teacher will give a surprise (b) for all values of and test during any class meeting is 3/5. If a student is (c) for all values of and only two values of absent on two days, the probability that he will miss (d) for only one value of and all values of . at least one test is 10
MATHEMATICS TODAY | FEBRUARY ‘17
12. Orthogonal trajectories of family of parabolas y2 = 4a(x + a) where ‘a’ is an arbitrary constant is (a) ax2 = 4cy (b) x2 + y2 = a2 −
x 2a
(c) y = ce (d) axy = c2, where c is a constant. 13. If
5z1 2z1 + 3z2 is purely imaginary, then = 7 z2 2z1 − 3z2
(a) 5/7
(b) 7/5
(c) 25/29 (d) 1
14. The mean daily profit made by a shopkeeper in a month of 30 days was ` 350. If the mean profit for the first twenty days was ` 400, then the mean profit for the last 10 days would be (a) ` 200 (b) ` 250 (c) ` 800 (d) ` 300 15. Consider points A(3, 4) and B(7, 13). If P be a point on the line y = x such that PA + PB is minimum, then coordinates of P are 13 13 12 12 (a) , (b) , 7 7 7 7 31 31 (c) , (d) (0, 0) 7 7 16. lim
log(2 + x 2 ) − log(2 − x 2 ) x2 (b) 2
x →0
(a) –1
17. The sum of the series is equal to (a) 1
(b) 0
= k , the value of k is
(c) 1 3 12 ⋅ 22
+
(d) 0 5 22 ⋅ 32
(c) –1
+
7 32 ⋅ 42
+ ...∞
(d) 2
18. If f : R → R, f (x + y) = f (x) · f (y). If f (1) = 1, then n
∑ f (k) = k =1
(a) n (c) 1
(b) n(n + 1)(2n + 1)/6 (d) n(n + 1)/2
sin(a + 1)x + sin x , for x c , for 19. If f (x ) = 2 1/2 1/2 (x + bx ) − x , forr bx 3/2
x <0 x =0 x >0
is continuous at x = 0, then a = (a) 3/2 (b) –3/2 (c) 1/4 (d) –1/4
12
mathematics today | February ‘17
20. If f (x) = xn, then the value of
2
2 2 n f ′(1) f ′′(1) f (1) { f (1)} + + + .... + 1! 2 ! n ! is given by (a) 2nCn (b) 2nCn – 1 2n (c) Cn + 1 (d) 2nC1 2
21. For the three events A, B and C, P(exactly one of the events A or B occurs) = P (exactly one of the events B or C occurs) = P(exactly one of the events C or A occurs) = p and P(all the three events occurs simultaneously) = p2, where 0 < p < 1/2. Then the probability of at least one of the three events A, B and C occurring is 1 1 2 (a) (3 p + 2 p ) (b) ( p + 3 p2 ) 2 4 1 1 2 (c) ( p + 3 p2 ) (d) (3 p + 2 p ) 4 2 22. The area of the smaller region bounded by the curves x2 + y2 = 5 and y2 = 4x is π 1 π 1 + (a) (b) − 4 6 4 6 1 5 −1 2 1 5 −1 2 (c) 2 − sin (d) 2 + sin 3 2 3 2 5 5 x + a2 ab ac 23. If a, b, c are real, then f (x ) = is decreasing in 2 (a) − (a2 + b2 + c 2 ), 3
ab
x + b2
bc
ac
bc
x + c2
0
2 (b) 0, (a2 + b2 + c 2 ) 3 a 2 + b2 + c 2 (c) , 0 (d) None of these 3
24. If P(n) : n2 + n is an odd integer. It is seen that truth of P(n) ⇒ the truth of P(n + 1). Therefore, P(n) is true for all (a) n > 1 (b) n (c) n > 2 (d) none of these 2 2 2 25. If y = cos x + cos x + cos x + ...... to ∞ ,
then (a)
dy is dx − sin x 2 x(2 y − 1)
(b)
−2 x sin x 2 2y −1
(c)
− sin x 2y −1
26. Area lying between the curves y2 = 4x and y = 2x is (a) 2/3 (b) 1/3 (c) 1/4 (d) none of these 27. If f(x) is a continuous function satisfying f(x) f(1/x) = f(x) + f(1/x) and f (1) > 0, then lim f ( x) x →1
is equal to (a) 2 (c) 3
(b) 1 (d) none of these
28. For any integer n, the integral π
∫e
cos 2 x
0
cos 3 x( 2n + 1)x dx has the value
(a) π (c) 0
(b) 1 (d) none of these
29. Let E and F be two independent events. The probability that both E and F happen is 1/12 and the probability that neither E nor F happens is 1/2. Then (a) P(E) = 1/3, P(F) = 1/4 (b) P(E) = 1/2, P(F) = 1/6 (c) P(E) = 1/6, P(F) = 1/2 (d) P(E) = 1/8, P(F) = 1/3 30. If nCr denotes the number of combinations of n things taken r at a time, then the expression n
n −1
C0 + ∑
n+ k
k =0
(a) (c)
2nC 2nC
n–1 n
5/2 4/3 5/3 5/4
(d) none of these
Ck +1 equals (b) 2nCn + 1 (d) nCn – 1 SOLUTIONS
1. (c): Let A be the event that the numbers of the triplets form a G.P. Since 3 numbers can be selected from 30 numbers in 30C ways, therefore total number of triplets = 30C 3 3 Now we count the triplets (arranged in increasing order) whose terms form a G.P. by listing them as follows. Common ratio Triplet 2 {(i, 2i, 4i), 1 ≤ i ≤ 7} 3 {(i, 3i, 9i), 1 ≤ i ≤ 3} 4 (1, 4, 16) 5 (1, 5, 25) 3/2 (4, 6, 9), (8, 12, 18), (12, 18, 27)
(4, 10, 25) (9, 12, 16) (9, 15, 25) (16, 20, 25)
Thus among 30C3 triplets, there are 19 triplets, whose terms form a G.P. 19 19 \ Required probability P ( A) = 30 = C3 4060
2. (c): The total number of ways of distributing n balls into m boxes is mn which are assumed to be equally likely. Since r balls which should go to the specified box can be chosen in n C r ways and for any such way the remaining (n – r) balls can be distributed to the (m – 1) boxes in (m – 1)n – r ways. n \ The number of favourable cases = (m − 1)n−r r n n−r r (m − 1) \ P ( A) = mn [where A is the event that a specified box will contain r balls] 3. \ or or or
(c): Given that P(A ∪ B) = 0.4 and P(A ∩ B) = 0.1 P(A) + P(B) – P(A ∩ B) = 0.4 P(A) + P(B) = 0.4 + 0.1 = 0.5 1 – P(AC) + 1 – P(BC) = 0.5 P(AC) + P(BC) = 2 – 0.5 = 1.5
4. (d) : Let A denote the event of a successful landing of the rocket, B1 denote the event of the monitoring system indicating it correctly and B2 denote the event of its indicating unsuccessful landing. \ P(A) = 0.4, P(B1 |A) = 0.9, P(B2|AC) = 0.9 P (B1 | A) P ( A) The required probability = P(A|B1) = P (B1 ) Now, P(B1) = P(A ∩ B1) + P(AC ∩ B1) = P(B1|A)P(A) + P(B1|AC) P(AC) = P(B1|A)P(A) + [1 – P(B1C|AC)] P(AC ) = P(B1|A) P(A) + [1 – P(B2|AC)] P(AC ) = 0.9 × 0.4 + (1 – 0.9) × 0.6 = 0.42 0. 9 × 0. 4 6 Hence the required probability = = 0.42 7 5. (a) : Let Ai be the event of choosing ith urn and B the event of choosing n black balls in succession and C the event of drawing (n + 1)th ball as black. The required probability is mathematics today | February‘17
13
P (B ∩ C ) P ( B)
P (C | B) =
1 3 4 As P ( A) > , we must have P ( A) = and P (B) = 2 5 5 7 \ P ( A) + P (B) = 5
N
Now, P (B) = ∑ P ( Ai ) ⋅ P (B |Ai) i =0
N
=∑ i =0
1
n
1 i n = ∫ x dx when N → ∞ N +1 n 0
Similarly, N
1 i N +1 N
P (B ∩ C ) = ∑ i =0
n +1
1
=∫x
n +1
dx when N → ∞
=
i ∑ n i =0 N
1
n +1
i ∑ n
n
i =0
=
∫x 0
1
=
n +1
dx if n is large
∫ x dx n
0
n +1 = n+2 6. (d) : Required probability = 1 – P (no test is missed) = 1 – P (no test on his first day of absence and no test of his second day of absence) = 1 – P (no test of his first day of absence) × P (no test of his second day of absence) 2 2 21 = 1− = 5 5 25 7. (c): Given that P ( A ∩ BC ) = 8 1 , P ( A) > , 25 2 Let P(A) = x, P(B) = y
3 25
and P ( AC ∩ B) =
\ We are given that P(A ∩ BC) = P(A) – P(A ∩ B) 3 = x − P ( A)P (B) 25 3 = x − xy [Q A and B are independent] 25 Also, P(AC ∩ B) = P(B) – P(A ∩ B) 8 1 ⇒ = y − xy. Hence y = x + 25 5 1 2 \ Solving x and y we get x = , y = 5 5 3 4 x = and y = or 5 5 14
=
0
Hence the required probability N
8. (a) : P(AC) = 0.3, P(B) = 0.4, P(ABC) = 0.5 \ P(A ∪ BC) = P(A) + P(BC) – P(A ∩ BC) = 0.7 + 0.6 – 0.5 = 0.8 Now, P(B |A ∪ BC)
mathematics today | February ‘17
=
P[B ∩ ( A ∪ BC )] P ( A ∪ BC ) P ( A ∩ B)
P ( A ∪ BC )
=
=
P[(B ∩ A) ∪ (B ∩ BC )] P ( A ∪ BC )
P ( A) − P ( A ∩ BC ) P ( A ∪ BC )
0. 7 − 0 . 5 0 . 2 1 = = 0.8 0.8 4
1 1 1 9. (c): Given that P ( A) = , P (B) = , P (C ) = 2 3 4 \ Probability that only two of them can solve the problem = P(A ∩ B ∩ CC) + P(A ∩ BC ∩ C) + P(AC ∩ B ∩ C) 1 1 3 1 2 1 1 1 1 6 1 = ⋅ ⋅ + ⋅ ⋅ + ⋅ ⋅ = = 2 3 4 2 3 4 2 3 4 24 4 1 10. (a) : P (E1 ) = P (E2 ) = 2 3 5 and P (R |E1) = , P (R |E2 ) = 5 9 By Bayes' theorem P (E2 ) ⋅ P (R | E2 ) P (E2 | R) = P (E1 ) ⋅ P (R | E1 ) + P (E2 ) ⋅ P (R | E2 ) 1 5 5 ⋅ 25 2 9 = 9 = = 1 3 1 5 3 5 27 + 25 ⋅ + ⋅ + 2 5 2 9 5 9 25 P (E2 | R) = 52 11. (a) : Determinant of coefficients λ −1 cos θ 1 2 = cos θ − cos2 θ + 6 = 3 2 cos θ 1 and this is positive for all θ since |cosθ| ≤ 1. The only solution is therefore the trivial solution. dy dy 2a = 4a ⇒ = dx dx y For orthogonal trajectory, 12. (c): 2 y ⋅
mathematics today | February‘17
15
−
dx 2a = ⇒ dy y
dy 1 ∫ y = ∫ − 2a dx x
− x ⇒ ln y = − + k ⇒ y = ce 2a 2a
5z1 z 7K = Ki (K ∈ R), then 1 = i 7 z2 z2 5 z 2 1 + 3 14 K i + 3 2z1 + 3z2 z2 =1 = 5 = Consider 14 K 2z1 − 3z2 z1 i−3 2 −3 5 z2 14. (b) : 30 × 350 = 20 × 400 + 10 × x ⇒ 10x = 10500 – 8000 ⇒ 10x = 2500 ⇒ x = 250 13. (d) : Let
15. (c): Consider a point A′, the image of A through y=x
19. (b) : f (0 − 0) = lim f (0 − h) h→0
(a + 1 + 1)(−h) sin(a + 1)(−h) + sin(−h) = lim =a+2 = lim h→0 ( − h) h→0 0−h f (0 + 0) = lim f (0 + h) = lim h→0
(h + bh2 )1/2 − h1/2
h→0
bh b1/2 1 + − 1 =1 2 = lim 1/2 h→0 2 bh ⋅ h
bh3/2
\ The given function is continuous f (0 – 0) = f (0 + 0). 1 3 \ a+2= , a = − 2 2 f r (1) n = Cr 20. (a) : f (x ) = x n ⇒ r! n
∑ (n Cr )2 = 2nCn \ Coordinates of A′ = (4, 3) r =0 [Notice that A and B lie to 21. (a) : P( A ∪ B ∪ C ) = ∑ P( A) − P( AB) − P(BC ) − P( AC ) + P( A ∩ B ∩ C ) the same side with respect P( A ∪ B ∪ C ) = ∑ P( A) − P( AB) − P(BC ) − P( AC ) + P( A ∩ B ∩ C ) to y = x]. P ( A) − P ( AB) + P (B) − P (BC ) + P (B) Then PA = PA′ 1 Thus, PA + PB is minimum, = − P ( AB) + P (C ) − P (BC ) + P ( A) − P ( AC ) + P ( ABC ) 2 if PA′ + PB is minimum, if P, A′, B are collinear. + P (C ) − P ( AC ) 13 − 3 (x − 4) ⇒ 3 y − 10 x + 31 = 0 Now, AB is y − 3 = 1 7−4 = {P ( AB ) + P ( AB) + P ( AC ) + P (BC ) + P (BC ) + P ( AC )} 2 31 31 It intersects y = x at , , which is the required + P(ABC) 7 7 1 1 point P. = (3 p) + p2 = (3 p + 2 p2 ) 2 2 1 16. (c): lim 2 [log(2 + x 2 ) − log(2 − x 2 )] x →0 x 2 − x2 1 2 + x2 log log − 2 x →0 x 2 2
= lim
1/ x 2 1/ x 2 2 2 x x = lim log 1 + − log 1 − x →0 2 2 1 1 = log e1/2 − log e −1/2 = + = 1 2 2 17. (a) : The given series is, ∞
(2r + 1)
∞
1
∞
1
∑ r 2 (r + 1)2 = ∑ r 2 − ∑ (r + 1)2 = 1 r =1
r =1
r =1
18. (a) : f(1) = 1 ⇒ f(2) = 1 ⇒ f(3) = 1 f(r) = 1 ∀ r = 1, 2, ...., n n
∑ f (k) = n k =1
16
mathematics today | February ‘17
WEST BENGAL at
mathematics today | February‘17
17
(2x)2 = 4x ⇒ 4x2 = 4x ⇒ 4x2 – 4x = 0 ⇒ 4x(x – 1) = 0 ⇒ x = 0, 1 1
∫ (2
Required area =
22. (d) :
x − 2 x )dx
0
1
4 1 2 = 2 ⋅ x 3/ 2 − x 2 = − 1 = 3 3 3 0 Area =
2
∫
−2
5 − y2 −
27. (a) : Since f is continuous function so,
y 2 dy 4
lim f ( x ) = f (1).
x →1
Put x = 1 in the given equation we have (f (1))2 = 2f (1), so f(1) = 0 or 2. Since f (1) > 0, so f (1) = 2.
2
2 1 5 2 = 2 ∫ 5 − y 2 − y dy = 2 + sin −1 5 3 2 4 0
28. (c)
23. (a) :
1
0
f ′(x ) = ab x + b2 ac
bc
0
x + a2 ab bc + 0 1 x + c2
ac +
ac 0
bc x + c 2 x + a2
ab
ac
ab 0
x + b2 0
bc 1
= (x + b2)(x + c2) – b2c2 + (x + a2)(x + c2) – a2c2 + (x + a2)(x + b2) – a2b2 2 2 2 2 = 3x + 2x(a + b + c ) f (x) will be decreasing when f ′(x) < 0 ⇒ 3x2 + 2x(a2 + b2 + c2) < 0 2 ⇒ x ∈ − (a2 + b2 + c 2 ), 0 3 24. (d) : The sufficient condition for the statement : P(n) to be true for n ≥ a is (i) truth of P(n) ⇒ truth of P(n + 1) (ii) P(a) is true. Hence answer is none of these. 2 2 2 25. (b) : y = cos x + y ⇒ y − y = cos x dy dy \ 2y − = − sin x 2 ⋅ 2 x dx dx dy 2 x sin x 2 ⇒ =− dx 2y −1
1 12 1 1 (1 − x )(1 − y ) = , or 1 − (x + y ) + xy = 2 2 1 1 7 7 1 x+y= + = x + y = , xy = ; 2 12 12 12 12 1 1 P (E ) = and P (F ) = 3 4 1 1 or, P (E ) = and P (F ) = 4 3 29. (a) : Let P(E) = x, P(F) = y, xy =
n −1
n 30. (c): C0 + ∑
n+ k
= ( n + 2C 2 +
3)
k =0
+ n + 2C3 + .... + n + n – 1Cn = (n + 1C1 + n + 1C2) + n + 2C3 + .... + 2n – 1Cn Similarly,
n + 2C
2n – 1C
18
JEE MAIN
n–1 +
2n – 1C
2n – 1C
n
=
n 2nC
n
1st April to 30th April (Online) 2nd April (Offline) 8th & 9th April (Online)
VITEEE
5th April to 16th April (Online)
NATA
16th April
WBJEE
23rd April
Karnataka CET
mathematics today | February ‘17
+ .... +
Exam Dates 2017 SRMJEEE
Kerala PET
26. (b) :
Ck +1 = (n C0 + nC1 ) + n+1C2
24th April (Physics & Chemistry) 25th April (Mathematics) 2nd May (Biology & Mathematics) 3rd May (Physics & Chemistry)
BITSAT
16th May to 30th May (Online)
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21st May
SECTION-1 Single OPTiOn CORReCT TyPe
1. Consider the two statements. S 1 : The number of solutions to the equation ex = x2 is 1. S 2 : The number of solutions to the equation ex = x3 is 1. (a) S1 is true, S2 is false. (b) S1 is true, S2 is true. (c) S1 is false, S2 is true. (d) S1 is false, S2 is also false.
2. The probability that a random chosen divisor of 3039 is a multiple of 3029 is 11 (a) 40
3
21 (b) 40
3
11 (c) 40
5
21 (d) 40
5
1 (c) 2 (d) 4 3 4. In a certain lottery, 7 balls are drawn at random from n balls numbered 1 through n. If the probability that no pair of consecutive numbers is drawn equals the probability of drawing exactly one pair of consecutive numbers, then n = (a) 50 (b) 60 (c) 54 (d) 64 8 27
(b)
7. An ellipse has foci at (9, 20) and (49, 55) in the xy plane and is tangent to x-axis. The length of its major axis is (a) 70 (b) 75 (c) 80 (d) 85 8. A rectangular box has faces parallel to co-ordinate planes. If two of its vertices are (1, –1, 0) and (2, 3, 6) then volume of the box is (a) 20 (b) 22 (c) 24 (d) 26 9. The number of distinct real number pairs (a, b) such that a + b ∈ integers and a2 + b2 = 2 is/are (a) 10 (b) 6 (c) 2 (d) 4
3. 4 persons are playing a game of toy-gun shooting. At exactly midnight, each person randomly chooses one of the other three and shoots him. The probability that exactly 2 persons are shot is (a)
6. Let A, B, C, D be distinct points on a circle with centre O. If there exists non-zero real numbers x and y such that | x OA + y OB | = | x OB + y OC | = | x OC + y OD | = | x OD + y OA |, then ABCD is (a) a square (b) a trapezium (c) a rectangle (d) none of these
4 9
5. If B is an idempotent matrix satisfying the condition (I – aB)–1 = I – 3B, where I is unit matrix of same order as B and a ∈ R then 2a = (a) 1 (b) 2 (c) –1 (d) 3
2p 10. Let n ≥ 3 be an integer and z = cis . Consider n the sets A = {1, z, z2, ....., zn –1} and B = {1, 1 + z, 1 + z + z2, ..., 1 + z + z2 + ... + zn – 1}. The number of elements in the set (A ∩ B) when n is even is (a) 0 (b) 1 (c) 2 (d) 3 11. Let a and b be positive real numbers such that a[a] = 17 and b[b] = 11 then (a – b) = (a) 3/7 (b) 4/7 (c) 7/12 (d) 5/12 ([·] denotes greatest integer function] 12. Let a function f(x), x ≠ 0 be such that 1 1 f (x) + f = f (x) ⋅ f then f (x) can be x x (a) 1 – x4 + x3 (c)
p 2 tan −1 | x |
(b)
| x | −x +1 2x (d) 1 + λ log | x |
By : Tapas Kr. Yogi, Mob : 9533632105. mathematics today | february ‘17
19
13. Assume that a (> 1) is a root of the equation x3 – x – 1 = 0 then (a) 1 (b) 2
3
3a2 − 4a + 3 3a2 + 4a + 2 = (c) 2a (d) a1/3
14. The maximum and minimum of the function 6sinx cosy + 2sinx siny + 3cosx is (a) 7 and –7 (b) 5 and –5 (c) 6 + 13 and 6 − 13 (d) 5 + 2 and 5 − 2 15. For a real number a, let p
I(a) = ∫ log(1 − 2a cos x + a2 ) dx then I(a) = 0
1 I(a2 ) 2 (d) I(–a2)
(a) I(a2)
(b)
(c) 2I(a2)
16. The minimum of the expression 4 2 16 x + 2 − 2 (1 + cos q)x + (1 + sin q) x x + (3 + 2cosq + 2sinq) for x > 0 and q ∈ [0, 2p] is (a) 2 2 − 1 (b) 2 2 + 3 (c) 2 2 + 1 (d) none of these 17. The area of the region contained by all the points (x, y) such that x2 + y2 ≤ 100 and sin(x + y) ≥ 0 is (a) 10p (b) 25p (c) 50p (d) 100p SECTION-2 COmPRehenSiOn TyPe
Passage-1 Let a be a fixed real number, a ∈(0, p) and Ir =
a
∫
1 − r cos u
2 −a 1 − 2r cos u + r
18. I1 = (a) 0
(b) a
du (c) 2a
(d) –a
19. lim Ir = + r →1
(a) a – p (b) a + p (c) a
(d) –a
20. lim Ir = − r →1
(a) a – p (b) a + p (c) a
(d) –a
Passage-2
Let r1 < 0 < r2 < r3 be the real roots of 8x 3 − 6x + 3 = 0 and let s 1 < 0 < s 2 < s 3 b e t he re a l ro ots of 8x3 – 6x + 1 = 0, then 20
mathematics today | february ‘17
21. 4r22 – 4r24 = (a) r12 (b) r22 22. r12 + s22 = (a) 1
(b) 2
(c) r32
(d) r32 – r12
(c) 4
(d) 8
SOLUTIONS
1. (a) : For x < 0, e x = x 2 has one solution and ex = x3 has no solution. x For x > 0, consider the function f (x) = log x log x − 1 \ f ′(x) = (log x)2 So, for x > e, f(x) increases and for x < e, f (x) decreases and moreover x = 1 is an asymptote. So, graph of f (x) is
Now, y = 2 does not intersect the graph but y = 3 meets the graph at 2 different points. Hence, in total ex = x2 intersects once and ex = x3 intersects twice. 2. (a) : 3039 = 239 × 339 × 539. So, 40 × 40 × 40 divisors in total. 2 a 3 b 5 c to be a multiple of 30 29 , we must have a ∈ [29, 39], b ∈ [29, 39], c ∈ [29, 39] i.e. 11 choices for each a, b, c. 3 11 × 11 × 11 11 = Hence, required probability = 40 × 40 × 40 40 3. (a) : Let the 4 persons were A, B, C, D. First, we choose which two were shot = 4C2. Let C, D were shot. Then C D and A → C or D and B → C or D. Any person shooting any other person has probability = 1/3. So, probability that C, D were shot 1 1 2 2 4 = × × × = 3 3 3 3 81 [for C, D] [for A, B]
Hence, probability that exactly 2 persons were shot 4 8 = 4C2 × = 81 27 4. (c) : There are n – 6C7 ways to draw 7 balls so that no two balls are consecutive and (n – 6) × n – 7C5 ways to draw 7 balls so that there is exactly one pair of consecutive balls. Hence, n – 6C7 = (n – 6) × n – 7C5 gives n = 54
5. (d) : From given relation, I = (I – 3B)(I – aB) i.e., I = I – aB – 3B + 3aB = I + 2aB – 3B Hence, 2a = 3 6. (a) : Squaring, | x OA + y OB |2 = | x OB + y OC |2 ⇒ x 2OA ⋅ OA + y 2OB ⋅ OB + 2xy OA ⋅ OB = x 2OB ⋅ OB + y 2OC ⋅ OC + 2xy OB ⋅ OC i.e., x 2r 2 + y 2r 2 + 2xy OA ⋅ OB = x 2r 2 + y 2r 2 + 2xy OB ⋅ OC Hence, ABCD is a square. 7. (d) : Let F1 and F2 be the foci and P be the point of tangency. Let F1′ be the image of F1 in x-axis then F1P + F2P = F1′P + F2P = F1′F2 = 2a = (40)2 + (75)2 = 85 8. (c) : Vertices are (1, –1, 0) and (2, 3, 6). Notice that these must be diagonal vertices. Making (1, –1, 0) as origin, we have (2, 3, 6) as (1, 4, 6). Hence, dimensions of the box are 6 × 4 × 1 i.e. Volume = 24 cubic units
12. (2) : Rearranging the given fractional equation, we have 1 1 , f −1 = x f (x) − 1 p which is satisfied by only f (x) = , out of the 2 tan −1 x four options. 13. (b) : Since, a3 – a – 1 = 0, we have 3
3a2 − 4a = 3 3a2 − 4a − (a3 − a − 1) = 3 (1 − a)3 = 1 − a
and 3 3a2 + 4a + 2 + a3 − a − 1 = 3 (1 + a)3 = 1 + a Hence, given expression = 2 14. (a) : Using Cauchy-Schwarz inequality \ (x1y1 + x2y2 + x3y3)2 ≤ (x12 + x22 + x32) (y12 + y22 + y32) We have, (6sinx cosy + 2sinx siny + 3cosx)2 ≤ (62 + 22 + 32) ((sinxcosy)2 + (sinx siny)2 + cos2x) i.e. (6sinx cosy + 2sinx siny + 3cosx)2 ≤ 49 Hence, maximum = 7 and minimum = –7
9. (b) : Using, R.M.S ≥ A.M., we have |a + b| ≤ 2 So, a + b ∈ {–2, –1, 0, 1, 2} and a2 + b2 = 2 gives 1± 3 1 3 , (a, b) = (1, –1), (–1, 1), , 2 2 −1 ± 3 −1 3 , 2 2 10. (c) : Clearly, 1 ∈ A ∩ B. Let w ∈ A ∩ B, w ≠ 1 As an element of B, for some k = 1, 2, 3, ... (n – 1) 1 − z k +1 1− z For w ∈ A, we have |w| = 1 w = 1 + z + z2 + .... + zk =
(k + 1)p 1 − z k +1 p = 1 ⇒ sin = sin n 1− z n i.e. k = n – 2 1 − 1 / z −1 = So, w = 1− z z −1 So, A ∩ B = 1, z ⇒
{ }
11. (c) : a[a] = 17 ⇒ a ∈ (4, 5). So, [a] = 4 11 17 and a = . Similarly, b[b] = 11 ⇒ b = 3 4 7 Hence, a − b = 12 mathematics today | february ‘17
21
15. (b) : We have, I(a) = I(–a) by putting x = p – y. Now, we have (1 – 2acosx + a2)(1 + 2acosx + a2) = 1 – 2a2 cos2x + a4
Putting tan(u/2) = t and simplifying Jr =
1+ r a tan −1 tan 1− r 2 |1 − r | 4
p
So, I(a) + I(−a) = ∫ log(1 − 2a2 cos 2x + a 4 ) dx
So, Ir = a +
0
=
1 2p 2 4 ∫ log(1 − 2a cos y + a ) dy, ( y = 2x) 2 0
= a − 2⋅
Putting y = 2p – t we have
∫ log(1 − 2a
2
4 a (1 − r 2 ) 1+ r tan −1 tan ⋅ 2 2 1− r 2 |1 − r |
1+ r a So, lim Ir = a − 2 tan −1 tan + r −1 2 r →1 r →1
1 1 2p = I(a2 ) + ∫ log(1 − 2a2 cos y + a 4 ) dy 2 2 p
2p
2
p =a−p 2
a −1 1 + r tan 20. (b) : lim− Ir = a + 2 tan 1− r 2
cos y + a 4 ) dy = I(a2)
r →1
p
= a + 2⋅
1 1 Hence, I(a) + I(−a) = I(a2 ) + I(a2 ) 2 2
3 2 p 2p 7p 8p 13p 14p ⇒ q= , , , , , 9 9 9 9 9 9
sin 3q =
16. (d) : The given expression can be rearranged as 2
4 which is square of distance between point x, x on hyperbola xy = 4 and point (1 + cosq, 1 + sinq) on a circle centre (1, 1) and radius 1. Clearly the minimum distance occurs at (2, 2) on the hyperbola 1 1 and 1 + , 1+ on the circle. 2 2 2
1 So, minimum distance = 2 2 − 1 + = 2 −1 2 17. (c) : Using the symmetry, required area of the region 1 is × 100p = 50p sq. units 2 a a 1 1 − cos u 18. (b) : I1 = ∫ du = ∫ du = a 2 2 − 2 cos u −a −a 1 19. (a) : For r > 0, Ir = a + (1 − r 2 ) ⋅ 2 (1 − r 2) Ir = a + ⋅ J r (let) 2 Jr = 22
a
∫
du
−a 1 − 2r cos u + r
a
∫
du
−a 1 − 2r cos u + r
2
mathematics today | february ‘17
p =a+p 2
21. (c) : Putting x = sinq in 8x 3 − 6x + 3 = 0 , we have
1 So, I(a) = I(−a) = I(a2 ) 2 4 [x − (1 + cos q)]2 + − (1 + sin q) x
r →1
2
Hence, r1 = sin
in [0, 2p]
13p 4p p 2p = − sin , r2 = sin , r3 = sin 9 9 9 9
Similarly, putting x = cosq in 8x3 – 6x + 1 = 0, 1 we have, cos 3q = − 2 8p p So, s1 = cos = − cos , and 9 9 4p 2p s2 = cos , s3 = cos 9 9 p p 2p 2 Hence, 4r22 − 4r24 = 4 sin2 − 4 sin 4 = sin2 = r3 9 9 9 4p 4p 22. (a) : r12 + s22 = sin2 + cos2 =1 9 9
MPP-8 CLASS XII 1. 6. 11. 15. 20.
(c) 2. (d) (c) 7. (a,b,c) (a,b,c,d) (d) 16. (d) (3)
3. 8. 12. 17.
ANSWER KEY (d) (a,b,c) (a,b) (2)
4. 9. 13. 18.
(a) (c,d) (b,c) (1)
5. 10. 14. 19.
(b) (a,b) (a) (1)
Y U ASK
WE ANSWER Do you have a question that you just can’t get answered? Use the vast expertise of our mtg team to get to the bottom of the question. From the serious to the silly, the controversial to the trivial, the team will tackle the questions, easy and tough. the best questions and their solutions will be printed in this column each month.
1. The base of a D is divided into three equal parts. If t1, t2, t3 be the tangents of the angles subtended by these parts at the opposite vertex, prove that: 1 1 1 1 1 (Yogesh, Delhi) t + t t + t = 4 1 + 2 2 3 1 2 t2 A Ans. Let the points P and Q divide the side BC in three equal parts α γ β such that BP = PQ = QC = x Also let, ∠BAP = a, ∠PAQ = b, θ ∠QAC = g and ∠AQC = q B P Q C From question, tan a = t1, tan b = t2, tan g = t3. Applying, m : n rule in triangle ABC we get, (2x + x) cot q = 2x cot (a + b) – x cot g ...(1) From DAPC, we get (x + x) cot q = x cot b – x cot g ...(2) Dividing (1) by (2), we get 3 2 cot (a + b) − cot g = cot b − cot g 2 4 (cot a ⋅ cot b − 1) or 3 cot b – cot g = cot b + cot a or 4 + 4 cot2 b = cot2 b + cot a ⋅ cot b + cot b ⋅ cot g + cot g ⋅ cot a or 4(1 + cot2 b) = (cot b + cot a)(cot b + cot g) 1 1 1 1 t or 4 1 + 2 = + + t2 t1 t2 t2 t3 2. If a regular polygon of n sides has the circumradius R and inradius r then prove that each side of the polygon is equal to π 2(R + r)tan . (Raman, Gujarat) 2n Ans. Let A1A2 be a side and O be the centre. Let OB ⊥ A1A2. Clearly, OA1 = R and OB = r. 2π π Also, ∠A1OA2 = and ∠A1OB = . n n
r π = cos ∠A1OB = cos ...(1) R n AB π π and 1 = tan , i.e., A1B = r tan n n OB π \ A1A2 = 2A1B = 2r tan ...(2) n π π r Now, 2(R + r) tan =2 + r tan , (from (1)) π 2n 2n cos n π π 1 + cos 1 − cos q 1 − cos q n ⋅ n tan = = 2r π π 2 sin q cos sin n n π π 1 − cos2 sin2 n = 2r ⋅ n = 2r ⋅ π π π π cos ⋅ sin cos ⋅ sin n n n n π = 2r tan = A1A2, (from (2)). n 3. In how many ways can two distinct subsets of the set A of k(k ≥ 3) elements be selected so that they have exactly two common elements? (Akhil, Assam) Ans. Let the two subsets be called P and Q. The elements for the two sets will be selected as follows: (i) 2 elements out of k elements for both the sets. This can be done in kC2 ways. (ii) r elements for the subset P from k – 2 elements and any number of elements for Q from the remaining k – 2 – r elements. Here r can vary from 0 to k – 2. For a fixed r, the number of selections = k – 2Cr ⋅ 2k – 2 – r, (because the number of selections of any number of things from n things is 2n.) If r varies from 0 to k – 2, the total number of selections In DA1OB,
k−2
= ∑
r =0
k−2
Cr ⋅ 2
k−2−r
− 1,
excluding the case when both the subsets are equal having only the two common elements. But every pair of P, Q is appearing twice like {a1, a2, a3}, {a1, a2, a4, a5} and {a1, a2, a4, a5}, {a1, a2, a3}. Hence, the required number of ways k−2 1 k−2 k−2−r Cr ⋅ 2 − 1 = k C2 × ∑ 2 r = 0 1 k(k − 1) k – 2 = ⋅ ⋅ [( C0 ⋅ 2k – 2 + k – 2C1 ⋅ 2k – 3 2 2 + k – 2C2 ⋅ 2k – 4 + ... + k – 2Ck – 2) – 1] k(k − 1) k(k − 1) k − 2 k−2 ⋅ {(2 + 1) − 1} = ⋅ (3 − 1). = 4 4 mathematics today | February‘17
23
*ALOK KUMAR, B.Tech, IIT Kanpur single option correct type
3p/ 4
∫
e p/ 4dx
=k
p /2
∫
sec xdx ,
1. If f(x) = 0 is a cubic equation with positive and distinct roots a, b, g such that b is the H.M. of the roots of f ′(x) = 0. Then a, b, g are in (a) A.P. (b) G.P. (c) H. P. (d) none of these
7. If
2. All the roots of the equation 11z10 + 10iz9 + 10iz – 11 = 0 lie (a) inside |z| = 1 (b) on |z| = 1 (c) outside |z| = 1 (d) none of these
8. A line meets the co-ordinate axes at A and B. A circle is circumscribed about DOAB. If the distances from A and B of the tangent to the circle at the origin be m and n respectively, then diameter of the circle is (a) m(m + n) (b) m + n (c) n(m + n) (d) m2 + n2
3. Let y = f(x) be a continuous and differentiable curve. The normals at (1, f (1)), (2, f (2)) and (3, f (3)) make angles p/3, p/4 and p/6 with positive x-axis respectively. Then value of 2
3
1
2
∫ ( f (x) + xf ′(x)) dx + ∫ f ′(x) f ′′(x) dx
(a) f ′(2) + f (1) + 1 (c) f (3) – f (2) + 1 4.
lim
(b) 2f (2) – f(1) + 1 (d) none of these
sin(p cos2(tan(sin x)))
x →0
(a) p (c) p/2
x
2
is equal to
is equal to
(b) p/4 (d) none of these
5. If a = x + y + z + w and b = (xy + yz + zw + wx + wy + xz), then which of the following statement(s), is/are true? (a) 8a2 ≥ 3b (b) 3a2 ≥ 8b 2 (c) a b ≥ 27 (d) none of these ^ ^ ^ ^ ^ 6. If a = i + j + k , a ⋅ b = 1 and a × b = j − k , then b is equal to ^ ^ ^ ^ (a) 2 i (b) i − j + k (c) i^
^
^
(d) 2 j − k
− p/ 4 (e
x
+ e p/ 4 )(sin x + cos x)
then the value of k is 1 1 1 1 (a) (b) (c) (d) − 2 2 2 2 2
9. If I1 =
101
dx
∫
−100 (5 + 2x − 2x 101
)(1 + e 2− 4x )
∫
2
| x |, for 0 < | x | ≤ 3 10. Let f (x) = , then at x = 0 f(x) x =0 1, for has (a) a local maximum (b) no local maximum (c) a local minimum (d) no extremum 15 5 3 11. If x 2 + 9 y 2 + 25z 2 = xyz + + , then x, y, z x y z are in (a) A.P. (b) G.P. (c) A.G.P. (d) H.P. 12. The set of values of ‘r’ for which 23 Cr + 2 ⋅ 23Cr +1+ 23 Cr +2 ≥ 25C15 contains (a) 3 elements (b) 4 elements (c) 5 elements (d) 6 elements
He trains IIT and Olympiad aspirants.
mathematics today | February ‘17
and
I , then 1 is I 2 −100 5 + 2x − 2x (a) 1/2 (b) 2 (c) 0 (d) none of these I2 =
dx
2
* Alok Kumar is a winner of INDIAN NATIONAL MATHEMATICS OLYMPIAD (INMO-91). 24
− p /2
13. In an equilateral triangle inradius(r), circumradius (R) and exradius (r1) are in (a) A.P. (b) G.P. (c) H.P. (d) none of these 14. If tanA, tanB are the roots of the quadratic equation abx2 – c2x + ab = 0 where a, b, c are sides of the DABC then sin2A + sin2B + sin2C is (a) 1 (b) 2 (c) 3 (d) 3/2 15. The real function f (x) = cos −1 x 2 + 3x + 1 + cos −1 x 2 + 3x is defined on the set (a) {0, 3} (b) (0, 3) (c) {0, –3} (d) (–3, 0) 16. If a function satisfies f (x + 1) + f (x − 1) = 2 f (x) , then the period of f(x) can be (a) 2 (b) 4 (c) 6 (d) 8 17. If (1 + x)n = nC0 + nC1x + nC2x2 + .... + nCn xn where nC , nC , nC , ... are binomial coefficients. Then the 0 1 2 value of 2(C0 + C3 + C6 + ...) + (C1 + C4 + C7 + ...)(1 + w) + (C2 + C5 + C8 + ...)(1 + w2), where w is the cube root of unity and n is a multiple of 3, is equal to (a) 2n + 1 (b) 2n – 1 + 1 n + 1 (c) 2 –1 (d) 2n – 1 18. The chords of contact of the pair of tangents drawn from each point on the line 2x + y = 4 to x2 + y2 = 1 pass through the point (a) (1, 2) (b) (1/2, 1/4) (c) (2, 4) (d) none of these 19. If
a bc
−2=
b c , where a, b, c > 0, then family + c b
of lines ax + b y + c = 0 passes through the fixed point given by (a) (1, 1) (b) (1, –2) (c) (–1, 2) (d) (–1, 1) 20. If f(x) is continuous such that |f(x)| ≤ 1 " x ∈ R and e f (x) − e f (x) , then range of g(x) is e f (x ) + e f (x ) e2 + 1 (a) [0, 1] (b) 0, 2 e − 1 g (x) =
e2 − 1 (c) 0, 2 e + 1
1 − e2 (d) , 0 2 1 + e
21. Let three non-collinear points A, B and C have position vector a , b and c , then the vector a × b + b × c + c × a is
(a) parallel to plane containing a, b and c (b) perpendicular to the plane formed with the vectors AB and AC (c) perpendicular to the plane containing a, b and c (d) none of these 22. 271 should be split into how many parts so as to maximize their product? (a) 99 (b) 10 (c) 105 (d) none of these more than one option correct type
p and cos a + cos b = 1 , then 2 p (a) a + b ≥ (b) cos(a + b) ≤ 0 2 p (c) a + b ≤ (d) cos(a + b) ≥ 0 2
23. If 0 ≤ a, b ≤
24. If lim
ae 2x − b cos 2x + ce −2x − x sin x
= 1 and x2 f (t) = (a + b)t2 + (a – b)t + c, then (a) a + b + c = 1 (b) a + b + c = 2 (c) f (1) = 3/4 (d) f (1) = 1 x →0
25. If the conics whose equations are S1 : (sin2q)x2 + (2htanq)xy + (cos2q)y2 + 32x + 16y + 19 = 0 S2 : (cos2q)x2 + (2h′cotq)xy + (sin2q)y2 + 16x + 32y + 19 = 0 p intersect in four concyclic points, where q ∈ 0, , 2 then the correct statement(s) can be (a) l = 0 (b) l = 0 (c) q = p/4 (d) none of these 26. ABCD is a square of side 1 unit. P and Q lie on the side AB and R lies on the side CD. The possible values for the circumradius of triangle PQR is ? (a) 0.5 (b) 0.6 (c) 0.7 (d) none of these ^ ^ ^ ^ ^ ^ 27. If a = i + j + k, a ⋅ b = 2 and a × b = 2 i + j − 3 k, then ^ ^ ^ (a) a + b = 5 i − 4 j + 2 k ^ ^ ^ (b) a + b = 3 i + 2 k (c) b = 2 ^i − ^j + k ^ (d) b = ^i − 2 ^j − 3 k 28. The solution of x1/3 + (2x – 3)1/3 = [3(x – 1)]1/3 is (a) 0 (b) 3/2 (c) 1 (d) none of these mathematics today | February ‘17
25
29. Let a1, a2, a3, ..., an are in G.P. such that 3a1 + 7a2 + 3a3 – 4a5 = 0, then common ratio of G.P. can be (a) 2 (b) 3/2 (c) 5/2 (d) –1/2 30. The diagonals of a square are along the pair represented by 2x2 – 3xy – 2y2 = 0. If (2, 1) is the vertex of the square, then the other vertices are (a) (–1, 2) (b) (1, –2) (c) (–2, –1)(d) (1, 2) x ∫ (5 + | 1 − t |) dt , if 31. Let f (x) = 0 5x + 1, if function is (a) continuous at x = 2 (b) differentiable at x = 2 (c) discontinuous at x = 2 (d) not differentiable at x = 2
x >2
then the
x ≤2
32. The length of other diagonal BD is 10 3
(b) 2
(c)
20 3
(d) none of these
33. The length of side AB is equal to (a)
4 58 3 58 2 58 5 58 (b) (c) (d) 9 9 3 9
34. The length of BC is equal to 4 10 2 10 8 10 (a) (b) (c) (d) none of these 3 3 3 matriX match type
35. Match the following. Column I sin 1 sin 5 − (A) sin 2 sin 6 (B)
3 9 tan − 2 4
Column II P.
positive
Q. negative
e x − 1 (C) lim (where [⋅] denotes R. 1 x →0 x the greatest integer function) S. d o e s n ot exist 26
Column I
Column II 24 (A) The area enclosed between the P. 5 2 curves |x| + |y| = 2 and x = y (in sq. units) is 7 (B) The maximum value of the function Q. 3 7 f (x) = 3 sin x − 4 cos x − 3 will be given by 16 (C) The length of common chord of R. 3 two circles of radii 3 and 4 units which intersect orthogonally is
comprehension type Let ABCD be a parallelogram whose diagonals equations are AC : x + 2y = 3; BD : 2x + y = 3. If length of diagonal AC = 4 units and area of ABCD = 8 sq. units.
(a)
36. Match the following.
mathematics today | February ‘17
(D) The length of chord intercepted by S. the parabola y2 = 4(x + 1) passing through its focus and inclined at 60° with positive x-axis is
8 3
integer type
37. If the normal to the curve x = t – 1, y = 3t2 – 6 at the point (1, 6) make intercepts a and b on x and y-axes a +12b respectively, then the value of is______ 146 38. Let S(n) denotes sum of first n terms of an A.P., n f (r ) then the value of S = lim ∑ , where n→∞ r =−n n S(3n) , is ______ f (n) = S(2n) − S(n) 39. If f (u, v, w, x) = u2 + v2 + w2 + x2 – 2(u + v + w + x) + 10 with u, w ∈ [–3, 3] and v, x ∈ [1, 3] then 1 max ⋅ f (u, v, w, x) is 23 40. If a = 2, b = 3, and c = 4 then 2 2 2 a −b + b −c + c −a cannot exceed _____ 87 solutions
1. (b) : f(x) = (x – a)(x – b)(x – g) ⇒ f ′(x) = 3x2 – 2x(a + b + g) + ab + bg + ga 2a1b1 ⇒ b= (where a1, b1 are the roots of f ′(x) = 0) a1 + b1 ⇒ b=
2(ab + bg + ga) ⇒ 2(a + b + g)
b2 = ga
11 − 10iz 2. (b) : z (11z + 10i) = 11 − 10iz ⇒ z = 11z + 10i i z + 11 10 ⇒ | z9 | = 11z + 10i 9
9
⇒ | 11i + 10z |2 − | 11z + 10i |2 = 21(1− | z |2 ) 9
⇒ if | z | < 1, then | z | > 1 (not possible)
Also a ⋅ b = 1 ⇒ a + b + g = 1 ⇒ b+1+b+b=1 ⇒ b=0 \ a = 1, g = 0 ^ \ b=i 7. (c) : I =
3p/ 4
⇒ |z| = 1 3. (b) : So,
=
∫ xf ′(x) dx
1
2 = xf (x)|12 − ∫ 1
2
3
1
2
f (x) dx
∫ ( f (x) + xf ′(x))dx + ∫ f ′(x) f ′′(x) dx =xf
I=
2 3 2 ( f ′(x)) (x) 1 + 2 2
1 = 2 f (2) − 1 f (1) + [( f ′(3))2 − ( f ′(2))2] 2 1 = 2 f (2) − f (1) + [3 − 1] = 2 f (2) − f (1) + 1 2 4. (a) : lim
(
sin p (1 − sin2 ( tan(sin x))) x
x →0
2
)
sin ( p sin (tan(sin x))) p sin ( tan(sin x)) x →0 p sin2 ( tan(sin x)) tan2(sin x) 2
= lim
2
tan2 (sin x ) sin2 x 2 = p 2 sin x x
2 − p/2 (e + 1)cos t
2 − p/2 (et + 1)cos t
Adding, 2I = \ I=
1 p /2
∫
2 − p /2
p /2
1
∫
2 2 − p /2
sec tdt
sec xdx
a b 8. (b) : Circumcentre of DOAB ≡ , 2 2 a 2 b2 + 4 4
Circumradius =
Equation of circle x2 + y2 – ax – by = 0 \ Equation of tangent at origin ax + by = 0 AL =
a2
⇒ 2I1 =
, BM =
a 2 + b2
(0, b) B m M L
b2
a 2 + b2
n
B(a, 0)
,
AL + BM = m + n = a2 + b2 = diameter of circle.
⇒ I1 =
101
dx
∫
−100 (5 + 2x − 2x
101
2
)(1 + e 2− 4x )
dx
∫
2 2 − 4(1− x) ) −100 (5 + 2(1 − x) − 2(1 − x) )(1 + e 101
I 1 = I2 ⇒ 1 = I2 2 −100 5 + 2x − 2x
∫
^
⇒ b – g = 0, a – g = 1, a – b = 1 ⇒ b = g, a = 1 + g, a = 1 + b,
dt
t
et dt
^ ^ ^ 6. (c) : Let b = a i + b j + g k, ^
∫
∫
9. (a) : I1 =
^
1 p /2
1 p /2
5. (b) : Since, 3(x2 + y2 + z2 + w2) – 2(xy + yz + zx + zw + wx + wy) = (x – y)2 + (x – z)2 + (x – w)2 + (y – z)2 + (y – w)2 + (z – w)2 ≥ 0 2 2 ⇒ 3Sx2 – 2Sxy ≥ 0 ⇒ Sx ≥ Sxy 3 2 2 2 Now, (x + y + z + w) = Sx + 2Sxy ≥ Sxy + 2Sxy 3 8 8 ⇒ (x + y + z + w)2 ≥ Sxy ⇒ a2 ≥ b 3 3
i j k ^ ^ ^ ^ a ×b = j −k ⇒ 1 1 1 = j −k a b g
p 2(e x − p/4 + 1)cos x − 4
− p/ 4
and if | z | > 1 ⇒ | z 9 | < 1 (not possible) 2
dx
∫
dx
2
(0, 1)
10. (a) : �3
3
x2
9y2
11. (d) : We have, + + 25z2 = 15yz + 5zx + 3xy 2 2 2 ⇒ (x) + (3y) + (5z) – (x)(3y) – (3y)(5z) – (x)(5z) = 0 mathematics today | February ‘17
27
f (x + 8) = –f (x + 4) = f (x) " x 1 = 0f(x) is periodic with period 8. 2(x)2 + 2(3 y)2 + 2(5z)2 − 2(x)(3 y) − 2(3 y)(5z) − 2(x)(5z)\ 2 1 17. (d) : (1 + w)n = C0 + C1w + C2w2 + ... + Cnwn 2 2 2 2(x) + 2(3 y) + 2(5z) − 2(x)(3 y) − 2(3 y)(5z) − 2(x)(5z) = 0 2 (1 + 1)n = C0 + C1 + C2 + ... + Cn 1 (1 + w)n + (1 + 1)n = 2C0 + C1(1 + w) + C2(1 + w2) ⇒ (x − 3 y)2 + (3 y − 5z)2 + (x − 5z)2 = 0 2 + C3(1 + w3) + C4(1 + w) + C5(1 + w2) + C6(1 + w3) ⇒ x – 3y = 0, 3y – 5z = 0, x – 5z = 0 + ... + Cn(1 + wn) 1 1 5 1 1 1 = , = and = ⇒ = 2(C0 + C3 + C6 + ....) + (C1 + C4 + C7 + ...)(1 + w) x 3y 3y z 5z x + (C2 + C5 + C8 + ...)(1 + w2) = –w2n + 2n 1 1 1 5 2 + = + = ⇒ ⇒ 2n – 1 (Q n is a multiple of 3, wn = 1) x z 3y 3y y 18. (b) : Let (h, k) be any point on the given line 1 1 1 ⇒\ , , are in A.P. ⇒ x, y, z are in H.P. \ 2h + k = 4 and the chord of contact hx + ky = 1 x y z hx + (4 – 2h)y = 1 ⇒ (4y – 1) + h(x – 2y) = 0, p + lq = 0, it 12. (d) : 23Cr + 2 · 23Cr + 1 + 23Cr + 2 = 24Cr + 1 + 24Cr + 2 passes through the intersection of p = 0, q = 0 or = 25Cr + 2 ≥ 25C15 (1/2, 1/4). ⇒
\ (r + 2) can be 10, 11, 12, 13, 14 and 15. So, 6 elements. 13. (a) : In an equilateral triangle r = R/2 A B C Also ex-radius r1 = 4R sin cos cos 2 2 2 1 3 3 3 = 4R . . = R 2 2 2 2 ⇒ r, R, r1, are in A.P. c2 and tan A·tan B = 1 ab tan A ⋅ tan B = 1 ⇒ tan A = cot B ⇒ A = 90° – B ⇒ A + B = 90° ⇒ C = 90° a b ⇒ sin A = , sin B = c c 14. (b) : Here, tan A + tan B =
a 2 b2 \ sin2 A + sin2 B + sin2 C = 2 + 2 + 1 = 2 c c 2 15. (c) : Obviously, 0 ≤ x + 3x + 1 ≤ 1 and 0 ≤ x2 + 3x ≤ 1 ⇒ x2 + 3x = 0 ⇒ x = 0, –3 16. (d) : By replacing x = x + 1 and x = x – 1, we get ...(1) f (x + 2) + f (x) = 2 f (x + 1) ...(2) f (x) + f (x − 2) = 2 f (x − 1) Adding (1) & (2) gives, f (x + 2) + f (x − 2) + 2 f (x) = 2 [ f (x + 1) + f (x − 1)] = 2 2 f (x) \ f (x + 2) + f (x – 2) = 0 On replacing x by x + 2 we get f(x + 4) + f(x) = 0 Finally replacing x by x + 4, we get f (x + 8) + f (x + 4) = 0, 28
mathematics today | February ‘17
19. (d) :
a
bc
−2=
b c + ⇒ a = b + c + 2 bc c b
⇒ a = ( b + c )2 ⇒ ( a − b − c )( a + b + c ) = 0 ⇒ a − b − c = 0, a + b + c ≠ 0 (Q a, b, c > 0) Comparing with ax + b y + c = 0 Hence x = –1, y = 1. e f (x) − e| f (x)| 20. (d) : We have, g (x) = f (x) | f (x)| , −1 ≤ f (x) ≤ 1 e +e For 0 ≤ f(x) ≤ 1, g(x) = 0 For –1 ≤ f (x) < 0, e f (x ) − e − f (x ) e 2 f (x ) − 1 2 g (x) = f (x) − f (x) = 2 f (x) = 1 − 2 f (x ) e +e e +1 e +1 1 − e2 , 0 For –1 ≤ f (x) < 0, g (x) ∈ 2 1 + e 1 − e2 For –1 ≤ f (x) < 1, g (x) ∈ , 0 . 2 1 + e 21. (c) : Let r be the position vector of any point on the plane of a, b and c ⇒ (r − a) ⋅ (b − a) × (c − a) = 0 ⇒ (r − a) ⋅ (a × b + b × c + c × a) = 0 Thus a × b + b × c + c × a is perpendicular to r − a and thus is perpendicular to a, b and c . 22. (d) 23. (a, b) : 0 ≤ a, b ≤ p and cos a + cos b = 1 2 a −b 1 a +b cos = ⇒ cos 2 2 2
mathematics today | February ‘17
29
1 a −b a +b and cos ≥ ⇒ atleast one of cos 2 2 2 p ⇒ atleast one of a + b and a − b ≥ 2 p p but a − b ≥ and a + b ≥ and cos(a + b) ≤ 0 2 2 24. (a, c) : lim
ae 2x − b cos 2x + ce −2x − x sin x x2
x →0
=1
(2x)2 (2x)4 (2x)2 a 1 + 2x + + ... − b 1 + + − ... 2! 2! 4! (2x)2 +c 1 − 2x + − ... 2!
⇒ lim
x
x →0
⇒ a–b+c=0 2a – 2c = 0 2a – 2b + 2c = 2 ⇒ a = c and b = 2a ⇒
a + 2a + a = 1 ⇒ a =
2
\
2
b=
=2
3 1 1 3t − t + 1 Now, f (t ) = t 2 − t + = 4 4 4 4 25. (b, c) : Curve through the intersection of S1 and S2 is given by S1 + lS2 = 0 ⇒ x2(sin2 q + l cos2 q) + 2(h tan q – lh′ cot q)xy
+ (cos2 q + l sin2 q)y2 + (32 + 16l)x + (16 + 32l)y + 19(1 + l) = 0 The above equation will represent a circle if sin2 q + l cos2 q = cos2 q + l sin2 q ⇒ sin2 q – l sin2 q = cos2 q – l cos2 q ⇒ (1 – l)sin2 q = (1 – l)cos2 q p ⇒ (1 – l)(sin2 q – cos2q) = 0 ⇒ l = 1 or q = 4 h tan q – lh′ cot q = 0 ⇒ h tan q = lh′ cot q which is satisfied if l = 1 and q = p/4 ⇒ h = h′ 26. (a, b, c) : Let O be the circumcentre. Then OP + OR ≥ PR ≥ AD = 1, so the radius is at least 1/2. P, Q, R always lie inside or on the circle through A, B, C, D 1 1 which has radius , so the radius is at most . 2 2
30
mathematics today | February ‘17
28. (a, b, c) : a1/3 + b1/3 = (a + b)1/3 is true, when either a = 0 or b = 0 or a + b = 0 3 ⇒ x = 0, , 1 2 ⇒ 7(a1 + a2 + a3) = 4(a1 + a3 + a5)
1 2
27. (b, c) : We have (a × b ) × a = (a ⋅ a) b − (b ⋅ a) a = | a |2 b − 2a
(4 ^i − 5 ^j + k^) + 2(^i + ^j + k^) ^ ^ ^ \ b= = 2 i − j + k. 3
29. (b, d) : Given 3a1 + 7a2 + 3a3 – 4a5 = 0
...(1) ...(2) ...(3) 1 =c 4
(a × b ) × a + 2a \ b= | a |2 ^ ^ ^ Now, (a × b ) × a = 4 i − 5 j + k and | a | = 3
⇒ 7(1 + r + r2) = 4(1 + r2 + r4)
⇒ 7 = 4(r2 – r + 1) ⇒ 4r2 – 4r + 1 = 4 ⇒ (2r – 1)2 = 4 3 1 r= ,− 2 2 30. (a, b, c) : Since the diagonals intersect at origin and are at right angles. Let B and D be the points adjacent to A. ⇒ 2r – 1 = ±2 ⇒
2
D O A
2
Also OA = 2 + 1 = 5
(0, 0) B
(2, 1)
Let affix of B and D are z2 and z1 respectively. z p \ arg 2 = 2+i 2
⇒ z2 = (2 + i)i = –1 + 2i ⇒ B(–1, 2) p z Also, arg 4 = − 2+i 2 ⇒ z4 = (2 + i)(–i) = 1 – 2i ⇒ D(1, –2) x
1
0
0
x
31. (c, d) : x > 2∫ (5 + | 1 − t |) dt = ∫ (6 − t )dt + ∫ (4 + t ) dt = 1+ 4 +
2
1
x 2
x2 1 + 4x + , x > 2 ⇒ f (x) = 2 5x + 1, x ≤2 4 + x, x > 2 f ′(x) = x ≤2 5, f(2+) = f(2–) = 11 continuous at x = 2 f ′(2+) ≠ f ′(2–) ⇒ not differentiable at x = 2
A
1 − +2 3 3 32. (c) : tan θ = 2 = , sin θ = , 1+1 4 5 1 20 area of ∆CPB = × PC × PB sin θ = 2 ⇒ BD = 2 3 AP 2 + PB2 − AB2 33. (a) : cos(π − θ) = 2AP ⋅ PB 100 2 4 4 + 9 − AB 2 58 ⇒ AB = ⇒ − = 10 5 3 2×2× 3 PC 2 + PB2 − BC 2 34. (a) : ln ∆CPB, cos θ = 2PC ⋅ PB ⇒ BC =
2 10 3
35. (A)-(Q), (B)-(P), (C)-(S) sin1 sin 5 sin x (A) − , Take f (x) = sin 2 sin 6 sin(x + 1) sin(x + 1)cos x − cos(x + 1)sin x sin1 f ′(x) = = 2 >0 2 sin (x + 1) sin (x + 1) ⇒ f(x) is increasing ⇒ f(1) < f(5). (B) Take f (x) = tanx – x2 f ′(x) = sec2x – 2x, f ′′(x) = 2sec2x tanx – 2 f ′(1) > 0 and f ′′(x) > 0 for x > 1 3 9 π ⇒ f (x) is increasing in 1, ⇒ tan > 2 2 4 e x − 1 (C) lim does not exist as left hand and right x →0 x hand limits are not equal. 36. (A) - (Q), (B)-(S), (C)-(P), (D)-(R) 1 1 (A) Required area = 2 (2 + 1) × 1 − ∫ x 2dx 2 0 3 1 7 = 2 − = sq. units 2 3 3 7 8 (B) 5 − = 3 3 (C) Length of chord =
2r1r2
r12 + r22
(D) Parabola is y2 = 4(x + 1)
=
2 × 3 × 4 24 = 5 5
... (i), focus is (0, 0) x −0 y −0 = =r Equation of AB is 1/ 2 3 /2
60° (1, 0) B
Substituting parametric coordinates in (1) 2
2 3 r 3r r = + , − 2r − 4 = 0 4 1 2 2 4
Length of AB = PA − PB =
(PA + PB )2 − 4PAPB
2
16 16 8 = −4× = 3 3 3 37. (1) : Given point is corresponding to t = 2 and dy 1 = 6t ⇒ slope of normal at t = 2 is − 12 dx 1 \ Equation of normal is y − 6 = − (x − 1) 12 73 ⇒ a = 73, b = ⇒ a + 12b = 146 12 38. (6) : Let a be the first term of A.P. and d be the common difference. 3n S(3n) S(3n) = {2a + (3n − 1) d}, = 3 = f (n) 2 S(2n) − S(n) . 1 n 2n + 1 \ S = lim = 6 ∑ f (r) = lim 3 n n→∞ n r =−n n→∞ 39. (2) : f(u, v, w, x) = (u – 1)2 + (v – 1)2 + (w – 1)2 + (x – 1)2 + 6 Clearly f is max when u = w = –3 and v = x = 3 ⇒ fmax =(–3 – 1)2 + (–3 – 1)2 + (3 – 1)2 + (3 – 1)2 + 6 = 46 2 40. (1) : | a − b |2 + | b − c |2 + c − a = 2(a2 + b2 + c 2) −2(a ⋅ b + b ⋅ c + c ⋅ a) = 2(4 + 9 + 16) − 2(a ⋅ b + b ⋅ c + c ⋅ a) = 58 − 2(a ⋅ b + b ⋅ c + c ⋅ a) Now (a + b + c )2 ≥ 0 ⇒ a2 + b2 + c 2 ≥ −2(a ⋅ b + b ⋅ c + c ⋅ a) ⇒ − 2(a ⋅ b + b ⋅ c + c ⋅ a) ≤ 29 ⇒ | a − b |2 + | b − c |2 + | c − a |2 ≤ 87 mathematics today | February ‘17
31
Series-7 The entire syllabus of Mathematics of WB-JEE is being divided in to eight units, on each unit there will be a Mock Test Paper (MTP) which will be published in the subsequent issues. The syllabus for module break-up is given bellow: Unit No.
Topic
Syllabus In Details
UNIT NO. 7
Co-ordinate Geometry-3D
Direction ratios and direction cosines. Angle between two intersecting lines. Skew lines, the shortest distance between them and its equation. Equation of a line and a plane in different forms, intersection of a line and a plane, coplanar lines. Differential Rolle’s and Lagrange’s Mean value theorem theorems, Applications of derivatives: Rate of calculus change of quantities, monotonic-increasing and decreasing functions, Maxima and minima of functions of one variable, tangents and normals. Integral calculus Integral as an anti-derivative. Fundamental integrals involving algebraic, trigonometric, exponential and logarithmic functions. Integration by substitution, by parts and by partial fractions. Integration using trigonometric identities. evaluation of simple integrals of type: dx dx dx dx (px + q)dx dx dx ,∫ ,∫ 2 ∫ 2 2 ,∫ 2 2 ,∫ 2 2 ,∫ 2 2, ∫ 2 2 x ±a a −x + bx + c ax + bx + c ax x ±a a −x ax + bx + c
Time : 1 hr 15 min.
Full marks : 50
CATEGORY-I For each correct answer one mark will be awarded, whereas, for each wrong answer, 25% of total marks (1/4) will be deducted. If candidates mark more than one answer, negative marking will be done.
p p and with positive 4 3 direction of x-axis and z-axis respectively. Then the acute angle made by the line with y-axis is (a) p/6 (b) p/4 (c) p/3 (d) cos–1(1/3)
1. A straight line makes angles
2. The angle between the lines x +1 y − 2 z + 4 x − 3 2y + 3 z − 2 = = and = = is 3 −2 1 1 5 2 (a) p/3 (b) p/2 (c) cos–1(3/5) (d) cos–1(4/5) 3. The coordinates of the foot of perpendicular drawn from the point A(2, 4, –1) on the line x +5 y +3 z −6 = = are 1 4 −9 (a) (1, –3, 4) (b) (–4, 1, –3) (c) (4, 1, 3) (d) none of these
4. The equation of the line joining the points (2, –1, 4) and (1, 1, –2) is x − 2 y +1 z − 4 = = (a) 1 2 6 x − 2 y +1 z − 4 (b) = = −1 2 −6 x −1 y −1 z + 2 (c) = = 1 2 6 x −1 y +1 z + 2 (d) = = 2 −6 −1 5. A plane meets the coordinate axes at P, Q, R such that the centroid of the triangle PQR is (a, b, c). If x y z the equation of the plane is + + = m , then the a b c value of m is (a) 2 (b) 3 (c) 1 (d) 6 6. The equation of the plane passing through the points (0, 1, 0) and (3, 4, 1) and parallel to the line x +3 y −3 z −2 is = = 2 7 5
By : Sankar Ghosh, S.G.M.C, Kolkata, Ph: 09831244397.
32
mathematics today | February ‘17
(a) 4x – 13y + 15z = 13 (b) 8x – 13y + 15z = 15 (c) 8x – 13y + 15z + 13 = 0 (d) none of these 7. The vector equation of the line passing through the points (2, –3, 1) and (–4, 3, 6) is ^ ^ ^ ^ ^ ^ (a) r = −4 i + 3 j + 6 k + l(2 i − 3 j + k) ^ ^ ^ ^ ^ ^ (b) r = 2 i − 3 j + k + l(−4 i + 3 j + 6 k) ^ ^ ^ ^ ^ ^ (c) r = −6 i + 6 j + 5 k + l(2 i − 6 j − 5 k) ^ ^ ^ ^ ^ ^ (d) r = 2 i − 3 j + k + l(−6 i + 6 j + 5 k) ^ ^ ^ ^ ^ ^ 8. If the line r = 2 i − j + 3 k + l(2 i + j + 2 k) is parallel ^ ^ ^ to the plane r ⋅ (3 i − 2 j + p k ) = 4, then the value of p is (a) 2 (b) –2 (c) 3 (d) –3 9. The vector equation of a plane passing through the point (1, –1, 2) and having 2, 3, 2 as direction number of normal to the plane is ^ ^ ^ (a) r ⋅ (2 i + 3 j + 2 k ) + 4 = 0 ^ ^ ^ ^ ^ ^ ^ ^ ^ (b) r ⋅ (i − j + 2 k ) + (i − j + 2 k ) ⋅ (2 i + 3 j + 2 k ) = 0 ^ ^ ^ (c) r ⋅ (2 i + 3 j + 2 k ) + 5 = 0 ^ ^ ^ (d) r ⋅ (2 i + 3 j + 2 k ) = 3 10. The distance of the point (–3, 2, 4) from the plane ^ ^ ^ r ⋅ (2 i − 3 j + 6 k ) + 7 = 0 is 29 (a) 5 units (b) units 7 18 19 (c) (d) units units 7 7 11. In the mean value theorem f(b) – f(a) = (b – a)f ′(c) (a < c < b), if a = 4, b = 9 and f (x) = x , then the value of c is (a) 8 (b) 5.25 (c) 4 (d) 6.25 12. If the function f (x) = 4x3 + ax2 + bx – 1 satisfies 1 all the conditions of Rolle’s theorem in − ≤ x ≤ 1 4 1 and if f ′ = 0 , then the values of a and b are 2 (a) a = 2, b = –3 (b) a = 1, b = –4 (c) a = –1, b = 4 (d) a = –4, b = –1 13. Let f (x) be a continuous in (–∞, ∞) and f ′(x) exists in (–∞, ∞). If f (3) = –6 and f ′(x) ≥ 6 for all x ∈ [3, 6] then (a) f (6) ≥ 24 (b) f (6) ≥ 12 (c) f (6) ≤ 24 (d) f (6) ≤ 12 14. A spherical balloon is being inflated at the rate of 35 cm3/min. Then the rate of increase of surface area of the balloon when its diameter is 14 cm, is
(a) 7 cm2/min (c) 17.5 cm2/min
(b) 10 cm2/min (d) 28 cm2/min
15. If h(x) = f(x) – (f (x))2 + (f (x))3 and f ′(x) > 0 " x then (a) h(x) is an increasing function of x in some specified interval (b) h(x) is an increasing function " x ∈ R (c) h(x) is a decreasing function " x ∈ R (d) h(x) is a decreasing function of x in some specified interval . 16. Electric current C, measured by a galvanometer, is given by the equation C = k tanq, where k is a constant. Then the percentage error in the current corresponding to an error 0.7 percent in the measurement of q when q = 45° is (a) 1.4 (b) 2.8 (c) 1.1 (d) 2.2 17. The point on the parabola 2y = x2, which is nearest to the point (0, 3) is 1 (a) (± 4, 8) (b) ± 1, 2 9 (d) ± 3, 2 18. If m be the slope of the tangent to the curve ey = 1 + x2, then (a) |m| > 1 (b) |m| < 1 (c) m < 1 (d) |m| ≤ 1 (c) (±2, 2)
19. The normal to the curve x = a(cosq + qsinq), y = a(sinq – q cosq) at any point q is such that (a) it passes through the origin (b) it passes through (a, –a) (c) it is at a constant distance from origin p (d) it makes angle + q with the x-axis. 4 20. values of a for which the function (a + 2)x3 – 3ax2 + 9ax – 1 decreases monotonically throughout for all real values of x, are (a) a < –2 (b) a > –2 (c) –3 < a < 0 (d) a ≤ –3 sin x dx = Ax + B log |sin(x − a)| + c, then the sin(x − a) value of (A, B) is (a) (cosa, sina) (b) (–sina, cosa) (c) (sina, cosa) (d) (–cosa, sina)
21. If ∫
22.
∫e
x
dx is
(a) e x + c 1 (c) e x + c 2
(b) 2( x − 1)e x + c (d) 2( x + 1)e x + c
mathematics today | February ‘17
33
23.
∫
−1 e tan x
dx is 1 + x2
(a) tan–1x + C
24.
1
(b)
−1 (c) e tan x + C
dx 30. If ∫ = f (a)log cos x + cos a
1 + x2 2xe
+C
+C (1 + x 2 )2
(d)
2
∫ f (x)dx = f (x), then ∫ { f (x)} 1 { f (x)}2 + C 2
(b) {f (x)}3 + C
(c)
{ f (x)}3 +C 3
(d) {f (x)}2 + C
(a)
∫2
dx x (x + 1)
1 tan −1( x ) + C 2
(c) tan −1( x ) + C
Every correct answer will yield 2 marks. For incorrect response, 25% of full mark (1/2) would be deducted. If candidates mark more than one answer, negative marking will be done.
x y z x −1 y − 2 z − 3 and = = , = = 1 2 3 3 −1 4 x + k y −1 z − 2 are concurrent. Then, = = 3 2 h (a) h = –2, k = –6 (b) h = 1/2, k = 2 (c) h = 6, k = 2 (d) h = 2, k = 1/2
31. If the lines
is
−1 (b) 2 tan ( x ) + C −1 (d) tan (2 x ) + C
26. The value of the integral
∫
dx
2
x + 4x + 13
is
(a)
x +2 1 + C (b) log(x2 + 4x + 13) + C tan −1 3 3
(c)
1 x +5 log +C 6 x −1
(d)
2
x +2
(x + 4x + 13)2
+C
32. The distance between the plane whose equation is ^ ^ ^ r ⋅ (2 i + j − 3 k) = 5 and the line whose equation is ^ ^ ^ ^ r = i + l (2 i + 5 j + 3 k) , is 3 5 (a) unit (b) units 14 14 (c) 5 units (d) none of these 33. The minimum area of the triangle formed by a x2 y2 tangent to the ellipse + = 1 and the coordinate a 2 b2 axes, is (a) ab
1 + sin x 27. The value of ∫ e x dx = 1 + cos x x +C 2 x x (c) e tan + C 2
x (b) e x sec + C 2
(a) e x sec2
(d) ex tanx + C
n
28. In = ∫ ( log x ) dx, then the value of (In + nIn−1) is (a) (logx)n–1 + C (b) n(logx)n + C n (c) (xlogx) + C (d) x(logx)n + C 29.
∫
e x dx
(e x + 2)(e x + 1) x
is equal to
e +1 +C (a) log x e +2 (c) 34
ex + 1
ex + 2
+C
e +2 +C (b) log x e +1 (d)
(c)
ex + 2 ex + 1
+C
mathematics today | February ‘17
(b)
(a + b)2 2
(d)
a 2 + b2 2
a2 + ab + b2 3
34. The four common tangents to the ellipses x2 y2 x2 y2 + = 1 form + = 1 and 4 9 9 4 (a) a rectangle of area 13 unit2 (b) a parallelogram which is neither a square nor a rectangle (c) a rhombus (d) none of these 35.
x
(c) coseca (d) seca
CATEGORY-II
dx is equal to
(a)
25. The value of
the value of f (a) is (a) sina (b) cosa
tan −1 x
x−a 2 + C , then x+a cos 2 cos
x2 − 1
dx = x 3 ⋅ 2x 4 − 2x 2 + 1 1 4 2 (a) 2 ⋅ 2x − 2x + 1 + C x 1 (b) 3 ⋅ 2x 4 − 2x 2 + 1 + C x
∫
1 . 2x 4 − 2x 2 + 1 + C x 1 . 2x 4 − 2x 2 + 1 + C (d) 2 2x (c)
(c)
36. Let A be a vector parallel to the line of intersection
of the planes P1 and P2 through the origin. The plane ^
^
^
^
P1 is parallel to the vectors 2 j + 3 k and 4 j − 3 k ^
^
while the plane P2 is parallel to the vectors j − k and ^ ^ ^ ^ ^ 3 i + 3 j . The angle between A and 2 i + j − 2 k is (a) p/2 (b) p/4 (c) p/6 (d) 3p/4 37. f (x) is a polynomial of third degree which has a local maxima at x = –1. If f (1) = –1, f (2) = 18 and f ′(x) has a local minimum at x = 0 then (a) f (0) = 5 (b) f (x) has local minimum at x = 1 (c) f (x) is increasing in [1, 2 5] (d) The distance between (–1, 2) and (a, f (a)) is 2 5, where a is a point of local minimum. 38. A tangent to the curve y = f (x) at P(x, y) cuts the x-axis and y-axis at A and B, respectively, such that BP : AP = 3 : 1. If f (1) = 1 then dy (a) the equation of the curve is x + 3 y = 0 dx 1 (b) the curve passes through 2, 8 dy (c) the equation of the curve is x − 3 y = 0 dx (d) the normal at (1, 1) is x + 3y = 4
∫
11 cos x − 16 sin x dx 2 cos x + 5 sin x = −lx + m log l cos x + d sin x + C, then
(a) l = 2 (c) d = l + m 40. If f(x) = ∫ (a) m = 2
(b) m = 3 (d) d = m – l
−1 e 2 tan x (1 + x)2
1+ x
2
mp
3 + 1 e 3 (d) f′( 3) = m SOLUTIONS
CATEGORY-III In this section more than 1 answer can be correct. Candidates will have to mark all the correct answers, for which 2 marks will be awarded. If, candidates marks one correct and one incorrect answer then no marks will be awarded. But if, candidate makes only correct, without making any incorrect, formula below will be used to allot marks. 2×(no. of correct response/total no. of correct options)
39. If
p m f′(1) = me
−1 dx =xe m tan x + C then
(b) f′(0) = 1
1. (c): Let the line makes an acute angle q with the y-axis. Therefore, the direction cosines are p p l = cos , m = cos and n = cos q 4 3 But l2 + m2 + n2 = 1 p p ⇒ cos2 + cos2 + cos2 q = 1 4 3 3 1 1 1 ⇒ cos2 q = 1 − + = 1 − = 2 4 4 4 1 ⇒ cos q = [ q is acute] 2 So, q = p/3 2. (b) : The given equations of the lines are x +1 y − 2 z + 4 x − 3 2y + 3 z − 2 = = and = = 3 1 1 5 2 −2 Let the angle between them be q. Then 5 3 ⋅1 + (−2) ⋅ + 1 ⋅ 2 2 =0 cos q = 2
5 32 + (−2)2 + 12 12 + + 22 2 Hence, q = p/2 3. (b) : The general point on the straight line x +5 y +3 z −6 ... (1) = = = l (say) 1 4 −9 is (l – 5, 4l – 3, –9l + 6) Let the foot of perpendicular L ≡ (l – 5, 4l – 3, –9l + 6) \ The direction ratios of AL are l – 7, 4l – 7, –9l + 7 Since AL is perpendicular to the line (1), therefore 1(l – 7) + 4(4l – 7) – 9(–9l + 7) = 0 ⇒ l = 1 Thus coordinates of foot of perpendicular are (–4, 1, –3) 4. (b) : The equation of the line joining the points (2, –1, 4) and (1, 1, –2) is x − 2 y +1 z − 4 x − 2 y +1 z − 4 = = = i.e., = 2 − 1 −1 − 1 4 + 2 1 −2 6 x − 2 y +1 z − 4 i.e., = = −1 2 −6 5. (b) : Let the vertices of the triangle PQR be P(x, 0, 0), Q(0, y, 0) and R(0, 0, z) x y \ a = ⇒ x = 3a, b = ⇒ y = 3b 3 3 z and c = ⇒ z = 3c. 3 mathematics today | February ‘17
35
Thus, the equation of the plane is x y z x y z + + =1⇒ + + = 3 3a 3b 3c a b c
=
6. (c) : The equation of the plane passing through the point (0, 1, 0) is ax + b(y – 1) + cz = 0 ... (1) Since (1) is also passing through the point (3, 4, 1) \ 3a + 3b + c = 0 ... (2) Again the plane (1) is parallel to the line x +3 y −3 z −2 = = 2 7 5 \ 2a + 7b + 5c = 0 ... (3) Now solving (2) and (3), we get a b c a b c = = ⇒ = = = l (say) 15 − 7 2 − 15 21 − 6 8 −13 15 \ a = 8l, b = –13l, c = 15l Now (1) becomes 8lx –13ly + 15lz = –13l ⇒ 8x – 13y + 15z + 13 = 0 is the required equation of plane 7. (d) : The vector equation of the line passing through the points (2, –3, 1) and (–4, 3, 6) ^ ^ ^ ^ ^ ^ i.e. a = 2 i − 3 j + k and b = −4 i + 3 j + 6 k is r = a + l(b − a) ^ ^ ^ ^ ^ ^ ⇒ r = 2 i − 3 j + k + l(−6 i + 6 j + 5 k) 8. (b) : Here the given equation of the line is ^ ^ ^ ^ ^ ^ ... (1) r = 2 i − j + 3 k + l(2 i + j + 2 k) ^ ^ ^ And equation of the plane is r ⋅ (3 i − 2 j + p k) = 4 ... (2) Since the line (1) is parallel to the plane (2) \ 2(3) + 1(–2) + 2 × p = 0 ⇒ p = –2 9. (d) : The position vector of the point having ^
^
^
coordinates (1, –1, 2) is i − j + 2 k and normal to the ^
^
^
plane is 2 i + 3 j + 2 k Thus, the equation of the plane is ^ ^ ^ ^ ^ ^ ^ ^ ^ r ⋅ (2 i + 3 j + 2 k ) = (i − j + 2 k) ⋅ (2 i + 3 j + 2 k) ^ ^ ^ i.e. r ⋅ (2 i + 3 j + 2 k ) = 3 10. (d) : The position vector of the point having ^
^
^
coordinates (–3, 2, 4) is −3 i + 2 j + 4 k ^ ^ ^ ^ ^ ^ Let a = −3 i + 2 j + 4 k and n = 2 i − 3 j + 6 k Then the distance of the point (–3, 2, 4) from the plane ^ ^ ^ r ⋅ (2 i − 3 j + 6 k) + 7 = 0 is ^ ^ ^ ^ ^ ^ a ⋅ n + 7 |(−3 i + 2 j + 4 k) ⋅ (2 i − 3 j + 6 k) + 7 | = n 22 + (−3)2 + 62 36
mathematics today | February ‘17
−6 − 6 + 24 + 7 4 + 9 + 36
=
19 units 7
11. (d) : Here f (x) = x 1 1 ⇒ f ′(x) = ⇒ f ′(c) = 2 x 2 c Now by mean value theorem, we have f (b) − f (a) f ′(c) = b−a ⇒
1 2 c
=
f (9) − f (4) 3 − 2 1 = = 9−4 5 5
⇒ 2 c = 5 ⇒ 4c = 25 ⇒ c =
25 = 6.25 (4 < 6.25 < 9) 4
12. (b) : Here, the given function is f (x) = 4x3 + ax2 + bx – 1 ⇒ f ′(x) = 12x2 + 2ax + b 1 Given that f ′ = 0 ⇒ 3 + a + b = 0 2 ⇒ a + b = –3 ... (1) Since f (x) satisfies all the conditions of Rolle’s Theorem 1 a b 1 \ f (1) = f − ⇒ 4 + a + b − 1 = − + − − 1 4 16 16 4 a b 65 15 5 65 +b+ = − ⇒ a+ b=− 16 4 4 16 4 16 ⇒ 3a + 4b = –13 Solving (1) and (2) we get a = 1 and b = –4 ⇒ a−
... (2)
13. (b) : Given that f (x) is continuous in (–∞, ∞) and f ′(x) exists in (–∞, ∞) f (6) − f (3) \ f ′(c) = where c ∈(3, 6) 6−3 ⇒
f (6) − f (3) ≥ 6 ⇒ f (6) − (−6) ≥ 18 ⇒ f (6) ≥ 12 3
14. (b) : Let the volume of the spherical balloon be V 4 dV dr \ V = pr 3 ⇒ = 4pr 2 3 dt dt dS dr Again S = 4pr 2 ⇒ = 8pr dt dt dS 2 dS 2 dV 2 Now dt = ⇒ = ⋅ = × 35 = 10 dV r dt r dt 7 dt Thus, the rate of increase of the surface area is 10 cm2/min
15. (b) : h(x) = f (x) − ( f (x))2 + ( f (x))3 h′(x) = f ′(x) − 2 f (x) f ′(x) + 3( f (x))2 f ′(x)
\
2 1 1 1 1 = 3 f ′(x) ( f (x))2 − 2 f (x) ⋅ + + − 3 3 3 9 2 1 2 = 3 f ′(x) f (x) − + 3 3 It is given that f ′(x) > 0 " x Thus h′(x) > 0 for all real values of x. Therefore h(x) is an increasing function for all real values of x. 16. (c) : We have, C = k tanq dC dC ⇒ = k sec2 q ⇒ dq = k sec2 qdq dq dq dC ⇒ dC = k sec2 qdq dC = dq dq
k sec2 qdq sec2 q dC × 100 = × 100 = dq × 100 tan q C k tan q p dq p =2× × 100 = × 0.7 = 1.57 × 0.7 = 1.099 4 q 2 \ Percentage error in C is 1.1 17. (c) : Let the point on the parabola 2y = x2 be x2 x, 2 Now,
x2 Now the distance between the point x, and 2 (0, 3) be L (say) x2 So, L = x + − 3 2 2
2
2
x2 1 dL2 = 2x + 2 − 3 × × 2x dx 2 2 x2 x2 = 2x 1 + − 3 = 2x − 2 2 2 d 2(L2 ) dx 2
= 3x 2 − 4 2
dL = 0 gives x3 – 4x = 0 dx ⇒ x(x2 – 4) = 0 ⇒ x = 0 or x = ±2 d 2(L2 ) \ =8>0 dx 2 x =±2 Now
So L2 is minimum when x = ±2 Therefore the required point is (±2, 2). 18. (d) : The given equation of the curve is e y = 1 + x2 ... (1) Differentiating both sides w.r.t. x, we get dy dy 2x 2x ey = 2x ⇒ = = = m (using (1)) dx dx e y 1 + x 2 ⇒ |m| =
2|x |
... (2)
1+ | x |2
But (1 – |x|)2 ≥ 0 ⇒ 1 + |x|2 – 2|x| ≥ 0 ⇒ 1 + |x|2 ≥ 2|x| Thus, from (2) we get |m| ≤ 1 19. (c) : Given that x = a(cosq + qsinq) dx ⇒ = a(− sin q + sin q + q cos q) = aq cos q dq and y = a(sinq – qcosq) dy ⇒ = a(cos q − cos q + q sin q) = aq sin q dq dx cos q \ = dy sin q Therefore, the equation of the normal to the given curve at q is cos q y − a(sin q − q cos q) = − {x − a(cos q + q sin q)} sin q ⇒ xcosq + ysinq = a(cos2q + qsinqcosq + sin2q – qsinq cosq) ⇒ xcosq + ysinq = a ⇒ xcosq + ysinq – a = 0 ... (1) Now the distance of the normal from the origin is |0 − a | = a(constant) cos2 q + sin2 q Therefore the normal (1) is at a constant distance from the origin. 20. (d) : Given that f (x) = (a + 2)x3 – 3ax2 + 9ax – 1 \ f ′(x) = 3(a + 2)x2 – 6ax + 9a 2 a a2 a = 3(a + 2) x 2 − 2 ⋅ x ⋅ + − + 9a a + 2 a + 2 (a + 2)2 2
a 3a2 = 3(a + 2) x − + 9 a − a+2 (a + 2) 2
a 6a(a + 3) = 3(a + 2) x − + a+2 a+2 Given that the function decreases monotonically throughout for all real values of x. mathematics today | February ‘17
37
2
a 6a(a + 3) <0 \ f ′(x) < 0 ⇒ 3(a + 2) x − + a+2 a+2 Clearly the above condition is satisfied for all real values of x when a ≤ –3 sin x dx 21. (a) : Let I = ∫ sin(x − a) Put x – a = z ⇒ dx = dz sin(z + a) sin z cos a + cos z sin a \ I=∫ dz = ∫ dz sin z sin z = cos a∫ dz + sin a∫ cot zdz = zcosa + sina⋅log(sinz) + c = (x – a)cosa + sina⋅log|sin(x – a)| + c = xcosa + sina⋅log|sin(x – a)| + k (where k = c – acosa) \ (A, B) = (cosa, sina) 22. (b) : Let I = ∫ e x dx Put x = z ⇒
1
2 x
dx = dz ⇒ dx = 2zdz
n 28. (d) : We have, In = ∫ (log x) dx
1 = x(log x)n − ∫ n(log x)n−1 ⋅ x dx x = x(log x)n − n∫ (log x)n−1dx + C
In = x(log x)n − nIn−1 + C ⇒ In + nIn−1 = x(log x)n + C 29. (a) : Let I =
23. (c) :
∫
−1 e tan x
1+ x
2
dx put tan −1x = z ⇒
1 1+ x
2
24. (a) :
∫ f (x) dx = f (x) 2
∫ { f (x)}
Now
\ f (x) = f ′(x)
dx =∫ f (x) f (x)dx =∫ f (x) f ′(x) dx 1 { f (x)}2 + C 2 dx
= ∫ f (x) d { f (x)} = 25. (c) : Let I =
∫2
x (x + 1)
Put x = z ⇒ dx = 2zdz 2z dz \ I =∫ dz = ∫ = tan −1 z + C = tan −1 x + C 2 2z(z + 1) 1+ z 2 dx =∫ 2 x + 4x + 13 x + 4x + 4 + 9 dx 1 x +2 +C =∫ = tan −1 2 2 3 3 (x + 2) + 3
26. (a) : We have,
∫
2
dx
x 1 + sin x dx 27. (c) : We have ∫ e 1 + cos x
x x 2 sin cos 1 2 2 dx = ∫e + x x 2 2 cos2 2 cos 2 2 x
mathematics today | February ‘17
=∫
dx = dz
−1 I = ∫ e zdz = e z + C = e tan x + C
\
∫
e x dx
(e x + 2)(e x + 1)
Put e x = z ⇒ e x dx = dz dz (z + 2) − (z + 1) \ I=∫ =∫ dz (z + 2)(z + 1) (z + 2)(z + 1)
I = ∫ 2ze zdz = 2(ze z − e z ) + c = 2e x ( x − 1) + c
\
38
x x 1 = ∫ e x sec2 + tan dx 2 2 2 x = e x tan + C ∫ e x { f (x) + f ′(x)} dx = e x f (x) + C 2
dz dz −∫ = log | z + 1 | − log | z + 2 | + C z +1 z +2
= log
ex + 1
z +1 + C = log z +2
30. (c) : Let I =
ex + 2
+C
dx
∫ cos x + cos a
2dz x = z ⇒ dx = 2 1 + z2 2dz 2dz \ I=∫ =∫ 2 (1 + z )cos a + 1 − z 2 1− z2 2 (z + 1) cos a + 1 + z 2 Put tan
= 2∫
dz (1 + cos a) − (1 − cos a)z
2
=
dz 2 ∫ (1 − cos a) a 2 2 cot 2 − z
a 2 +C log = ⋅ a a 2a sin 2 cot 1 − z tan 2 2 2 x a 1 + tan tan 2 2 +C = coseca ⋅ log x a 1 − tan tan 2 2 x − a cos 2 = coseca log +C x+a cos 2 \ f(a) = cosec a 1
1
1 + z tan
31. (d) : The coordinates of any point on the line x y z = = = l (say) 1 2 3 are given by x = l, y = 2l and z = 3l Thus the coordinates of a general point on first line are (l, 2l, 3l) The coordinates of any point on the line x −1 y − 2 z − 3 = = = m (say) 3 −1 4 are given by x = 3m + 1, y = –m + 2 and z = 4m + 3 Thus the coordinates of a general point on second line are (3m + 1, –m + 2, 4m + 3) If the lines intersect, then they have a common point. So, for some values of l and m, we must have l = 3m + 1, 2l = –m + 2 and 3l = 4m + 3 or l – 3m = 1, 2l + m = 2, 3l – 4m = 3 Solving two of these equations, we get l = 1 and m = 0. So, the point of intersection of the first two lines is (1, 2, 3). It is also lies on the third line. 1+ k 2 −1 3 − 2 So, = = 3 2 h 1+ k 1 1 1 1 ⇒ = ⇒ k= and = ⇒h=2 3 2 2 2 h 32. (a) : The vector equations of the plane and the straight line are respectively ^ ^ ^ ... (1) r ⋅ (2 i + j − 3 k) = 5 ^ ^ ^ ^ ... (2) and r = i + l(2 i + 5 j + 3 k) Here, b ⋅ n = (2 ^i + 5 ^j + 3 k^) ⋅ (2 ^i + ^j − 3 k^) = 0 So, b is perpendicular to n . Hence, the given line is parallel to the given plane. Let the required distance be d ^
d=
^
^
^
^
2
2
2
^
(i + 0 j + 0 k) ⋅ (2 i + j − 3 k) − 5 2 +1 + 3
=
3 unit 14
33. (a) : The equation of tangent to the ellipse x2
y2 x cos q y sin q + = 1 at (acosq, bsinq) is + = 1. 2 2 a b a b The intercepts of the tangent on the axes are asecq and bcosecq. 1 \ The area of the triangle = × a sec q × bcosecq 2 ab = ≥ ab sin2q
34. (d)
1
x2 − 1
35. (d) : Let I = ∫
x 3 2x 4 − 2x 2 + 1
dx = ∫
1 − 5 x x dx 2 1 2− 2 + 4 x x 3
2 1 1 1 Put 2 − 2 + 4 = z 2 ⇒ 4 3 − 5 dx = 2zdz x x x x 1 zdz 1 1 1 2 1 = ∫ dz = z + C = 2− 2 + 4 +C ∫ 2 z 2 2 2 x x 1 = 2 2x 4 − 2x 2 + 1 + C 2x 36. (a, d) : A vector along the line of intersection of two planes one of them being parallel to the vectors a and b and theother plane being parallel to the vectors c and d , is (a × b ) × (c × d ). ^ ^ ^ ^ ^ ^ ^ ^ \ A = {(2 j + 3 k) × (4 j − 3 k)} × {(j − k) × (3 i + 3 j)} \
I=
^
^
^
^
^
^
= −18 i × (3 i − 3 j − 3 k) = 54(k − j) Let q be an angle between them, then ^
cos q =
^
54(k − j )
^
^
^
.(2 i + j − 2 k) = −
1
54 2 2 p 3p So, angle , 4 4 37. (b, c) : Let f (x) = ax3 + bx2 + cx + d f(1) = –1 ⇒ a + b + c + d = –1 f(2) = 18 ⇒ 8a + 4b + 2c + d = 18 f ′′(0) = 0 ⇒ b = 0 and f ′(1) = 0 ⇒ 3a + 2b + c = 0 19 57 17 Solving, a = , b = 0, c = − , d = 4 4 2 1 \ f (x) = (19x 3 − 57x + 34) 4
=
1 2
f ′(x) = 0 ⇒ 57x 2 − 57 = 0 ⇒ x = 1, −1. f ′′(x) = 57 × 2x, f ′′(−1) < 0 and f ′′(1) > 0. \ f (x) has a local minimum at x = 1. f (x) is increasing ⇒ f ′(x) ≥ 0 ⇒ x2 – 1 ≥ 0 ⇒ x ≤ –1 or x ≥ 1 \ In [1, 2 5], f (x) is increasing. The distance of (–1, 2) from (1, –1) is not 2 5. 38. (a, b) : The equation of tangent to the curve at P(x, y) is dy dy dy Y − y = ( X − x) ⇒ X −Y = x − y dx dx dx mathematics today | February ‘17
39
X Y =1 + dy dy x − y y−x dx dx dy dx dx dy \ A x − y , 0 and B 0, y − x dy dx BP 3 Now, = . AP 1 3 dx 1 dy So, P = x − y , y − x = (x, y) 4 4 dy dx ⇒
1 dy dy =0 y − x = y ⇒ 3 y + x 4 dx dx dx dy dx dy ⇒ 3 + = 0 ⇒ 3∫ + ∫ = 0 x y x y ⇒ 3logx + logy = logc ⇒ x3y = c Now, f (1) = 1 ⇒ (1, 1) is on it. \ c = 1. Hence, x 3 y = 1 1 Also 2, lies on it. 8 The normal to x3y = 1 at (1, 1) has the equation −1 y −1 = (x − 1) dy dx (11 ,) ⇒
−1 (x − 1) = x − 3 y + 2 = 0 −3 11 cos x − 16 sin x dx 39. (a, b, c) : Let I = ∫ 2 cos x + 5 sin x ⇒
40
y −1 =
mathematics today | February ‘17
Let, 11cosx – 16sinx = l(5cosx – 2sinx) + m(2cosx + 5sinx) Now equating the coefficient of cosx and sinx, we get 5l + 2m = 11 .... (1) and –2l + 5m = –16 ⇒ 2l – 5m = 16 .... (2) Solving (1) and (2), we get l = 3 and m = –2 3(5 cos x − 2 sin x) − 2(2 cos x + 5 sin x) \ I=∫ dx 2 cos x + 5 sin x (5 cos x − 2 sin x) = 3∫ dx − 2∫ dx 2 cos x + 5 sin x = 3log|2cosx + 5sinx| – 2x + C = –2x + 3log|2cosx + 5sinx| + C \ l = 2, m = 3 and d = 5 40. (a, b, c, d) : We have, f(x) = ∫
−1 e 2 tan x (1 + x)2
1 + x2
dx
−1 −1 e2 tan x (1 + x 2 + 2x) 2xe 2 tan x 2 tan −1 x =∫ dx = ∫ e dx + ∫ dx 1 + x2 1 + x2
= xe
2 tan −1 x −1
− ∫ 2e
tan −1 x
×
= xe 2 tan x + C \ m=2 −1 Now f′(x) = xe 2 tan x ×
1
1 + x2 2
1+ x
2
xdx + ∫
−1 2xe 2 tan x
1 + x2
dx + C
−1 + e 2 tan x
\ f′(0) = 1, f′(1) = 2e p/2 = me p/m 2p
mp
3 3 and f′( 3) = + 1 e 3 = + 1 e 3 2 m
10 BEST
PROBLE
MS
Math archives, as the title itself suggests, is a collection of various challenging problems related to the topics of IIT-Jee Syllabus. This section is basically aimed at providing an extra insight and knowledge to the candidates preparing for IIT-Jee. In every issue of MT, challenging problems are offered with detailed solution. The readers’ comments and suggestions regarding the problems and solutions offered are always welcome.
−x 3 + x 2 sin6 – xsin4· sin8 – 5sin–1 3 (a2 – 8a + 17) then f ′(sin8) is (a) < 0 (b) > 0 (c) 12 (d) –12 1.
If f ( x ) =
{ }
1 If 2 < x < 3, and if x 2 = , where {⋅} represents x 1 fractional part function, then the value of x − is x (a) –1 (b) 0 (c) 1 (d) 2 2.
3. If f(x) = a + bcos–1x , (b > 0) has same domain and range then the value of a + b is 2 2 p p (a) 1 − (b) 1 − (c) −1 − 1 (d) p p 2 2 4. If x2 + pxy + 4y2 + x – 2y + q = 0 represent the pair of parallel lines then p + q may be (a) –5 (b) 1 (c) 2 (d) 5 5. Consider the circle x2 + y2 = 8. If from point P(2, 2), two chords PA and PB drawn of length 1 unit then the equation of AB is (a) 4x + 4y + 15 = 0 (b) 4x + 4y – 15 =0 (c) x + y + 15 = 0 (d) x + y – 15 = 0 6.
x3 − 1 is x −1 3 (b) , ∞ 4
Range of the function f ( x ) =
(a) 0, ∞ ) 3 (c) , ∞ − {3} 4
(d) R
7. Let A and B be two sets containing 3 and 4 elements respectively. The number of subsets of A × B having 2 or more elements is (a) 4083 (b) 4096 (c) 4046 (d) 4076 ^ ^ ^ ^ ^ 8. If the vectors AB = 2 i − 2 k and AC = 4 i + j+ 2 k are sides of a DABC, then the length of the median through A is 33 4
(a) 9.
(b)
33 2
(c)
37 4
(d)
37 2
5
∫1 (x − 1) (x − 2) (x − 3) (x − 4) (x − 5) dx =
(a) 3
(b) –3
(c) 0
(d) 1
10. The number of all possible positive triplets(x, y, z) such that (2y – x)sinq + (x – z)cos2q – 2xcos2q = 0 for all q is (a) infinite (b) one (c) zero (d) greater than 10 but less than 100 soLUtioNs 3
−x + x 2 sin6 – xsin4 · sin8 – 5sin–1 3 (a2 – 8a + 17) 2 f ′(x) = – x + 2x · sin6 – sin4 · sin8 f ′(sin8) = –(sin8)2 + 2sin8 · sin6 – sin4 · sin8 = –sin28 + 2sin8 · sin6 – sin4 · sin8 = –sin8[sin8 + sin4 – 2sin6] = –sin8[2sin6 · cos2 – 2sin6] = –2sin6 · sin8[cos2 – 1] = 2sin8 · sin6(1 – cos2) > 0 1. (b) : f ( x ) =
By : Prof. Shyam Bhushan, Director, Narayana IIT Academy, Jamshedpur. Mob. : 09334870021
mathematics today | February ‘17
41
A
1 2. (c) : Since, 2 < x < 3 and {x 2} = x 1 ⇒ x 2 − 2 = ⇒ x3 – 2x – 1 = 0 x ⇒ x3 + x2 – x2 – x – x – 1 = 0 ⇒ x2(x + 1) – x(x + 1) – 1(x + 1) = 0 ⇒ (x + 1)(x2 – x – 1) = 0 x ≠ –1, x2 – x – 1 = 0 ⇒ x= ⇒
8. (c) :
B
1± 5 1− 5 1+ 5 ⇒ x= But x ≠ 2 2 2
1 2 5 −1 5 −1 = × = x 2 5 +1 5 −1
1 5 +1 5 −1 5 +1− 5 +1 = − = =1 x 2 2 2 3. (d) : Domain : x ∈ [–1, 1] Q 0 ≤ cos–1x ≤ p ⇒ 0 ≤ bcos–1x ≤ bp ⇒ a ≤ a + bcos–1x ≤ a + bp From (i) and (ii), a = –1 2 ⇒ a + bp = 1 ⇒ bp = 2 ⇒ b = p \ x−
...(i) ...(ii)
4. (a) : For pair of lines to be parallel, h2 – ab = 0 ⇒ p = –4 Now (x – 2y)2 + (x – 2y) + q = 0, lines to be parallel root is real and distinct 1 ⇒ 1 − 4q > 0 ⇒ q < 4 ⇒ p+q=–5 5. (b) : Required equation of AB is S – S′ = 0 ((x – 2)2 + (y – 2)2 – 1) – (x2 + y2 – 8) = 0 ⇒ 4x + 4y – 15 = 0
(
)
2 x 3 − 1 ( x − 1) x + x + 1 = x −1 ( x − 1) 2 ⇒ f(x) = x + x + 1, x ≠ 1
6. (b) : f ( x ) =
f(x) = x2 + x + 1
3
3
3/4
x=1
7. (a) : n (A × B) = 12 No. of subset each contain 2 or more elements is = (12C2 + 12C3 + ... + 12C12)
= (12C0 + 12C1 + 12C2 + 12C3 + ... + 12C12) – (12C0 + 12C1) = 212 – (1 + 12) = 4096 – 13 = 4083 42
mathematics today | February ‘17
C
M Since, AB + BC + CA = 0 AC − AB ⇒ BC = AC − AB ⇒ BM = 2 Again, AB + BM + MA = 0 AC − AB \ AM = AB + BM = AB + 2 ^ ^ ^ AB + AC 6 i + j ^ j = = = 3 i+ 2 2 2 1 37 AM = 9 + = 4 4 9. (c) : Put x – 3 = t ⇒ dx = dt ⇒
\
5
∫ (x − 1)(x − 2)(x − 3)(x − 4)(x − 5)dx 1
=
2
∫ (t + 2)(t + 1)t ⋅ (t − 1) ⋅ (t − 2) dt
−2
= 0 (since function is odd). 10. (c) : 2ysinq – xsinq + (x – z)(1 – sin2q) – 2x (1 – 2sin2q) = 0 ⇒ 2ysinq – xsinq + x – xsin2q – z + zsin2q – 2x + 4xsin2q = 0 2 ⇒ (3x + z)sin q + (2y –x)sinq – x – z = 0 ⇒ (3x + z)sin2q + (2y – x)sinq – (x + z) = 0 This equation is zero for all values of q. Thus, 3x + z = 0, 2y – x = 0, x + z = 0 −z x So, x = , y = , x = −z 3 2 −z −z −z So the triplet formed is −z, , z and , , z 3 6 2 Thus no positive triplet is formed.
ANSWER KEY
MPP-8 CLASS XI 1.
(a)
2.
(a)
6.
(d)
7.
(b, c) 8.
10. (b) 15. (d) 20. (4)
3.
(a)
4.
(a, b) 9 .
(a)
5.
(d)
(a, b, c, d)
11. (a, b, c) 12. (a, c) 13. (a, b, c)14. (b) 16. (c) 17. (5) 18. (4) 19. (0)
ELLIPSE This article is a collection of shortcut methods, important formulas and MCQs along with their detailed solutions which provides an extra edge to the readers who are preparing for various competitive exams like Jee(Main & advanced) and other PeTs.
Definition anD StanDarD equation of ellipSe An ellipse is the locus of a point in a plane which moves in such a way that the ratio of its distance from a fixed point (called focus) in the same plane to its distance from a fixed straight line (called directrix) is always constant which is less than unity. The constant ratio is generally denoted by e (0 < e < 1) and is known as the eccentricity of the ellipse. The general equation ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 represents an ellipse, if D = abc + 2fgh – af 2 – bg2 – ch2 ≠ 0 and h2 < ab. equation of the ellipSe The equation of the ellipse, whose focus is the point (h, k) and directrix is lx + my + n = 0 and having eccentricity e, is (lh + mk + n)2 (x − h)2 + ( y − k)2 = e 2 ⋅ (l 2 + m2 )
z
Equation of the ellipse whose focus is (ae, 0) and directrix is x = a/e and having eccentricity e is given by x
2
a
2
+
y 2
2 2
a (1 − e )
= 1 or
x
2
a
2
+
y b
2
2
=1
X
Directrix
(0, b)B P(x1, y1)
M
Y
x2
Ellipse y2 b
2
a2
= 1−
(PN )2 b2
=
+
x2 a
+
2
y2 b2 =
= 1 can also be defined as
a2 − x 2 a
a2
⇒
2
A ′N ⋅ AN
y2 b
2
=
(a + x )(a − x ) a2
(PN )2 b2 (BC )2 = = AN ⋅ A ′N a2 ( AC )2
or Y
B p2
M
C A S S A (–ae, 0) (ae, 0) (a, 0) (–a, 0) B(0, –b) x = –a/e
z
or Y
y2
...(1) =1 a 2 b2 The foci S and S′ are (ae, 0) and (–ae, 0). The equation of its directrices are x = a/e and x = –a/e. Let P(x1, y1) be any point on (1), Now, SP = ePM = e (a/e – x1) = a – ex1 and S′P = ePM′ = e(a/e + x1) = a + ex1 \ SP + S′P = (a – ex1) + (a + ex1) = 2a = AA′ = Length of major axis The sum of focal distances of any point on the ellipse is equal to the length of major axis. note : (a) Two ellipses are said to be similar, if they have the same value of eccentricity. (b) Distance of every focus from the extremity of minor axis is equal to a, as b2 + a2e2 = a2.
⇒
where b2 = a2 · (1 – e2) Directrix
z
x2
The ellipse is
x = a/e
X
X
A (–a, 0)
P p1 N
C
A (a, 0)
X
B Y
Sanjay Singh Mathematics Classes, Chandigarh, Ph : 9888228231, 9216338231 mathematics today | February‘17
43
equation of ellipse referred to two perpendicular lines
z
(a) Ellipse
x2
a2
+
y2
b2
= 1 can also be defined as
Let Q be a point on auxiliary circle x2 + y2 = a2 such that QP produced is perpendicular to the x-axis. Then P and Q are the corresponding points on the ellipse and the auxiliary circle respectively. Y
(L.P. from any point on ellipseto the major axis)2
Q
2
P
(Length of semi minor axis) +
(L.P. from any point on ellipseto the minor axis)2 2
(Length of semi major axis)
X
=1
x2 + y2 = a2
(where L.P. means length of perpendicular) ⇒
p12
p2 + 2 =1 b2 a2
Y
(b) If L1 : a1x + b1y + c1 = 0 and L2 : b1x – a1y + c2 = 0, then
the equation
a x + b1 y + c1 1 a 2 + b 2 1 1
2
b x − a1 y + c2 1 a 2 + b 2 1 1
2
+ =1 a2 b2 represents an ellipse in the plane such that (i) The cent re of t he el l ip s e is t he p oi nt of intersection of the lines L1 = 0 and L2 = 0. (ii) The major axis lies along L2 = 0 and the minor axis along L1 = 0, if a > b. If a < b, then the major axis is along L1 = 0 and minor axis is along L2 = 0.
p2
P(x, y) p1
b
L1
a
L2
(iii) If a > b, then the lengths of the major and minor axes are 2a and 2b respectively and if a < b, then the lengths of major and minor axes are 2b and 2a respectively. parametric equation of the ellipse The circle described on the major axis of an ellipse as diameter is called the auxiliary circle of the ellipse. z
Let the equation of ellipse be \
x2
+ = 1 (a > b). a 2 b2 Equation of its auxiliary circle is x 2 + y 2 = a 2
mathematics today | February‘17
NS
X A (a, 0)
x2 + y 2 = 1 a2 b2
Let ∠QCA = q (0 ≤ q < 2p) i.e., the eccentric angle of P on an ellipse is the angle which the radius (or radius vector) through the corresponding point on the auxiliary circle makes with the major axis. \ Q ≡ (a cosq, a sinq) and P ≡ (a cosq , b sinq) The equation x = a cosq and y = b sinq taken together 2 2 are called parametric equation of ellipse x + y = 1 a 2 b2 where q is real parameter and P ≡ (a cosq , b sinq ) is any point on the ellipse also known as P(q) or ‘q’ point on the ellipse. note: (i) Eccentric angles of the extremities of latus
x2 y2 rectum of the ellipse + = 1 are given by 2 2 a b b tan −1 ± . ae x2 y2 (ii) Area of the ellipse + = 1 is pab sq. units. a 2 b2 position of points w.r.t. an ellipse x2
y2
= 1 and P ≡ (h, k) be any a 2 b2 point then P will lie outside, on or inside the ellipse Let the ellipse be x2
a
2
+
+
y2
2 2 = 1 according as h + k >, =, < 1 b a 2 b2 2
Y
P(outside)
y2
( AA′ is diameter of the circle) 44
C
A (–a, 0)
X
A
B P(inside)
C
B P(on) Y
A
X
important points The equation of the ellipse whose axes are parallel to the coordinate axes and whose centre is at the origin, is x2
a2
+
y2
b2
= 1 with the following properties :
S. no.
x2
ellipse
(i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix)
Basic fundamentals Coordinates of the centre Coordinates of the vertices Coordinates of the foci Length of the major axis Length of the minor axis Equation of the major axis Equation of the minor axis Equations of the directrices Eccentricity
(x) (xi) (xii) (xiii)
Ends of the latus rectum Length of the latus rectum Parametric coordinates Auxilliary circle
a2
x
2
a
2
+
y b
2
2
=1
If the chord joining two points whose eccentric angles are a and b, cut the major axis of the ellipse a2 tan
z
+
y2 b2
= 1 (a > b) at the point (ae, 0), then
a b e −1 tan = . z z e +1
If a and b are the eccentric angles of extremities of a focal chord of the ellipse
x
2
a
2
+
y
2
b2
a2
+
= 1 (a > b) , then
a −b cos 2 a b e −1 e +1 ⇒ tan ⋅ tan = or . e= 2 2 e +1 e −1 a +b cos 2
z
y2 b2
= 1, a < b
(0, 0) (0, ±b) (0, ±be) 2b 2a x=0 y=0 y = ± b/e
b2 e = 1− a2
x a −b a +b y a +b cos + sin = cos . 2 b 2 2 a
x2
b2
x2
= 1, a > b
a2 e = 1− a2
(± ae, ± b2/a) 2b2/a (a cosq, b sinq) x2 + y2 = a2
eccentric angles are a and b on the ellipse
z
y2
(0, 0) (±a, 0) (±ae, 0) 2a 2b y=0 x=0 x = ±a/e
equation of Chord z Equation of the chord joining the points whose is given by
+
(± a2/b, ± be) 2a2/b (a cosq, b sinq) x2 + y2 = b2
Locus of mid-points of focal chords of an ellipse x2
a
2
+
y2
b
2
= 1 (a > b) is
x2
a
2
+
y2
b
2
=±
ex . a
equation of tangents in Different forms point form : Equation of tangent to the ellipse xx yy x2 y2 + = 1 at the point (x1, y1) is 1 + 1 = 1. 2 2 2 a b2 a b parametric form : Equation of tangent to the x2
a2
the ellipse
+
y2
= 1 at the point (a cosq, b sinq) b2 y is x cos q + sin q = 1. a b Slope form : Equation of tangent of slope m to ellipse
x2
a2
+
y2
b2
= 1 are given by
y = mx ± (a2m2 + b2 ) and the coordinates of the points of contact are a2 m b2 ,± . (a2 m2 + b2 ) (a2 m2 + b2 ) mathematics today | February‘17
45
results related to tangents (i) The equations of the tangents to the ellipse at points P(a cosq1, b sinq1) and Q(a cosq2, b sinq2) y y x x are cos q1 + sin q1 = 1 and cos q2 + sin q2 = 1 a b a b and these two intersect at the point q1 + q2 q1 + q2 a cos 2 b sin 2 . , q1 − q2 q1 − q2 cos 2 cos 2 (ii) The line joining two points on an ellipse, the difference of whose eccentric angle is constant, touches an other ellipse of same eccentricity. (iii) The locus of the point of intersection of tangents to an ellipse at the points whose eccentric angles differs by constant is an ellipse of the same eccentricity. If the eccentric angles differs by a 2 2 right angle then the locus is x + y = 2. a 2 b2 (iv) The locus of the point of intersection of tangents
to the ellipse
x2
y2
= 1 at the points the sum a 2 b2 of whose eccentric angles is constant is a straight line passing through the centre of the ellipse. (v) If the tangent at P on an ellipse meets the directrix in F, then PF will subtend a right angle at the corresponding focus i.e., ∠PSF = p/2. (vi) If SM and S′M′ are perpendiculars from the foci upon the tangent at any point of the ellipse, then SM ·S′M′ = b2 and the point M & M′ lie on the auxiliary circle (where a > b). (vii) The tangents drawn from any point on the director circle of a given ellipse to the ellipse are always at right angle. +
equations of normals in Different forms poi nt for m : E quat ion of nor ma l to t he y2
+ = 1 at t h e p oi nt ( x 1 , y 1 ) i s a 2 b2 b2 y − = a 2 − b2 . y1
e l l ip s e a2 x x1
x2
parametric form : Equation of normal to x2
y2
+ = 1 at (a cosq, b sinq) is a 2 b2 ax sec q – by cosecq = a2 – b2. the ellipse
46
mathematics today | February‘17
Slope form : Equation of normal of slope m to the 2 2 m(a2 − b2 ) ellipse x + y = 1 are given y = mx (a2 + b2m2 ) a 2 b2 a2 b2 m ,± at the points ± a2 + b2m2 a2 + b2m2 results related to normals (i) If the line lx + my + n = 0 be a normal to the ellipse x2
+
y2
= 1 then
a2
+
b2
=
(a2 − b2 )2
l 2 m2 a 2 b2 n2 (ii) Locus of mid-points of normal chords of an ellipse x2
2
y2
x 2 y 2 a 6 b6 = 1 is + + = (a2 − b2 )2 a 2 b2 a 2 b2 x 2 y 2 (iii) Four normals can be drawn from a point to an ellipse. intersection of a Circle and an ellipse (i) The circle x2 + y2 + 2gx + 2fy + c = 0 and ellipse x2 y2 + = 1 can intersect each other at maximum a 2 b2 of four points. (ii) The maximum number of common chord of a circle and an ellipse is six. (iii) The maximum number of common tangent of a circle and an ellipse is four. Director Circle The locus of the point of intersection of the tangents to +
an ellipse
x2
y2
= 1 which are perpendicular to each a 2 b2 other is called director circle and its equation is given by x2 + y2 = a2 + b2. Sub-tangent and Sub-normal +
Length of sub-tangent NT = Length of sub-normal GN = Y
a2 − x1 x1 b2 a
T X
C
Y
2
|x1| = (1 – e2) |x1|
P(x1, y1) G
N
T
X
reflection property of an ellipse If an incoming light ray passes through the focus (S) strike the concave side of the ellipse then it will get reflected towards other focus (S′) and ∠SPS′ = ∠SQS′ Y
Q X
S
A
Y C
B
Tang ent P
N S Normal B
A
X
+
y
2
= 1 (|b| > |a|) , provided a b2 (a) c2 = a2 + b2m2 (b) c2 = b2 + a2m2 2 2 2 2 (c) c = (a + b )m (d) c2 = (b2 – a2)m2 2
2. Tangents are drawn to the ellipse x2 + 2y2 = 4 from any arbitrary point on the line x + y = 4, the corresponding chord of contact will always pass through a fixed point, whose coordinates are 1 1 (a) 1, (b) , 1 2 2 1 1 (c) 1, − (d) , − 1 2 2 3. Equation of the ellipse whose axes are along the coordinate axes and whose length of latus rectum and eccentricity are equal and equal to 1/2 each , is (a) 6x2 + 12y2 = 1 (b) 12x2 + 6y2 = 1 2 2 (c) 3x + 12y = 1 (d) 9x2 + 12y2 = 1 The line y = x - 1 touches the ellipse 3x2 + 4y2 = 12, at 1 1 (a) , − (b) (3, 2) 2 2 (c) (–1, –2) (d) None of these 4.
5. One foot of normal of the ellipse 4x2 + 9y2 = 36, that is parallel to the line 2x + y = 3, is 9 8 (a) , 5 5 9 8 (c) − , − 5 5
(a)
x2
9 8 (b) − , 5 5 (d) None of these
x2
y2
= 1 at a 2 b2 point ‘P’ meets the coordinate axes at points A and B respectively. Locus of mid-point of segment AB is a2 a2
(c)
The tangent and normal at any point of an ellipse bisect the external and internal angles between the focal radii to the point. problems 1. The line y = mx + c, will be a tangent to the ellipse x
Tangent drawn to the ellipse
Light ray
Re ected ray Y
2
6.
7.
x2
+ +
y2 b2 b2 y2
=2
(b)
=4
(d)
a2
x2 x2 a2
+ +
b2
y2 y2 b2
Normals drawn to the ellipse
+
=2 =4 x2
y2
= 1 at a 2 b2 point ‘P’ meets the coordinate axes at points A and B respectively. Locus of mid point of segment AB is (a) 4x2a2 + 4y2b2 = (a2 – b2)2 (b) 4x2b2 + 4y2a2 = (a2 – b2) (c) 16x2a2 + 16y2b2 = (a2 – b2)2 (d) 16x2b2 + 16y2a2 = (a2 – b2) +
8. If the line y = mx + c is a tangent to the ellipse x2 y2 + = 1 then corresponding point of contact is a 2 b2 2 a2 m 2 b2 (a) a m , b (b) − ,− c c c c a2 m b2 (c) ,− c c 9. x
a 2 m b2 (d) − , c c
If the line y = mx + c is a normal to the ellipse 2
a2
+
y2
b2
= 1, then corresponding foot of normal is
a2 c b2 c (a) , m(b2 − a2 ) (b2 − a2 ) a2 c b2 c (b) , m(a2 − b2 ) (a2 − b2 ) a2 m b2 c (c) , c(b2 − a2 ) (b2 − a2 ) a2 m b2 c (d) , c(a2 − b2 ) (a2 − b2 ) 10. Locus of the mid-point of chords of the ellipse x2
+
y2
= 1 that are parallel to the line y = 2x + c, is a 2 b2 (a) 2b2y – a2x = 0 (b) 2a2y – b2 x = 0 2 2 (c) 2b y + a x = 0 (d) 2a2y + b2 x = 0 mathematics today | February‘17
51
11. The locus of the moving point P(x, y) satisfying (x − 1)2 + y 2 + (x + 1)2 + ( y − 12 )2 = a will be an ellipse if (a) a < 4 (b) a > 2
(c) a > 4
(d) a < 2
y2 x2 12. The equation + = 1 will represent an 6 − − 2 a a ellipse if (a) a ∈ (1, 3) (b) a ∈ (1, 6) (c) a ∈ (–∞, 2) ∪ (6, ∞) (d) a ∈ (2, 6) ~ {4} 13. Foci of the ellipse 16x2 + 25y2 = 400 are (a) (± 2, 0) (b) (±3, 0) 48 (d) ± , 0 5 14. Equation of an ellipse whose focus is S ≡ (1, 0), corresponding directrix being the line x + y = 2 and 1 eccentricity being is 2 (a) 3x2 + 3y2 – 2xy + 4x – 4y = 0 (b) 3x2 + 3y2 + 2xy – 4x + 4y = 0 (c) 3x2 + 3y2 + 2xy + 4x – 4y = 0 (d) 3x2 + 3y2 – 2xy – 4x + 4y = 0 15. The equation 3(x + y – 5)2 + 2(x – y + 7)2 = 6 represents an ellipse, whose centre is (a) (–1, 6) (b) (6, –1) (c) (1, –6) (d) (–6, 1) x2 2
+
y2 2
= 1, centered at
a b point ‘O’ and having AB and CD as it’s major and minor axes respectively. If S1 be one of the focus of the ellipse, radius of incircle of triangle OCS1 be 1 unit and OS1 = 6 units, then area of the ellipse is equal to 65 (a) 16 p sq. units (b) p sq. units 4 (c) 65 p sq. units (d) 65 p sq. units 2 x2 y2 17. Normal drawn to the ellipse + = 1 at the 42 32 point P(q) meets the ellipse again at point Q(2q), then value of cos q can be 1 16 (a) − (b) − (c) − 19 (d) − 21 2 23 23 23 18. P is any variable point on the ellipse
x2
y2
+ =1 a 2 b2 having the points S1 and S2 as its foci. Then maximum 52
mathematics today | February‘17
2 y2 19. P is any variable point on the ellipse x + =1 a 2 b2 having the points S1 and S2 as its foci. Then Locus of
incentre of triangle PS1S2 will be (a) a straight line (b) a circle (c) a parabola (d) an ellipse 20. P1 and P2 are the foot of altitudes drawn from the
24 (c) ± , 0 5
16. Consider an ellipse
area of the triangle PS1S2 is equal to (a) b2e sq. units (b) a2e sq. units (c) ab sq. units (d) abe sq. units
foci S1 and S2 respectively of the ellipse
x2
y2
=1 a 2 b2 on one of its variable tangent. Maximum value of (S1P1) (S2P2) is equal to (a) b2
(b) b
(c) a2
+
(d) a
21. The equation of the ellipse centered at (1, 2) having the point (6, 2) as one of its focus and passing through the point (4, 6) is 2 3( y − 2)2 (a) (x − 1) + =1 36 64
(b)
(x − 1)2 ( y − 2)2 + =1 18 32
(c)
(x − 1)2 7( y − 2)2 + =1 72 128
2 2 (d) (x − 1) + ( y − 2) = 1 45 20 22. The angle between the tangents drawn to the ellipse 3x2 + 2y2 = 5, from the point P(1, 2) is equal to
(a) tan −1 (24 5 )
(b) tan −1 (12 5 )
24 (c) tan −1 5
12 (d) tan −1 5
23. The line 5x – 3y = 8 2 is a normal to the ellipse
x2 y2 + = 1. If ‘q’ be the eccentric angle of the foot of 25 9 this normal then ‘q’ is equal to p 3
(a)
p 6
(b)
(c)
p 4
(d) None of these
24. The distance of a point P (lying in the first quadrant) on the ellipse x2 + 3y2 = 6 from the center of the ellipse is 2 units. Eccentric angle of the point ‘P’ is (a) p (b) p 3 6 p (c) (d) None of these 4 x 2 11 y 2 25. The tangent drawn to the ellipse + = 1 at 16 256 the point P(q), touches the circle (x – 1)2 + y2 = 16. ‘q’ is equal to p (a) p (b) 3 6 p (c) (d) None of these 4 26. There are exactly two points on the ellipse x2
30. Length of the focal chord of the ellipse
27. For all admissible values of the parameter ‘a’ the straight line 2ax + y 1 − a2 = 1 will touch an ellipse whose eccentricity is equal to 1 3 2 (a) (b) (c) 1 (d) 3 2 2 3 28. The normal to the ellipse 4x2 + 5y2 = 20 at the point ‘P’ touches the parabola y2 = 4x, the eccentric angle of the point ‘P’ is 1 p 1 p (a) (b) + tan −1 + cos −1 5 2 5 2 1 (c) p − tan −1 5
1 (d) p − cos −1 5
29. The tangent drawn to the ellipse
x2
y2
= 1 , at a 2 b2 the point ‘P’ meets the circle x2 + y2 = a2, (a > b) at the points A and B. The line segment AB subtends an angle p/2 at the center of the ellipse. If ‘e’ be the eccentricity of the ellipse and ‘q’ be the eccentric angle of point ‘P’ then (a) e2(1 + sin2q) = 1 (b) e2(1 + cos2q) = 1 2 2 (c) e (2 + sin q) = 1 (d) e2(2 + cos2q) = 1 +
2b2a
(a)
a2 sin2 q + b2 cos2 q 2a2b
(c)
a2 sin2 q + b2 cos2 q
(b) (d)
+
y2
= 1, a 2 b2 that is inclined at an angle ‘q’ with the x-axis, is equal to 2b2a
a2 cos2 q + b2 sin2 q 2a2b a2 cos2 q + b2 sin2 q
31. The line y = mx + c, will be a normal to the ellipse , x2
a2 (a)
y2
+ = 1 whose distance from the center of the a 2 b2 a2 + 2b2 ellipse are equal and equal to . Eccentricity 2 of this ellipse is equal to 1 1 3 2 (a) (b) (c) (d) 2 3 3 2
x2
+
y2
b2 a2 m2
= 1, if + b2 =
2 (c) a +
b2 m2
=
2 (b2 − a2 )2 (b2 − a2 )2 (b) c + b2 = m2 c2 c2
(b2 − a2 )2 c2
b2 (b2 − a2 )2 (d) c 2 + = m2 a2
32. Locus of the point of intersection of tangents
2 2 drawn to the given ellipse x + y = 1, at points P(q1) a 2 b2 2p and Q(q2), such that q1 – q2 = , is 3
(a) (c)
a2 x2 x2 a2
+ +
b2 y2 y2 b2
=4
(b)
=4
(d)
33. Consider the ellipse
a2 x2 x2 a2 x2
+ +
b2 y2 y2 b2
=2 =2
y2
= 1, having it’s a 2 b2 eccentricity equal to e. P is any variable point on it and P1, P2 are the foot of perpendiculars drawn from P to the x and y-axis respectively. The line P1P2 will always be a normal to an ellipse whose eccentricity is equal to (a) e2
(b)
e
(c)
+
2e (d) e 1+ e
34. The normal drawn to the ellipse
x2
+
y2
=1 a 2 b2 (a > b) at the extremity of the latus rectum passes through the extremity of the minor axis. Eccentricity of this ellipse is equal to mathematics today | February‘17
53
(a)
5 −1 2
(b)
5 −1 2
(c)
3 −1 2
(d)
3 −1 2
35. The equation of common x2 + 4y2 = 8 and y2 = 4x are (a) y – 2x – 4 = 0, y + 2x + 4 (b) y – 2x – 2 = 0, y + 2x + 2 (c) 2y – x – 4 = 0, 2y + x + 4 (d) None of these
tangents of the curves =0 =0 =0
36. If L is the length of perpendicular drawn from the 2 2 origin to any normal of the ellipse x + y = 1, then 25 16 maximum value of L is (a) 5 (b) 4 (c) 1 (d) None of these
37. The maximum distance of the centre of the ellipse x2 y2 + = 1 from the chord of contact of mutually 9 4 perpendicular tangents of the ellipse is 3 6 36 9 (a) (b) (c) (d) 13 13 13 13 38. Tangents PA and PB are drawn to the ellipse x2 y2 + = 1 from the point P(0, 5). Area of triangle 16 9 PAB is equal to 256 16 sq. units (a) (b) sq. units 25 5
1024 32 (d) sq. units sq. units 25 5 39. The straight line x – 2y + 4 = 0 is one of the common x2 y2 tangents of the parabola y2 = 4x and + = 1. The 4 b2 equation of another common tangent of these curves is (a) x + 2y + 4 = 0 (b) x + 2y – 4 = 0 (c) x + 2y + 2 = 0 (d) x + 2y – 2 = 0 (c)
x2 y2 + = 1 at the 36 9 point ‘P’ meets the y-axis at A and normal drawn to the ellipse at point ‘P’ meets the x-axis at B. If area of 27 triangle OAB is sq. units, then eccentric angle of 4 point ‘P’ is 40. Tangent drawn to the ellipse
54
mathematics today | February‘17
p (b) p 3 6 (c) p (d) None of these 4 41. The tangent and normal drawn to the ellipse x2 + 4y2 = 4 at the point P, meet the x-axis at A and B respectively. If AB = 2 then cosq is equal to, (‘q’ being the eccentric angle of point P) (a)
8 2 5 (a) 1 (b) (c) (d) 3 3 3 3 42. The chord joining the points P1(a cosq1, b sinq1) and P2(a cosq2, b sinq2) meets the x-axis at A. If OA = c(‘O’ being the origin), then the value of q q tan 1 ⋅ tan 2 is equal to 2 2 c−a c −b c−a (a) (b) (c) c − b (d) c +b c+a c+a c +b 2 2 y x 43. A tangent to the ellipse + = 1, having 18 32 4 slope − cuts the x and y-axis at the points A and B 3 respectively. If O is the origin then area of triangle OAB is equal to (a) 12 sq. units (b) 24 sq. units (c) 36 sq. units (d) None of these 44. Which of the following is a common tangent to the ellipses
x2
a 2 + b2
+
y2
b2
= 1 and
x2
a2
+
y2
a 2 + b2
= 1?
(a) by = ax − a 4 + a2b2 + b 4 (b) ay = bx − a 4 + a2b2 + b 4 (c) by = ax − a 4 − a2b2 + b 4 (d) ay = bx − a 4 − a2b2 + b 4 45. Maximum length of the chord of the ellipse x2 y2 + = 1, such that eccentric angles of its extremities a 2 b2 p differ by is (a > b) 2 (a) a 2 (b) b 2 (c) ab 2 (d) None of these 46. Consider an ellipse having its axes along the coordinate axes and passing through the point (4, –1).
If the line x + 4y – 10 = 0 is one of its tangent, then area of the ellipse is equal to (a) 20 p sq. units (b) 30 p sq. units (c) 15 p sq. units (d) 10 p sq. units 47. S1 and S2 are the foci of an ellipse. ‘B’ be one of the extremity of its minor axis. If triangle S1S2B is right angled then eccentricity of the ellipse is equal to 1 3 (a) (b) 2 2 3 (d) None of these 2 48. If the chord of contact of tangents drawn from a x2 y2 point P on the ellipse + = 1 touches the circles a 2 b2 x2 + y2 = c2, then locus of P is (c)
(a) (c)
x2 a2 x2 a4
+ +
y2 b2 y2 b4
= =
a2
(b)
c2 1
(d)
c2
x2
a2 x2
+ +
y2
b2 y2
= =
b4 c4
a4
a 2 b2 c 4 49. The point on the ellipse 4x2 + 9y2 = 1 such that tangent drawn to the ellipse at this point is perpendicular to the line 8x – 9y = 0, is 27 16 27 16 ,− , (a) (b) − 2 985 3 985 2 985 2 985 27 16 , (c) 2 985 3 985
(d) None of these
x2 y2 + = 1, from 16 9 the point ‘P’, meets the coordinate axes at concyclic points, then locus of point ‘P ’ is (a) x2 + y2 = 7 (b) x2 – y2 = 7 2 2 (c) x + y = 25 (d) x2 – y2 = 25 50. Tangents drawn to the ellipse
solutions 1. (b) : If the point of contact is P(a cosq, b sinq), y x then equation of tangent cos q + sin q = 1 a b and given line y = mx + c will be identical. Comparing the coefficients, we get cos q sin q 1 = =− am −b c am b ⇒ cos q = − , sin q = ⇒ c c
a2m2 + b2 = c2
2. (a) : Let the aribitrary point be (a, 4 – a), then equation of corresponding chord of contact is, x · a + 2y (4 – a) = 4 ⇒ a(x – 2y) + 8y – 4 = 0 It is indeed a family of concurrent lines. The point of concurrency being the intersection point of the lines x – 2y = 0 and 8y – 4 = 0 1 \ Required point is 1, 2 3. (d) : e = 1−
2b2 1 b2 1 = ⇒ = a 2 a 4 b2 a
2
=
...(i)
(Given)
1 2
...(ii)
1 2 1 From (i) and (ii), we get, a = , b = 3 12 x2 y2 + =1 \ Equation of ellipse is 1 1 9 12 i.e., 9x2 + 12y2 = 1 2 2 4. (d) : Given ellipse is, x + y = 1 4 3 and line is y = x – 1 Let the point of contact be (2 cosq, 3 sinq).
\
Equation of tangent becomes
y x cos q + sin q = 1 2 3
cos q sin q = =1 2 − 3 ⇒ cosq = 2, sinq = – 3 which is not possible. Thus, the given line can’t be a tangent to the given ellipse. ⇒
2 2 5. (b) : Given ellipse is, x + y = 1 9 4 Let the foot of normal be (3 cosq, 2 sinq) Equation of normal at this point is, 2y 3x − =5 cos q sin q 3 Slope = tanq = – 2 2 4 ⇒ tanq = – 3 4 3 or sin q = , cos q = − 5 5 9 8 Thus foot of normal be − , . 5 5 mathematics today | February‘17
55
6. (c) : Let P ≡ (a cosq, b sinq) y x \ Equation of tangent is cos q + sin q = 1 a b ⇒ A ≡ (a secq, 0), B ≡ (0, b cosecq) If the mid-point of segment AB be Q(h, k), then 2h = a secq, 2k = b cosecq ⇒
a2
b2
+
=1 4h 2 4 k 2 a 2 b2 + =4 \ Locus of Q is x2 y2 7. (a) : Let P ≡ (a cosq, b sinq) by ax \ Equation of normal is − = a 2 − b2 cos q sin q a 2 − b2 a 2 − b2 ⇒ A≡ cos q, 0 and B ≡ 0, sin q b a If (h, k) be the mid-point of segment AB, then 2
2
2
2
a −b a −b cos q, 2k = sin q a b ⇒ 4h2a2 + 4k2b2 = (a2 – b2)2 Thus, locus of the point is 4a2x2 + 4b2y2 = (a2 – b2)2 2h =
8. (d) : Let the point of contact be P(a cosq, b sinq). y x Then equation of tangent is cos q + sin q = 1 a b cos q sin q 1 Thus on comparing, = =− am −b c ⇒ cos q = −
am b , sin q = c c
a 2 m b2 Thus the point of contact is − , c c 9. (a) : Let the foot of normal be P(a cosq, b sinq) by ax Equation of normal at P is − = a 2 − b2 cos q sin q m cos q sin q c \ = = 2 a b (b − a2 ) ⇒ cos q =
ac 2
2
m(b − a )
, sin q =
bc 2
(b − a2 )
a2 c b2 c Thus the foot of normal is , m(b2 − a2 ) (b2 − a2 ) 10. (d) : Let the mid-point of chord be P(h, k). Equation of this chord will be T = S1 56
mathematics today | February‘17
i.e.,
xh a2
+
yk b2
=
h2 a2
+
k2
b2 h2 k 2 2+ 2 b a h k b2 ⇒ =− =− ⇒ k =− ⋅h c 2a2 b2 2a2 b2 Thus the required locus is y = − x or 2a2y + b2x = 0 2a2 11. (c) : Let A ≡ (1, 0), B ≡ (–1, 12 ) \ Given expression is, PA + PB = a Here, AB = 4 + 12 = 4 Thus P(x, y) will lie on an ellipse having foci as A and B, provided a > 4 12. (d) : Clearly, 6 – a > 0, a – 2 > 0 and 6 – a ≠ a – 2 ⇒ a ∈ (2, 6) ~ {4} 13. (b) : Equation of the ellipse is
x2 y2 + =1 25 16
16 9 3 = ⇒ e= 25 25 5 3 3 Thus, foci are 5 ⋅ , 0 and −5 ⋅ , 0 5 5 i.e. (± 3, 0) ⇒
e2 = 1 −
14. (d) : If P(x, y) be any point on the ellipse, then PS = e · PM, where PM is perpendicular distance of P from directrix. 1 x + y −2 ⇒ (x − 1)2 + y 2 = 2 2
2
Thus required equation is 3x2 + 3y2 – 2xy – 4x + 4y = 0 15. (a) : Given equation can be rewritten as 2
2
x + y −5 x − y +7 2 2 + =1 1 3/2 It is clearly an ellipse whose major and minor axes are along the lines x – y + 7 = 0 and x + y – 5 = 0 Thus, centre of this ellipse is the intersection point of these lines i.e., the point (–1, 6) 16. (b) : OS1 = ae = 6, OC = b (say), CS1 = a 1 DOCS = ⋅ (OS1 )(OC ) = 3b 1 2 Semi perimeter of triangle OCS1 1 1 = (OS1 + OC + CS1) = (6 + a + b) 2 2
Now, inradius of triangle OCS1 3b = = 1 (given) 1 (6 + a + b) 2 ⇒ 5b = 6 + a Also b2 = a2 (1 – e2) = a2 – 36 Thus, 25b2 = 36 + a2 + 12a ⇒ 25(a2 – 36) = 36 + a2 + 12a 13 5 ⇒ 2a2 – a – 78 = 0 ⇒ a = , b = 2 2 65p Hence, area of ellipse = pab = sq. units. 4 17. (a) : Equation of normal at P(q) is, 3y 4x − =7 cos q sin q Since it passes through the point Q(2q), thus 16 ⋅ cos 2q 9 ⋅ sin 2q − =7 cos q sin q 16(2 cos2 q − 1) − 18 cos q = 7 cos q ⇒ 23 cos2q – 7 cosq – 16 = 0 16 ⇒ cosq = – [ as cosq ≠ 1] 23 18. (d) : In this case, base S1S2 is fixed and PS1 + PS2 is fixed. Hence area will be maximum when PS1 = PS2 1 \ Maximum area = (2ae)b = abe sq. units. 2 19. (d) : Here, S1S2 = 2ae, PS1 + PS2 = 2a If the normal to ellipse at P, meets the x-axis at Q, then incentre of DPS1S2 will lie on segment PQ and PS + PS2 1 will divide it in the ratio 1 i.e., S S e 1 2 Let P ≡ (a cosq, b sinq) ⇒
Equation of normal at ‘P’ is by ax − = a 2 − b2 = a 2 e 2 cos q sin q ⇒ Q ≡ (ae2 cosq , 0) Let R(h, k) be the incenter, then h=
a cos q ⋅ e + ae 2 cos q b sin q ⋅ e + 0 ⋅1 , k= 1 1 1+ 1+ e e
1 1 ⇒ h 1 + = (ae + ae 2 ) cos q and be ⋅ sin q = k 1 + e e Thus locus of R is
1 x 1 + e 2
2
1 y 1 + e 2
+
a2e 2 (1 + e)2
2
b2 e 2
=1
which is clearly an ellipse. 20. (a) : S1 = (ae, 0), S2 = (–ae, 0) Let the tangent be xb cosq + ay sinq – ab = 0 S1P1 = ⇒ S1P1 =
| abe cos q − ab | b2 cos2 q + a2 sin2 q ab(1 − e cos q) b2 cos2 q + a2 sin2 q ab(1 + e cos q)
Similarly, S2P2 = ⇒ S1P1·S2P2 = =
2
b cos2 q + a2 sin2 q a2b2 (1 − e 2 cos2 q)
b2 cos2 q + a2 sin2 q
a2b2 (1 − e 2 cos2 q) 2
2
2
2
a + (b − a )cos q
=
a2b2 (1 − e 2 cos2 q) 2
2 2
2
a − a e cos q
= b2
which is clearly a constant Thus, (S1P1) (S2P2) = b2 21. (d) : Let the equation of ellipse having centre (1, 2) be
(x − 1)2
( y − 2)2
=1 a2 b2 We have, ae = 5 ⇒ a2e2 = 25 Also,
9 a
2
+
16 b2
+
=1
Now, b2 = a2(1 – e2) = a2 – 25 9 16 ⇒ + = 1 ⇒ a4 – 50a2 + 225 = 0 2 2 a a − 25 ⇒ a2 = 5, 45 But a2 = 5 is clearly rejected \ a2 = 45 ⇒ b2 = 20 (x − 1)2 ( y − 2)2 Thus, required equation is + =1 45 20 22. (d) : Tangent to the ellipse is 5m2 5 + 3 2 5m2 5 If it passes through P(1, 2), then (2 – m)2 = + 3 2 ⇒ 4m2 + 24m – 9 = 0 If it’s roots are m1 and m2 then y = mx ±
mathematics today | February‘17
57
m1 + m2 = – 6, m1m2 = –
9 4
⇒ | m1 − m2 | = 36 + 9 = 3 5 \
⇒
m1 − m2 3 5 12 = = 9 1 + m1m2 5 1− 4 12 q = tan −1 5
⇒ A2 ⋅
23. (c) : Equation of normal at point ‘q’ is 3y 5x − = 16 cos q sin q On comparing with 5x – 3y = 8 2, we get p 1 ⇒ q= ⇒ cos q = sin q = 4 2
⇒
25. (b) : Equation of tangent at point ‘q’ is, x 11 cos q + y sin q = 1 4 16 It will touch the given circle, if cos q −1 4 =4 cos2 q 11sin2 q + 16 256 2 ⇒ 4 cos q + 8 cosq – 5 = 0 1 ⇒ cos q = ⇒ q = p , 5p 2 3 3 26. (c) : The given distance is clearly the length of semi major axis a2 + 2b2 Thus, =a 2 ⇒ 2b2 = a2 ⇒ 2a2 (1 – e2) = a2 1 1 ⇒ e2 = ⇒ e = 2 2
mathematics today | February‘17
1 − a2
x2 A
2
+
y2 B
2
4a 2
1− a
2
+ B2 =
1 1 − a2
a 2 ( 4 A2 − B 2 ) + B 2 2
=
1 2
⇒ A2 =
1 4
1− a 1− a If ‘e’ be the eccentricity of the ellipse, then A2 = B2(1 – e2) 1 3 ⇒ 1 − e2 = ⇒ e= 4 2 28. (d) : Any normal to the ellipse is 2y 5x − =1 cos q sin q
24. (c) : Let the point ‘P’ be ( 6 cos q, 2 sin q) A.T.Q. 6 cos2q + 2 sin2q = 4 1 ⇒ 6 – 4 sin2q = 4 ⇒ sin2q = 2 p ⇒ q = ( q lies in first quadrant) 4
58
1 − a2
1
x+
2 2 2 Comparing it with y = mx ± A m + B , we get 2a m=− 1 − a2 1 A2m2 + B2 = 1 − a2
tan q =
27. (a) : Let the equation of ellipse be
2a
We have, y = −
sin q i.e., y = x . 5 tanq – 2 2 2 Since (i) is tangent to y = 4x 1 \ y = mx + m ⇒ m= ⇒
...(i)
5 1 sin q tan q, = − 2 m 2
1 5 sin2 q , and cos q = 5 ⋅ = −1 ⇒ cos q = − 5 4 cos q
1 1 ⇒ q = p − cos −1 or p + cos −1 5 5 29. (a) : Let line AB be bx cosq + ay sinq – ab = 0 Now, combined equation of lines OA and OB is x2
+
y2
=
y a cos q + b sin q
x a2
2
a2 a i.e., x2 · sin2q + y2 1 − sin2 q − sin 2q · xy = 0 b2 b Since the lines OA and OB are mutually perpendicular, hence =1
sin2 q + 1 −
a2
b
2
sin2 q = 0
a2 1 ⇒ sin q − 1 = 1 ⇒ sin2 q − 1 = 1 2 2 1 − e b 2
⇒ e2 · sin2q = 1 – e2 ⇒ e2 (1 + sin2q ) = 1 30. (a) : Let the extremities of the focal chord be P1(q1) and P2(q2). Equation of chord P1P2 is x q +q y q +q q −q cos 1 2 + sin 1 2 = cos 1 2 2 b 2 2 a It shou ld p ass t hroug h (ae, 0) and its slop e should be tan q q +q q −q ⇒ e cos 1 2 = cos 1 2 2 2 b q +q and − cot 1 2 = tan q 2 a a tan q q +q ⇒ cot 1 2 = − 2 b Now, P1P2 = P1S1 + P2S1 = a – ea cosq1 + a– ea cosq2 = 2a – ea(cosq1 + cosq2) q −q q +q = 2a – 2ae cos 1 2 cos 1 2 2 2 q +q = 2a – 2ae2 cos2 1 2 2 2 2 = 2a – 2ae2 · a tan q a2 tan2 q + b2 3 2
= =
2
a2 sin2 q + b2 cos2 q 2a sin2 q(1 − e 2 ) + 2ab2 cos2 q 3
a2 sin2 q + b2 cos2 q
2ab2 sin2 q + 2ab2 cos2 q a2 sin2 q + b2 cos2 q
=
2ab2 a2 sin2 q + b2 cos2 q
31. (a) : Let the foot of normal is (a cosq, b sinq), by ax then − = a 2 − b2 cos q sin q and the given lines will be identical, m cos q sin q c Thus, = = 2 a b (b − a2 ) ac bc ⇒ cos q = , sin q = 2 2 2 m(b − a ) (b − a2 ) ⇒
a2 m2
+ b2 =
(b2 − a2 )2 c2
Thus,
h2 a2
+
k2 b2
=4
Hence, locus is,
x2 a2
+
y2 b2
=4
33. (d) : Let P ≡ (a cosq, b sinq) ⇒ P1 ≡ (a cosq, 0), P2 ≡ (0, b sinq) y x + =1 Thus, equation of line P1P2 is a cos q b sin q y /b x /a ⇒ − =1 cos(−q) sin(−q) which is clearly a normal to the ellipse of the form x2
2
+
y2
2
A
= 1 where,
2
2
=
B λ λ and = 2 2 a b A −B
A −B A B If a > b, then B > A Let the eccentricity of the second ellipse be e1 ⇒ 1 − e12 =
2a e sin q
= 2a −
32. (c) : If the point of intersection be R(h, k), then q +q q +q a cos 1 2 b sin 1 2 2 2 , k= h= q −q q −q cos 1 2 cos 1 2 2 2 q +q q +q b sin 1 2 a cos 1 2 2 2 , k= ⇒ h= 1 1 2 2
A2 B
2
=
b2 a
2
= 1 − e 2 ⇒ e1 = e
b2 34. (a) : Equation of normal at ae, is a a(x − ae) = (ay – b2) e It should pass through (0, – b) a(0 − ae) = – a2 – b2 e b2 b ⇒ a2 = ab + b2 ⇒ 2 + − 1 = 0 a a ⇒
⇒
b −1 + 5 b2 3 − 5 = ⇒ = = 1 − e2 2 a 2 2 a
⇒ e2 =
5 −1 ⇒ e= 2
5 −1 2
35. (c) : Tangent for y2 = 4x is y = mx +
1 m
mathematics today | February‘17
59
x2 y2 + = 1 is y = mx ± 8m2 + 2 8 2 For common tangents we must have 1 1 = ± 8m2 + 2 ⇒ m = ± m 2 Thus, common tangents are x x y = + 2 and y = − − 2 2 2 i.e. 2y – x – 4 = 0 and 2y + x + 4 = 0 36. (c) : Any normal to the ellipse is 4y 5x − = 9 i.e., 5 secq x – 4 cosecq y – 9 = 0 cos q sin q 9 \ L= 2 25 sec q + 16 cosec2q 9 = 2 25(1 + tan q) + 16(1 + cot 2 q) 9 = 2 25 tan q + 16 cot 2 q + 41 25 tan2 q + 16 cot 2 q Now, ≥ 20 2 ⇒ 25 tan2q + 16 cot2q + 41 ≥ 81 9 ⇒ ≤1 ⇒ L ≤ 1 2 25 tan q + 16 cot 2 q + 41 \ Maximum value of L is 1. 37. (a) : Clearly, the tangents have been drawn by taking any point on the director circle of the ellipse. Any point on the director circle can be taken as ( 13 cos q, 13 sin q). Equation of corresponding chord of contact is 13 13 .cos q.x + sin q y = 1 9 4 and for
\
Distance from the origin =
1
13 2 13 cos q + sin2 q 81 16 36 36 9 = = = 208 13 1053 − 845 cos2 q
38. (b) : Equation of AB is T = 0, 5y 9 i.e. =1 ⇒ y = 9 5 9 Putting y = in the equation of ellipse, we get 5 81 16 32 x2 + = 1 ⇒ x = ± . Thus, AB = 16 25 ⋅ 9 5 5
60
mathematics today | February‘17
9 16 = 5 5 1 32 16 256 Thus, area of D = ⋅ ⋅ = sq. units PAB 2 5 5 25 39. (a) : Other common tangent will clearly be the reflection of x – 2y + 4 = 0 in the x-axis. Thus other common tangent will be x + 2y + 4 = 0 40. (c) : Let the eccentric angle of point ‘P’ be q, then equation of tangent and normal at point are respectively 3y y 6x x − = 27 cos q + sin q = 1 and cos q sin q 6 3 Distance of P from AB = 5 –
27 3 Thus, A ≡ 0, and B ≡ cos q, 0 6 sin q 1 3 27 27 Area of DOAB = ⋅ (given) ⋅ ⋅ cos q = 2 sin q 6 4 p 5p ⇒ cotq = 1 ⇒ q = , 4 4 41. (c) : Let P ≡ (2 cosq, sinq) x Equation of tangent at ‘P’ is cosq + y sinq = 1 2 ⇒ A ≡ (2 secq, 0)
y 2x − =3 Equation of normal at ‘P’ is cos q sin q 3 ⇒ B ≡ cos q, 0 2 3 Now, AB = | 2 secq – cosq | = 2 (given) 2 ⇒ |4 – 3 cos2q| = 4 |cosq| ⇒ 3 cos2q ± 4 cosq – 4 = 0 2 2 ⇒ cosq = , − 3 3 42. (a) : Equation of chord is x q +q y q +q q −q cos 1 2 + sin 1 2 = cos 1 2 2 b 2 2 a q1 − q2 a cos 2 , 0 ⇒ A≡ q1 + q2 cos 2 Since OA = c, therefore, q −q q +q a cos 1 2 = c.cos 1 2 2 2 q1 q2 q q ⇒ cos .cos (a − c) = sin 1 .sin 2 (−c − a) 2 2 2 2 q q c−a ⇒ tan 1 . tan 2 = 2 2 c+a
mathematics today | February‘17
61
43. (b) : Equation of tangent at point y x cos q + sin q = 1 ( 18 cos q, 32 sin q) , is 18 32 32 4 Its slope = – . cotq = – (given) 3 18 ⇒ cotq = 1 Also, A ≡ ( 18 sec q, 0), B ≡ (0, 32 cosec q ) 1 ⇒ DOAB = 3 2 sec q ⋅ 4 2cosec q 2
Thus, OS1 = OB = OS2 ⇒ ae = b ⇒ a2e2 = a2 – a2 e2 1 1 ⇒ e= ⇒ e2 = 2 2
= 12 ⋅ 2 ⋅ 2 = 24 sq. units
=c b 4 h2 + a 4 k 2 Hence, locus of point ‘P ’ is
44. (a) : Any tangent to
x2
a 2 + b2
+
y2 b2
= 1 is,
y = mx ± (a2 + b2 )m2 + b2 It will also be a tangent to second ellipse if a2m2 + (a2 + b2) = (a2 + b2) m2 + b2 a ⇒ b2m2 = a2 ⇒ m = ± b Thus common tangents are y=±
2
a a x ± (a2 + b2 ) ⋅ + b2 b b2
i.e., by = ± ax ± a 4 + b 4 + a2b2 45. (a) : Let the extremities of the chord be P1 ≡ (a cosq, b sinq) and P2 ≡ (–a sinq, b cosq) Now, P1P22 = a2 (cosq + sinq)2 + b2(sinq – cosq)2 = a2 + b2 + (a2 – b2)2sinqcosq ≤ a2 + b2 + a2 – b2 = 2a2 ⇒ P 1P 2 ≤ a 2 46. (d) : Let the ellipse be
x
2
y
=1 a b2 If the line y = mx + c be its tangent, then c2 = a2m2 + b2 Tangent is given to be x + 4y – 10 = 0 2
+
2
m 1 c 1 5 25 a2 =− = ⇒m= − ,c = \ = + b2 1 4 −10 4 2 4 16 ⇒ a2 + 16b2 = 100 We also have P(4, –1) to be a point on the ellipse, 16 1 \ + =1 a 2 b2 ⇒ a2 + 16b2 = a2b2 ⇒ a2b2 = 100 ⇒ ab = 10 \ Area of ellipse = pab = 10p sq. units 47. (b) : If ‘O’ be the center of ellipse then for triangle S1S2B to be right angled, we must have p ∠OBS1 = ∠OBS2 = . 4 ⇒
62
mathematics today | February‘17
48. (c) : Let P ≡ (h, k)
xh
Equation of chord of contact is
a
+
2
i.e., xb2h + ya2k – a2b2 = 0 It will touch the circle x2 + y2 = c2 , if
yk b2
=1
a 2 b2
b4 x2 + a4y2 =
a 4b 4 c
2
i.e.,
x2 a
4
+
y2 b
4
=
1 c2
9 . 8 Slope of tangent at any point P(q) on the ellipse 49. (c) : Slope of tangent must be – x2
a
2
+
y2
= 1 is − b cot q a b 2
9 2 16 Thus, − = − cot q ⇒ tan q = 8 3 27 ⇒ cos q = − or cos q = −
27 985 27
, sin q = , sin q = −
985 Thus, required point is 27 16 , or 2 985 3 985
16 985 16 985
27 16 ,− − 2 985 3 985
50. (b) : Let P ≡ (h, k) Any tangent to the ellipse is y = mx ± 16m2 + 9 If it passes through ‘P’, then (k – mh)2 = 16m2 + 9 ⇒ m2(h2 – 16) – 2mhk + (k2 – 9) = 0 If its roots are m1 and m2, then m1m2 =
k2 − 9
h2 − 16 Since, the drawn tangents meet the coordinate axes at concyclic points, thus, m1m2 = 1 ⇒ k2 – 9 = h2 – 16 Hence required locus is, x2 – y2 = 7.
on
Conic Sections
*ALOK KUMAR, B.Tech, IIT Kanpur
important facts and formulae z
z
z
z
z
z
z
z
z
z
z
z
The area of the triangle inscribed in the parabola 1 y2 = 4ax is |( y − y )( y − y3)( y3 − y1)|, where 8a 1 2 2 y1, y2, y3 are the ordinates of the vertices. The length of the side of an equilateral triangle inscribed in the parabola y2 = 4ax is 8a 3. (one angular point is at the vertex). y2 = 4a(x + a) is the equation of the parabola whose focus is the origin and the axis is x-axis. y2 = 4a(x – a) is the equation of parabola whose axis is x-axis and y-axis is directrix. x2 = 4a(y + a) is the equation of parabola whose focus is the origin and the axis is y-axis. x2 = 4a(y – a) is the equation of parabola whose axis is y-axis and the x-axis is directrix. The equation of the parabola whose vertex and focus are on x-axis at a distance a and a′ respectively from the origin is y2 = 4(a′ – a)(x – a). The equation of the parabola whose axis is parallel to x-axis is x = Ay2 + By + C and y = Ax2 + Bx + C is a parabola with its axis parallel to y-axis. If the straight line lx + my + n = 0 touches the parabola y2 = 4ax, then ln = am2. If the line xcosa + ysina = p touches the parabola y2 = 4ax then pcosa + asin2a = 0 and point of contact is (atan2a, –2atana). x y If the line + = 1 touches the parabola l m y2 = 4a(x + b), then m2(l + b) + al2 = 0. If the two parabolas y2 = 4x and x2 = 4y intersect at point P, whose absiccae is not zero, then the tangent to each curve at P, make complementary angle with the x-axis.
z
z
z
z
z
Tangents at the extremities of any focal chord of a parabola meet at right angle on the directrix. Area of the triangle formed by three points on a parabola is twice the area of the triangle formed by the tangents at these points. If the tangents at the points P and Q on a parabola meet in T, then ST is the geometric mean between SP and SQ, i.e., ST2 = SP·SQ Tangent at one extremity of the focal chord of a parabola is parallel to the normal at the other extremity. The angle of intersection of two parabolas y2 = 4ax and x2 = 4by is given by tan −1
z
z
z
z
z
2(a2/3 + b2/3)
.
The equation of the common tangent to the parabolas y2
z
3a1/3b1/3
= 4ax and
x2
= 4by is
1 1 2 2 3 3 a x + b y + a 3b 3
=0.
The line lx + my + n = 0 is a normal to the parabola y2 = 4ax, if al(l2 + 2m2) + m2n = 0. If the normals at points (at12, 2at1) and (at22, 2at2) on the parabola y2 = 4ax meet on the parabola, then t1t2 = 2. If the normal at a point P(at2, 2at) to the parabola y2 = 4ax subtends a right angle at the vertex of the parabola then t2 = 2. If the normal to a parabola y2 = 4ax, makes an angle f with the axis, then it will cut the curve again at an 1 angle tan −1 tan f . 2 If the normal at two points P and Q of a parabola y2 = 4ax intersect at a third point R on the curve. Then the product of the ordinate of P and Q is 8a2.
* Alok Kumar is a winner of INDIAN NATIONAL MATHEMATICS OLYMPIAD (INMO-91). He trains IIT and Olympiad aspirants.
mathematics today | FEBRUARY ‘17
63
z
z
z
The chord of contact joining the point of contact of two perpendicular tangents always passes through focus. If tangents are drawn (at12 , 2at1) Q from the point (x1, y1) to the parabola y2 = 4ax, (x1, y1)P then the length of their (at22 , 2at2)R chord of contact is 1 ( y12 − 4ax1)( y12 + 4a2 ) |a | The area of the triangle formed by the tangents drawn from (x1, y1) to y2 = 4ax and their chord of ( y12 − 4ax1)3/2 . 2a If one extremity of a focal chord is (at12, 2at1), then a −2a the other extremity (at22, 2at2) becomes , t12 t1 by virtue of relation t1t2 = –1 and length of chord
z
x2
y2 + = 1 , if a2cos2a + b2sin2a = p2 and that a 2 b2 2 2 point of contact is a cos a , b sin a . p p z
y2 + = 1 , then the common tangent is inclined a 2 b2 r 2 − b2 to the major axis at an angle tan −1 2 2 . a −r z
z
z
on the parabola y2 = 4ax is a(t1 − t2 ) (t1 + t2 )2 + 4 The length of intercept made by line y = mx + c between 4 the parabola y2 = 4ax is a(1 + m2 )(a − mc) . m2 Locus of mid-points of all chords which is inscribed a right angle on the vertices of parabola is parabola. The focal chord of parabola y2 = 4ax making an angle a with the x-axis is of length 4acosec2a and perpendicular on it from the vertex is asina. The length of a focal chord of a parabola varies inversely as the square of its distance from the vertex. If l1 and l2 are the length of segments of a focal chord 4l l of a parabola, then its latus-rectum is 1 2 . l1 + l2 The semi-latus rectum of the parabola y2 = 4ax is
z
the harmonic mean between the segments of any focal chord of the parabola. The straight line lx + my + n = 0 touches the ellipse
z
z
z
z
z
x2
a2 64
y2
+ 2 = 1 , if b
a2l2
+
b2m2
=
n2.
mathematics today | FEBRUARY‘17
y2 + = 1 is (x2 + y2)2 = a2 x2 + b2y2 or r2 = a 2 b2 a2cos2q + b2sin2q. (In terms of polar coordinates) The locus of the mid points of the portion of the
y2 + = 1 intercepted a 2 b2 between the axes is a2y2 + b2x2 = 4x2y2. tangents to the ellipse
2
The length of the chord joining two points 't1' and 't2'
The locus of the foot of the perpendicular drawn from centre upon any tangent to the ellipse x2
1 = a t + . t z
A circle of radius r is concentric with the ellipse x2
contact is z
The line xcosa + ysina = p touches the ellipse
x2
x2
y2 + = 1, a 2 b2
z
If y = mx + c is the normal to the ellipse
z
m2(a2 − b2 )2 then condition of normality is c 2 = 2 2 2 . (a + b m ) The straight line lx + my + n = 0 is a normal to the
2 2 2 2 2 y2 if a + b = (a − b ) . + = 1 , a 2 b2 l 2 m2 n2 Four normals can be drawn from a point to an ellipse. Y Normal If S be the focus P(x1, y1) and G be the point where the normal at X X S C G S T P meets the axis of an el l ip s e, t he n Y SG = e. SP, and the tangent and normal at P bisect the external and internal angles between the focal distances of P. Any point P of an ellipse is joined to the extremities of the major axis then the portion of a directrix intercepted by them subtends a right angle at the corresponding focus. The equations to the normals at the end of the latus rectum and that each passes through an end of the minor axis, if e4 + e2 – 1 = 0.
ellipse z
z
z
z
x2
z
The area of the triangle formed by the three points, on 2
y + 2 = 1 , whose eccentric angles are q, a b q−f f − y y −q f and y is 2absin sin sin 2 2 2 the ellipse
z
x
2
z
2
z
If the point of intersection of the ellipses x2
y2 x2 y2 + = 1 and + = 1 be at the extremities a 2 b2 a 2 b2 of the conjugate diameters of the former, then a2
b2 + =2. a 2 b2
z
z
z
z
z
y2 − = 1 , then a2l2 – b2m2 = n2. a 2 b2 If t h e s t r ai g ht l i n e x c o s a + y s i n a = p hyperbola
z
x2
tou che s t he hy p e r b ol as a2 cos2 a − b2 sin2 a = p2 . z
x2
normal to the hyperbola a2
z
hyperbola from any point. If a, b, g are the eccentric angles of three points on
y2 − = 1 , the normals at which a 2 b2 are concurrent, then sin(a + b) + sin(b + g) + sin(g + a) = 0
q q 1− e y2 . − = 1 , then tan 1 tan 2 = 2 2 2 2 1+ e a b
If the polars of (x1, y1) and (x2, y2) with respect
then z
z
x2
y2 − = 1 are at right angles, a 2 b2
x1x2 a 4 + =0. y1 y2 b 4
The parallelogram formed by the tangents at the extremities of conjugate diameters of a hyperbola has its vertices lying on the asymptotes and is of constant area. The product of length of perpendiculars drawn from any point on the hyperbola the asymptotes is
z
y2 − = 1 , then a 2 b2
z
x2
to the hyperbola
x2
b2 (a2 + b2 )2 − = l 2 m2 n2 In general, four normals can be drawn to a
the hyperbola
z
z
z
x2
z
.
If the chord joining two points (asecq1, btanq1) (asecq2, btanq2) passes through the focus of the hyperbola
y2
If the line lx + my + n = 0 will be
x2 y2 The length of chord cut off by hyperbola 2 − 2 = 1 a b from the line y = mx + c is (b2 − a2m2 )
− 2 = 1 , t he n a b 2
y2 − = 1 from (h, k) a 2 b2 lie on a2y(x – h) + b2x(y – k) = 0.
2ab [c 2 − (a2m2 − b2 )](1 + m2 )
z
The sum of the squares of the reciprocal of two perpendicular diameters of an ellipse is constant. In an ellipse, the major axis bisects all chords parallel to the minor axis and vice-versa, therefore major and minor axes of an ellipse are conjugate diameters of the ellipse but they do not satisfy the condition m1 · m2 = –b2/a2 and are the only perpendicular conjugate diameters. The foci of a hyperbola and its conjugate are concyclic. Two tangents can be drawn from an outside point to a hyperbola. If the straight line lx + my + n = 0 touches the
x2
The feet of the normals to
a2b2
a 2 + b2
x2
y2 − = 1 to a 2 b2
.
c If the normal at ct , on the curve xy = c2 meets t 1 the curve again in 't', then t ′ = − 3 . t A triangle has its vertices on a rectangular hyperbola; then the orthocentre of the triangle also lies on the same hyperbola. All conics passing through the intersection of two rectangular hyperbolas are themselves rectangular hyperbolas. An infinite number of triangles can be inscribed in the rectangular hyperbola xy = c2 whose all sides touch the parabola y2 = 4ax. mathematics today | FEBRUARY ‘17
65
proBlems Single Correct Answer Type
1. The equation of the common tangent of the parabolas x2 = 108y and y2 = 32x, is (a) 2x + 3y = 36 (b) 2x + 3y + 36 = 0 (c) 3x + 2y = 36 (d) 3x + 2y + 36 = 0 2. The line xcosa + ysina = p will touch the parabola y2 = 4a(x + a), if (a) pcosa + a = 0 (b) pcosa – a = 0 (c) acosa + p = 0 (d) acosa – p = 0 3. The angle of intersection between the curves y2 = 4x and x2 = 32y at point (16, 8), is 4 3 (a) tan −1 (b) tan −1 5 5 p (c) p (d) 2 4. If y1, y2 are the ordinates of two points P and Q on the parabola and y3 is the ordinate of the point of intersection of tangents at P and Q, then (a) y1, y2, y3 are in A.P. (b) y1, y3, y2 are in A.P. (c) y1, y2, y3 are in G.P. (d) y1, y3, y2 are in G.P. 5. The equation of the common tangent touching the circle (x – 3)2 + y2 = 9 and the parabola y2 = 4x above the x-axis, is (a) (b) 3 y = 3x + 1 3 y = −(x + 3) (c)
3y = x + 3
(d)
3 y = −(3x + 1)
6. The angle between the tangents drawn from the points (1,4) to the parabola y2 = 4x is p p p p (a) (b) (c) (d) 2 3 6 4 7. Let a circle tangent to the directrix of a parabola y2 =2ax has its centre coinciding with the focus of the parabola. Then the point of intersection of the parabola and circle is (a) (a, –a) (b) (a/2, a/2) (c) (a/2, ± a) (d) (± a, a/2) 8.
An ellipse passes through the point (–3, 1) and its
2 The equation of the ellipse is . 5 (a) 3x 2 + 5 y 2 = 32 (b) 3x 2 + 5 y 2 = 25 (c) 3x 2 + y 2 = 4 (d) 3x 2 + y 2 = 9 eccentricity is
9.
The locus of the point of intersection of 2
2
perpendicular tangents to the ellipse x + y = 1 , is a 2 b2 66
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(a) x2 + y2 = a2 – b2 (c) x2 + y2 = a2 + b2
(b) x2 – y2 = a2 – b2 (d) x2 – y2 = a2 + b2
2 2 10. If the eccentricity of the two ellipse x + y = 1 169 25 2 2 and x + y = 1 are equal, then the value of a/b is a 2 b2 (a) 5/13 (b) 6/13 (c) 13/5 (d) 13/6
11. The eccentricity of the curve represented by the equation x2 + 2y2 – 2x + 3y + 2 = 0 is (a) 0 (b) 1/2 (c) 1 / 2 (d) 2 12. The equation of the hyperbola whose conjugate axis is 5 and the distance between the foci is 13, is (a) 25x2 – 144y2 = 900 (b) 144x2 – 25y2 = 900 (c) 144x2 + 25y2 = 900 (d) 25x2 + 144y2 = 900 13. The vertices of a hyperbola are at (0, 0) and (10, 0) and one of its foci is at (18, 0). The equation of the hyperbola is (a)
x2 y2 − =1 25 144
2 2 (b) (x − 5) − y = 1 25 144
2 2 x 2 ( y − 5)2 (d) (x − 5) − ( y − 5) = 1 − =1 25 144 25 144 14. The equation of the hyperbola whose foci are (6, 4) and (–4, 4) and eccentricity 2 is given by (a) 12x 2 − 4 y 2 − 24x + 32 y − 127 = 0
(c)
(b) (c)
12x 2 + 4 y 2 + 24x − 32 y − 127 = 0 12x 2 − 4 y 2 − 24x − 32 y + 127 = 0
(d) 12x 2 − 4 y 2 + 24x + 32 y + 127 = 0 15. The latus rectum of the hyperbola 9x2 – 16y2 + 72x – 32y – 16 = 0 is 32 9 32 9 (a) (b) − (c) (d) − 3 2 3 2 16. If m1 and m2 are the slopes of the tangents to the x2 y2 − = 1 which pass through the point 25 16 (6, 2), then hyperbola
24 11 48 (c) m1 + m2 = 11 (a) m1 + m2 =
(b) (d)
10 11 11 m1m2 = 20 m1m2 =
2 2 17. Let E be the ellipse x + y = 1 and C be the circle 9 4 x2 + y2 = 9. Let P and Q be the points (1, 2) and (2, 1)
respectively. Then (a) Q lies inside C but outside E (b) Q lies outside both C and E (c) P lies inside both C and E (d) P lies inside C but outside E
Comprehension Type
18. The value of m, for which the line y = mx + 2 2 is a normal to the conic x − y = 1, is 16 9 2 3 (a) (b) − (c) − (d) 1 3 3 2
25 3 , 3
19. If q is the acute angle of intersection at a real point of intersection of the circle x2 + y2 = 5 and the parabola y2 = 4x then tanq is equal to 1 (a) 1 (b) (c) 3 (d) 3 3 20. The equation of the hyperbola in the standard form (with transverse axis along the x-axis) having the length of the latus rectum = 9 units and eccentricity = 5/4 is (a)
2
2
x y − =1 16 18
(c)
x2 y2 − =1 64 36
(e)
x2 y2 − =1 16 9
(b) (d)
2
2
x y − =1 36 27 x2 y2 − =1 36 64
Paragraph for Q. No. 24 to 26 Consider the circles x2 + y2 = a2 and x2 + y2 = b2 where b > a > 0. Let A(–a, 0), B(a, 0). A parabola passes through the points A, B and its directrix is a tangent to x2 + y2 = b2. If the locus of focus of the parabola is a conic then 24. The eccentricity of the conic is (a) 2a/b (b) b/a (c) a/b
(d) 1
25. The foci of the conic are (a) (±2a, 0) (b) (0, ±a) (c) (±a, 2a) (d) (±a, 0) 26. Area of triangle formed by a latusrectum and the lines joining the end points of the latusrectum and the centre of the conic is a 2 2 (a) (b) 2ab (b − a ) b (c) ab/2 (d) 4ab/3 Paragraph for Q. No. 27 to 29 y2 + = 1. S and S′ are a 2 b2 foci and e is the eccentricity of ellipse. ∠PSS′ = a and ∠PS′S = b P is any point of an ellipse
27. tan
x2
a b tan is equal to 2 2
2e 1− e 1− e (c) 1+ e
1+ e 1− e 2e (d) 1+ e
21. The equations of the common tangents of the curves x2 + 4y2 = 8 and y2 = 4x are (a) x + 2y + 4 = 0 (b) x – 2y + 4 = 0 (c) 2x + y = 4 (d) 2x – y + 4 = 0
(a)
22. A straight line touches the rectangular hyperbola 9x2 – 9y2 = 8 and the parabola y2 = 32x. An equation of the line is (a) 9x + 3y – 8 = 0 (b) 9x – 3y + 8 = 0 (c) 9x + 3y + 8 = 0 (d) 9x – 3y – 8 = 0
28. Locus of incentre of triangle PSS′ is (a) an ellipse (b) hyperbola (c) parabola (c) circle
23. A hyperbola having the transverse axis of length 1 unit is confocal with the ellipse 3x2 + 4y2 = 12, then 2 x2 y2 1 (a) Equation of the hyperbola is − = 15 1 16 (b) Eccentricity of the hyperbola is 4 (c) Distance between the directrices of the hyperbola 1 is units 8 15 (d) Length of latus rectum of the hyperbola is 2 units
(b)
29. Eccentricity of conic, which is locus of incentre of triangle PSS′ (a)
e 1+ e
(b)
2e 1+ e
(c)
2e 1− e
(d)
e 1− e
Paragraph for Q. No. 30 to 32 A point P moves such that sum of the slopes of the normals drawn from it to the hyperbola xy = 16 is equal to the sum of ordinates of feet of normals. The locus of P is a curve C. mathematics today | FEBRUARY ‘17
67
30. The equation of the curve C is (a) x2 = 4y (b) x2 = 16y 2 (c) x = 12y (d) y2 = 8x
36. The equation of the curve on reflection of the ellipse
31. If the tangent to the curve C cuts the coordinate axis in A and B, then the locus of the middle point of AB is (a) x2 = 4y (b) x2 = 2y 2 (c) x + 2y = 0 (d) x2 + 4y = 0 32. Area of the equilateral triangle inscribed in a curve C having one vertex is the vertex of curve C. (a) 772 3 sq. units (b) 776 3 sq. units (c) 760 3 sq. units
(d) 768 3 sq. units
33. Match the following Consider the parabola y2 = 12x Column II
A. Tangent and normal at the extremities of the latus rectum intersect the x-axis at T & G respectively. The coordinates of middle point of T & G are
P.
B.
Variable chords of the parabola passing through a fixed point K on the axis, such that sum of the reciprocals of two parts of the chord through K, is a constant. Coordinates of K are
Q.
(3, 0)
C. AB and CD are the chords of a parabola which intersect at a point E on the axis. The radical axis of the two circles described on AB and CD as diameter always passes through the point
R.
(12, 0)
(0, 0)
2
circle
concentric
to
an
ellipse
cuts the ellipse at ‘P’ such that area of triangle P F1 F2 is 30 sq.units. If F1F2 = 13K where K ∈ Z then K = mathematics today | FEBRUARY‘17
x2
π y2 − = 1 is . Then the eccentricity of conjugate 2 2 3 a b hyperbola is x2
– 108y = 0
x2 T ≡ xx1 − 2a( y + y1) = 0 ⇒ xx1 − 54 y + 1 = 0 108 2 S2 ≡ y – 32x = 0 y2 T ≡ yy2 − 2a(x + x2 ) = 0 ⇒ yy2 − 16 x + 2 = 0 32 x1 54 −x12 54 = = 2 = r ⇒ x1 = 16r and y2 = r 16 y2 y2 2 −(16r) 9 \ =r ⇒ r =− 2 4 (54 / r) \
x1 = −36, y2 = −24, y1 =
4x 4y 17 + = 1 λ < passes through foci F1 and F2 289 λ2 2
68
x2 y2 + = 1 is 3K. Then K is equal to 9 5 38. The equation of Asymptotes of xy + 2x + 4y + 6 = 0 is xy + 2x + 4y + C = 0, then C = ___
1. (b) : S1 ≡
34. A line passing through (21, 30) and normal to the curve y = 2 x . If m is slope of the normal then m+6= 2
37. The area of the quadrilateral formed by the tangents at the end point of latus rectum to the ellipse
soLUtioNs
Integer Answer Type
35. A
is 16x2 + 9y2 + ax – 36y + b = 0 then the value of a + b – 125 =
39. If e is the eccentricity of the hyperbola (5x – 10)2 + 25e (5y + 15)2 = (12x – 5y + 1)2 then is equal to 13 40. If the angle between the asymptotes of hyperbola
Matrix – Match Type
Column I
(x − 4)2 ( y − 3)2 + = 1 about the line x – y – 2 = 0 16 9
(−36)2 (−24)2 = 12, x2 = = 18 108 32
\ Equation of common tangent is −36 ( y − 12) = (x + 36) ⇒ 2x + 3 y + 36 = 0 54 2. (a) : xcosa + ysina – p = 0 2ax – yy1 + 2a(x1 + 2a) = 0 −p cos a sin a From (i) and (ii), = = − y1 2a(x1 + 2a) 2a
...(i) ...(ii)
⇒ y1 = –2a tana and x1 = –pseca – 2a \ y2 = 4a(x + a) ⇒ 4a2 tan2a = –4a(pseca + a) ⇒ pcosa + a = 0
3. (a) : Using formula q = tan −1 \
3a1/3b1/3
2(a2/3 + b2/3)
q = tan −1
, where a = 1 and b = 8
6 3 = tan −1 5 2(1 + 4)
4. (b) : L e t t h e c o o r d i n a t e s o f P a n d Q b e (at12 , 2at1) and (at22 , 2at2 ) respectively. Then y1 = 2at1 and y2 = 2at2. The coordinates of the point of intersection of the tangents at P and Q are {at1t2, a(t1 + t2)} \ y3 = a(t1 + t2) y +y ⇒ y3 = 1 2 ⇒ y1, y3 and y2 are in A.P. 2 1 5. (c) : Any tangent to y2 = 4x is y = mx + . It touches m 1 3m + m the circle, if 3 = 1 + m2 1 ⇒ 9(1 + m ) = 3m + m 1 ⇒ 2 =3 m 2
\ m=±
2
1
Common tangent is, y =
1 3
x+ 3 ⇒
3y = x + 3
6. (b) 7. (c) : Given parabola is y2 = 2ax \ Focus (a/2, 0) and directrix is given by x = –a/2, as circle touches the directrix. \ Radius of circle = distance from the point (a/2, 0) y to the line x = –a/2 2
a a = + =a 2 2
(–a/2, 0) 2
(a/2, 0)
Equation of circle be x − a + y 2 = a2 2 Also y2 = 2ax a 3a Solving (i) and (ii) we get, x = , − 2 2 \
x2 y2 8. (a) : Let the equation of ellipse be 2 + 2 = 1 a b Q It passes through (– 3, 1) 9 1 a2 So, 2 + 2 = 1 ⇒ 9 + 2 = a2 a b b Given eccentricity is 2 / 5
...(i)
2 b2 b2 3 So, = 1 − 2 ⇒ 2 = 5 5 a a
...(ii)
32 32 From equation (i) and (ii), a2 = , b2 = 3 5 Hence, required equation of ellipse is 3x2 + 5y2 = 32 9. (c) : Let point be (h, k). Their pair of tangent will be x 2 y 2 h2 k 2 hx yk 2 2 + 2 − 1 2 + 2 − 1 = 2 + 2 − 1 b b a b a a
3 For the common tangent to be above the x-axis, 1 m= 3 \
Putting these values in y2 = 2ax, we get, a y = ±a at x = 2 3a and y = a −3 at x = − 2 [Which is imaginary value of y] \ Required points are (a/2, ±a)
x
...(i) ...(ii)
Pair of tangents will be perpendicular, if coefficient of x2 + coefficient of y2 = 0 ⇒
k2
h2 1 1 + = 2 + 2 ⇒ h 2 + k 2 = a 2 + b2 2 2 2 2 ab ab a b
Replace (h, k) by (x, y) ⇒ x2 + y2 = a2 + b2 10. (c) : In the first case, e = 1 − (25 / 169) In the second case, e′ = 1 − (b2 / a2) According to the given condition, 1 − b2 / a2 = 1 − (25 / 169) ⇒ b / a = 5 / 13 (Q a > 0, b > 0) ⇒ a / b = 13 / 5 11. (c) : Equation x2 + 2y2 – 2x + 3y + 2 = 0 can be written as 2
3 y+ 3 1 (x − 1) (x − 1) 4 = 1, +y+ = ⇒ + 2 4 16 (1 / 8) (1 / 16) 2
2
2
1 1 which is an ellipse with a2 = and b2 = 8 16 1 1 1 1 1 \ = (1 − e 2 ) ⇒ = (1 − e 2 ) ⇒ e = 16 8 16 8 2 mathematics today | FEBRUARY ‘17
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12. (a) : Conjugate axis is 5 and distance between foci = 13 ⇒ 2b = 5 and 2ae = 13 Now, also we know for hyperbola 25 (13)2 2 = 2 (e − 1) 4 4e 25 169 169 169 13 ⇒ = − 2 or e 2 = ⇒ e= 4 4 4e 144 12 5 ⇒ a = 6, b = 2
b2 = a2(e 2 − 1) ⇒
\
Required equation is 25x2 – 144y2 = 900
13. (b) : 2a = 10, ⇒ a = 5 8 13 ae − a = 8 or e = 1 + = 5 5 \ b=5
132 2
−1 = 5 ×
12 = 12 5
5 and centre of hyperbola ≡(5, 0) (x − 5)2 ( y − 0)2 \ Required equation of hyperbola is − =1 25 144
14. (a) : Foci are (6,4) and (–4,4), e = 2 \
6− 4 4+ 4 , Centre is = (1, 4) 2 2
5 5 ⇒ 6 = 1 + ae ⇒ ae = 5 ⇒ a = and b = ( 3) 2 2
2 2 Hence, the required equation is (x − 1) − ( y − 4) = 1 (25 / 4) (75 / 4) or 12x2 – 4y2 – 24x + 32y – 127 = 0
15. (a) : Given equation of hyperbola is 9x2 – 16y2 + 72x – 32y – 16 = 0 ⇒ 9(x2 + 8x) – 16(y2 + 2y) – 16 = 0 ⇒ 9(x + 4)2 – 16(y + 1)2 = 144 (x + 4)2 ( y + 1)2 − =1 16 9 2b2 9 9 =2× = Therefore, latus rectum = 4 2 a ⇒
16. (a) : The line through (6,2) is y – 2 = m(x – 6) ⇒ y = mx + 2 – 6m Now from condition of tangency, (2 – 6m)2 = 25m2 – 16 ⇒ 36m2 + 4 – 24m – 25m2 + 16 = 0 ⇒ 11m2 – 24m + 20 = 0 If m1, m2 are its roots then 24 20 m1 + m2 = and m1m2 = 11 11 70
mathematics today | FEBRUARY‘17
2 2 17. (d) : The given ellipse is x + y = 1. The value 9 4 x2 y2 of the expression + − 1 is positive for x = 1, 9 4 y = 2 and negative for x = 2, y = 1. Therefore P lies outside E and Q lies inside E. The value of the expression x2 + y2 – 9 is negative for both the points P and Q. Therefore P and Q both lie inside C. Hence P lies inside C but outside E. 18. (b) : Any normal to the hyperbola is ax by ...(i) + = a 2 + b2 sec q tan q But it is given by lx + m′y – n = 0 ...(ii) Comparing (i) and (ii), we get a n b n sec q = 2 2 and tan q = l a +b m′ a2 + b2
b2 (a2 + b2 )2 − = ...(iii) l 2 m′2 n2 25 3 Since a2 = 16, b2 = 9, l = m, m′ = –1 and n = 3 2 . \ On substituting in (iii), we get m = ± 3 19. (c) : Solving equations x2 + y2 = 5 and y2 = 4x we get, x2 + 4x – 5 = 0 i.e., x = 1, – 5 For x = 1; y2 = 4 ⇒ y = ± 2 For x = –5 ; y2 = –20 (imaginary values) \ Points are (1, 2)(1, –2) m1 for x2 + y2 = 5 at (1, 2) is dy x 1 =− =− dx y (1, 2) 2
Hence eliminating q, we get
a2
Similarly, m2 for y2 = 4x at (1, 2) is 1 1 − −1 m1 − m2 \ tan q = = 2 =3 1 1 + m1m2 1− 2 2 20. (c) : Q 2b = 9 ⇒ 2b2 = 9a2 a2 9 4 Now b2 = a2(e 2 − 1) = a2 ⇒ a = b 16 3
From (i) and (ii), b = 6, a = 8 Hence, equation of hyperbola
x2 y2 − =1 64 36
...(i) ...(ii) 5 Q e = 4
21. (a, b) :
x2 y2 + = 1, y 2 = 4x 8 2
Any tangent to parabola is y = mx +
1 m
If this line is tangent to ellipse then 1
m2
= 8m2 + 2 ⇒ 8m4 + 2m2 − 1 = 0
−2 ± 4 + 32 −2 ± 6 = 16 16 1 1 ⇒ m2 = ⇒ m = ± 4 2 x x y = + 2 or y = − − 2 2 2 x – 2y + 4 = 0 or x + 2y + 4 = 0 \
m2 =
22. (b, c) : y2 = 32x
8 Let equation of tangent be y = mx + m 64 8 2 8 = m − 9 m2 9 m = ± 3, y = ± 3x ± 8/3
2 2 23. (b, c, d) : Ellipse is x + y = 1 4 3 3 1 Here, = 1 − e 2 ⇒ e = 4 2 Foci are (± 1, 0) Now the hyperbola is having same focus i.e. (± 1, 0) Let e1 be the eccentricity of hyperbola 2ae1 = 2 1 But 2a = So, e1 = 4 2 1 15 b2 = a2 e 2 − 1 = (16 − 1) = 16 16 So, the equation of the hyperbola is
(
2
2
1
)
2
2
1 x y x y − =1⇒ − = 1 15 1 15 16 16 16 Its distance between the directrices 2a 1 1 = = = units e1 2 × 4 8 2b2 2 × 15 × 4 15 = = units \ Length of latusrectum = 16 × 1 2 a (24 - 26) : 24. (c) 25. (d) 26. (a) x 2 + y 2 = a 2 ; x 2 + y 2 = b 2 ; b > a > 0, A = (–a, 0); B = (a, 0)
Let (h, k) be a point on the locus. Any tangent to circle x2 + y2 = b2 is xcosq + ysinq = b \ Equation of parabola is (x − h)2 + ( y − k)2 =| x cos q + y sin q − b | i.e., (x – h)2 + (y – k)2 = (xcosq + ysinq – b)2 The points (±a, 0) satisfy this equation \ (a – h)2 + k2 = (acosq – b)2 (a + h)2 + k2 = (acosq + b)2 Subtracting (1) from (2), we get h = bcosq 2 ax \ Required locus is (a + x)2 + y 2 = + b b 2 2 x y i.e., + 2 2 = 1 which is an ellipse. 2 b b −a 27. (c) : or
PS PS′ 2ae = = sin b sin a sin(π − (a + b)
2a 2ae = sin a + sin b sin(a + b)
a +b a −b 1 2 sin 2 .cos 2 ⇒ = e 2 sin a + b cos a + b 2 2 1− e a b \ = tan tan 1+ e 2 2
y β
S′ O
...(1) ...(2)
P α S
b 28 (a) : y − 0 = tan (x + ae) 2 a y − 0 = − tan (x − ae) 2 1− e 2 2 2 or, y 2 = − [x − a e ] 1 + e
x
... (i) ... (ii)
2 y2 1− e 2 2 1 − e 2 2 or, x + =1 ⇒ + = x y a e 1 + e 1+ e a2e 2 1 − e a2e 2 1 + e
which is clearly an ellipse.
1− e 2e = 1+ e 1+ e 4 30. (b) : Any point on the hyperbola xy = 16 is 4t , t of the normal passes through P(h, k), then k – 4/t = t2(h – 4t) ⇒ 4t4 – t3h + tk – 4 = 0 h \ ∑ t1 = and St1t2 = 0 4 k ∑ t1t2t3 = − 4 and t1t2t3t 4 = −1 1 1 1 1 k \ + + + = ⇒ y1 + y2 + y3 + y 4 = k t1 t2 t3 t 4 4 29. (b) : e ′ = 1 −
mathematics today | FEBRUARY ‘17
71
h2 Now, = =k 16 ⇒ Locus of (h, k) is x2 = 16y. 31. (c) : x2 = 16y Equation of tangent of P is t12 + t22 + t32 + t 42
2
16( y + t ) 2 4tx = 8y + 8t2 tx = 2y + 2t2 A = (2t, 0), B = (0, –t2) M(h, k) is the middle point of AB x ⋅ 4t =
(4t, t2) P A B M
t2 ⇒ 2k = −h2 2 Locus of M(h, k) is x 2 + 2 y = 0.
h = t, k = −
4t 4 32. (d) : tan 30° = 21 = t1 t1 1 4 ⇒ = ⇒ t1 = 4 3 3 t1
(–4t,
t2
1
)A
(0, 3) B
O (0, 0)
3 Area of ∆OAB = × 32 3 × 32 3 = 768 3 sq.units 4 33. A–Q, B–Q, C–P (A) Equation of tangent at (3, 6) : y = x + 3 \ T( – 3, 0) Equation of normal at (3, 6) : y = –x + 9 \ G( 9, 0) Hence middle point is (3, 0) (B) Point is obviously focus (3, 0) (C) Let A(t1) and B(t2), C(t3) and D(t4) If AB and CD intersect at a point E on the axis, then by solving the equations of AB and CD we get the relation t1t2 = t3t4 Now equations of the circles with AB and CD as diameters are (x − at12 )(x − at22 ) + ( y − 2at1)( y − 2at2 ) = 0 (x − at32 )(x − at 42 ) + ( y − 2at3)( y − 2at 4 ) = 0 If we solve these two circles, then the equation of their radical axis is of the form y = mx. So it passes through the origin. 34. (1) : Equation of the normal is y = mx – 2m – m3 If it pass through (21, 30) we have 30 = 21m – 2m – m3 ⇒ m3 – 19m + 30 = 0 Then m = –5, 2, 3 But if m = 2 or 3 then the point where the normal mathematics today | FEBRUARY‘17
Y
B (+4t, t2 ) 1 30°
AB = 8t1 = 32 3
72
meets the curve will be (am2 , –2am) where the curve does not exist. Therefore m = –5 \ m+6=1 35. (1) : Since F1 and F2 are the ends of the diameter 1 1 Area of ∆PF1F2 = ( F1P )( F2P ) = x (17 − x ) = 30 2 2 ⇒ x = 5 or 12 ⇒ F1F2 = 13 \ K = 1 36. (7) : Let P(4, 0) and Q(0, 3) are two points on given ellipse E1 P1 and Q1 are images of P,Q w.r.t x – y – 2 = 0 \ P1(2, 2) Q1(5, –2) lies on E2 \ a = –160, b = 292 ⇒ a + b – 125 = 7 37. (9) : Eccentricity of given ellipse e = 2/3 2x y Equation of tangent at L is + = 1 . It meets 9 3
(–2, 0)
5 L (2, 3 )
(0, 0) (2, 0)
9 ( 2 , 0)
X
9 x-axis at A , 0 and y axis at B(0, 3) 2 1 9 \ Area = 4 ⋅ ⋅ 3 = 27 \ K = 9. 2 2 38. (8) : xy + 2x + 4y + C = 0 represents pair of lines ⇒C=8 39. (5) : Equation can be rewritten as 13 12x − 5 y + 1 13 (x − 2)2 + ( y + 3)2 = So, e = . 5 13 5 25e \ =5 13 b π 40. (2) : 2 tan −1 = a 3 b 1 , e2 = 1 + 1 = 4 = a 3 3 3 Let eccentricity of conjugate hyperbola be e′. 1 3 1 1 1 1 \ + 2 = 1 ⇒ 2 + = 1 ⇒ 2 = ⇒ e′ = 2 2 4 4 e′ e′ e′ e
8
T
Class XI
his specially designed column enables students to self analyse their extent of understanding of specified chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness.
STATISTICS AND PROBABILITY Total Marks : 80
Time Taken : 60 Min.
Only One Option Correct Type 1. A five digit number be selected at random. The probability that the digits in the odd places are odd and the even places are even (repetition is not allowed) 5 5 P2 × 5 P3 P2 × 5 P3 (a) (b) 104 × 9 105 5
5
C2 × 5C3
C2 × 5C3
×2 (d) 9 × 104 104 × 9 2. The mean of the data is 30 and data are given as (c)
C.I. 0-10 10-20 20-30 30-40 40-50 Frequency 8 10 f1 15 f2 If total of frequencies is 70, then the missing numbers are (a) 14, 23 (b) 25, 21 (c) 24, 20 (d) none of these 4 3. The probability that A speaks truth is , B speaks 5 truth is 3 . The probability they contradict each 4 other is 4 7 3 (a) (b) 1 (c) (d) 5 20 20 5 4. In a series there are 2n observations half of them equal to ‘a’ and half equal to – a. If the standard deviation of the observations is 2, then |a| equals 2 1 (d) 2 n n 5. If ten rupee coins which are 10 in number and five rupee coins which are 5 in number are to be placed in a line, then the probability that the extreme coins are five rupee coins is (a) 2
(b)
(c)
1 1 (b) 10! 15! 5!10 ! (c) (d) none of these 15! 6. In an experiment with 15 observations on x, the following results were available : Σx2 = 2830 and Σx = 170. One observation i.e., + 20 was found to be wrong and was replaced by the correct value 30, then the corrected variance is (a) 188.66 (b) 177.33 (c) 8.33 (d) 78.00 One or More Than One Option(s) Correct Type (a)
7. The probabilities that a student pass in Mathematics, Physics and Chemistry are a, b and g respectively. Of these subjects, student has a 75% chance of passing in atleast one, a 50% chance of passing in atleast two and a 40% chance of passing in exactly two subjects. Which of the following relations are true? (a) a + b +g = 19/20 (b) a + b + g = 27/20 (c) abg = 1/10 (d) abg = 1/4 8. The variable x takes two values x1 and x2 with frequencies f1 and f2, respectively. If s denotes the standard deviation of x, then f1x12 + f 2 x22 f1x1 + f 2 x2 − f1 + f 2 f1 + f 2 f1 f 2 (x − x )2 (b) s 2 = ( f1 + f 2 )2 1 2
2
2 (a) s =
(x1 − x2 )2 (d) None of these f1 + f 2 9. If A and B are two events such that P(A) = 1/2 and P(B) = 2/3, then (a) P(A∪B) ≥ 2/3 (b) P(A∩B′)≤1/3 (c) 1/6 ≤P(A∩B) ≤ 1/2 (d) 1/6 ≤P(A′∩B)≤1/2 (c) s 2 =
mathematics today | February‘17
73
10. For two data sets, each of size 5, the variance are given to be 4 and 5 and the corresponding means are given to be 2 and 4, respectively. The variance of the combined data set is 13 11 5 (a) (b) (c) 6 (d) 2 2 2 11. The letters of the word PROBABILITY are written down at random in a row. Let E1 denotes the event that two I’s are together and E2 denotes the event that two B’s are together, then 2 2 (a) P(E1) = P(E2) = (b) P(E1∩E2) = 11 55 18 (c) P(E1∪E2) = (d) none of these 55 12. There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test Marks 0 1 2 3 4 5 2 2 Frequency x–2 x x (x + 1) 2x x + 1 where, x is a positive integer. Then, (a) Mean = 2.8 (b) Variance = 1.12 (c) Standard Deviation = 1.12 (d) None of these
Matrix Match Type 16. Match the following: Column I Column II P. Two balls are drawn from an urn containing 8 2 white, 3 red and 4 black balls one by one 1. m =
Q. A bag contains 2 white and 4 black balls while 11 another bag contains 6 white and 4 black balls. 2. m = 12 A bag is selected at random and a ball is drawn. If l be the probability that the ball drawn is of white colour and m be the probability that the ball drawn is black colour, then R. Bag A contains 4 red and 5 black balls and
5 bag B contains 3 red and 7 black balls. One 3. l = 18 ball is drawn from bag A and two from bag B. If l be the probability that out of 3 balls drawn two are black and one is red and m be the probability that out of three balls drawn two are red and one is black, then 4. m =
13. For two events A and B, P(A∩B) is (a) not less than P(A) + P(B) – 1 (b) not greater than P(A) + P(B) (c) equal to P(A) + P(B) – P(A∪B) (d) equal to P(A) + P(B) + P(A∪B) Comprehension Type
(a) (b) (c) (d)
A JEE aspirant estimates that she will be successful with an 80% chance if she studies 10 hours per day, with a 60% chance if she studies 7 hours per day and with a 40% chance if she studies 4 hours per day. She further believes that she will study 10 hours, 7 hours and 4 hours per day with probabilities 0.1, 0.2 and 0.7, respectively: 14. Given that she is successful, the probability that she studied for 4 hours, is 6 7 8 9 (a) (b) (c) (d) 12 12 12 12 15. Given that she does not achieve success, the probability she studied for 4 hours, is 18 21 19 20 (a) (b) (c) (d) 26 26 26 26
15
without replacement. If the probability that both the balls are of same colour is l and at least one ball is red is m, then
P 2 3 3 1
11 45
Q R 1 3 4 2 1 4 2 3 Integer Answer Type
17. If four squares are chosen at random on a chess board. If the probability that they lie on a diagonal xyz line is 64 , then the value of x + y – z must be C4 18. If the mean deviation about the median of the numbers a, 2a, ......., 50a is 50, then |a| equals 19. Two integers x and y are chosen (without random) at random from the set {x : 0 ≤ x ≤ 10,} x is an integer}. If the probability for |x – y| ≤ 5 is p, then the value of 121 p – 91 must be 20. If the variance of first 50 even natural numbers is abc. Then find the value of a + b – c. Keys are published in this issue. Search now! J
Check your score! If your score is No. of questions attempted …… No. of questions correct …… Marks scored in percentage ……
74
> 90%
excellent work ! You are well prepared to take the challenge of final exam.
90-75%
Good work !
You can score good in the final exam.
74-60%
satisfactory !
You need to score more next time.
< 60%
mathematics today | February ‘17
not satisfactory! Revise thoroughly and strengthen your concepts.
mathematics today | February‘17
75
Exam on 20th March 2017
PRACTICE PAPER 2017 Time Allowed : 3 hours
Maximum Marks : 100 GENERAL INSTRUCTIONS
(i) (ii) (iii) (iv) (v) (vi)
All questions are compulsory. This question paper contains 29 questions. Questions 1-4 in Section-A are very short-answer type questions carrying 1 mark each. Questions 5-12 in Section-B are short-answer type questions carrying 2 marks each. Questions 13-23 in Section-C are long-answer-I type questions carrying 4 marks each. Questions 24-29 in Section-D are long-answer-II type questions carrying 6 marks each.
section-a
1. Write the vector equation of a line passing through the point (1, –1, 2) and parallel to the line x − 3 y −1 z +1 . = = 1 2 −2 2. Evaluate :
∫
x3 − x2 + x − 1 dx x −1
3. If the binary operation * on the set of integers Z is defined by a * b = a + 3b2, then find the value of 2 * 4. x sin q cos q 4. If − sin q − x 1 = 8 , then find the value of x. cos q 1 x section-B
5. Prove that : 1 1 1 1 p tan −1 + tan −1 + tan −1 + tan −1 = . 3 5 7 8 4
2 0 1 6. If A = 2 1 3 , find A2 – 5A + 4I. 1 −1 0 76
mathematics today | february ‘17
7. Show that f : N → N, defined by n + 1 , if n is odd f (n) = 2 n , if n is even 2 is a many-one onto function. 8. For what value of k, the function defined by k(x 2 + 2), if x ≤ 0 f (x ) = if x > 0 3x + 1, is continuous at x = 0 ? 9. Form the differential equation of the family of curves y = a cos (x + b), where a and b are arbitrary constants. 10. If the radius of a sphere is measured as 9 cm with an error of 0.03 cm, then find the approximate error in calculating its surface area. 11. Let * be a binary operation on the set of rational numbers given as a * b = (2a – b)2, a, b ∈ Q. Find 3 * 5 and 5 * 3. Is 3 * 5 = 5 * 3 ? 12. If xy + y2 = tan x + y, find
dy . dx
section-c
13. Using integration, find the area of the region bounded by the curves y = x2 and y = x. 14. Solve the differential equation
dy + x 2 y 2 = 0. dx 15. In a parliament election, a political party hired a public relations firm to promote its candidates in three ways-telephone, house calls and letters. The cost per contact (in paise) is given in matrix A as 140 Telephone A = 200 House call 150 Letters The number of contacts of each type made in two cities X and Y is given in matrix B as Telephone House call Letters cosec x log y
500 5000 City X 1000 B= 3000 1000 10000 City Y Find the total amount spent by the party in the two cities. What should one consider before casting his/ her vote-party’s promotional activity or their social activities?
OR 1 2 2 If A = 2 1 2 , then verify that A2 – 4A – 5I = O. 2 2 1
Hence find A–1. 16. If a ,b and c are three vectors such that each one is perpendicular to the vector obtained by sum of the other two and a = 3, b = 4 and c = 5, then prove that a + b + c = 5 2 . 17. An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be at least 4 successes. 18. If y = 3cos(logx) + 4sin(logx), then show that d2 y dy x2 2 + x + y = 0 dx dx OR For what value of l, the function defined by l(x 2 + 2), if x ≤ 0 f (x ) = is continuous at if x > 0 4 x + 6, x = 0? Hence check the differentiability of f(x) at x = 0.
19. Find the equation of the plane passing through the intersection of the planes x + 3y + 6 = 0 and 3x – y – 4z = 0 and whose perpendicular distance from the origin is unity. 20. Evaluate :
cos x
∫ (1 − sin x)(2 − sin x) dx
OR Using properties of definite integrals prove that p
x tan x p2 dx = ∫ sec x cosec x 4 0
21. Prove that :
9 p 9 −1 1 9 −1 2 2 − sin = sin . 3 4 3 8 4
22. Find the intervals in which the following function is (a) increasing (b) decreasing : f(x) = x4 – 8x3 + 22x2 – 24x + 21 23. Using properties of determinants, show that
x+y x x 5x + 4 y 4 x 2 x = x 3 . 10 x + 8 y 8 x 3x
section-d
24. Find the equation of the plane through the line of intersection of the planes 2x + y – z = 3 and 5x – 3y + 4z + 9 = 0 x −1 y − 3 z − 5 and parallel to the line = = . 2 4 5 Also, find the distance of the plane from origin. OR Find the equation of the plane passing through the point A(1, 2, 1) and perpendicular to the line joining the points P(1, 4, 2) and Q(2, 3, 5). Also, find the distance of this plane from the line x +3 y −5 z −7 = = . 2 −1 −1 25. Find the particular solution of the differential dy equation x + y – x + xy cot x = 0; x ≠ 0, dx p given that when x = , y = 0. 2 1 −1 0 2 2 −4 26. If A = 2 3 4 and B = −4 2 −4 are two 0 1 2 2 −1 5 square matrices, find AB and hence solve the system of equations x – y = 3, 2x + 3y + 4z = 17 and y + 2z = 7. mathematics today | february ‘17
77
OR Two schools P and Q want to award their selected students on the values of discipline, politeness and punctuality. The school P wants to award ` x each, ` y each and ` z each for the three respective values to its 3, 2 and 1 students with a total award money of ` 1,000. School Q wants to spend ` 1,500 to award its 4, 1 and 3 students on the respective values (by giving the same award money for the three values as before). If the total amount of awards for one prize on each value is ` 600, using matrices, find the award money for each value. 27. A dealer in rural area wishes to purchase a number of sewing machines. He has only ` 5760 to invest and has space for at most 20 items. An electronic sewing machine costs him ` 360 and a manually operated sewing machine costs ` 240. He can sell an electronic sewing machine at a profit of ` 22 and a manually operated sewing machine at a profit of ` 18. Assuming that, he can sell all the items that he can buy, how should he invest his money in order to maximize his profit. Make it as a linear programming problem and solve it graphically. Keeping the rural background in mind justify the ‘values’ to be promoted for the selection of manually operated machine. 28. If the length of three sides of a trapezium other than base is 10 cm each, then find the area of the trapezium when it is maximum. 29. From a well shuffled pack of 52 cards, 3 cards are drawn one-by-one with replacement. Find the probability distribution of number of queens. OR Suppose the probability for A to win a game against B is 0.4. If A has an option of playing a ‘best of 3 games’ or a ‘best of 5 games’ match against B, which option should A choose so that the probability of his winning the match a higher? (No game ends in draw). solutions
1. Vector equation of the line passing through (1, –1, 2) x − 3 y −1 z +1 and parallel to the line is = = 1 2 −2 r = (i − j + 2k ) + l (i + 2 j − 2k ) 78
mathematics today | february ‘17
2. Let I = ∫
x3 − x2 + x − 1 dx x −1
x 2 (x − 1) + 1(x − 1) (x 2 + 1)(x − 1) =∫ dx x −1 x −1 1 = ∫ (x 2 + 1)dx = x 3 + x + C 3
=∫
3. Here, a * b = a + 3b2 \ 2 * 4 = 2 + 3(4)2 = 2 + 3 × 16 = 50 x sin q cos q 4. Given, − sin q − x 1 =8 cos q 1 x ⇒ ⇒ ⇒ ⇒ ⇒ ⇒
x(–x2 – 1) –sin q(–xsin q – cos q) + cos q (–sin q + xcos q) = 8 –x3 – x + xsin2q + sin q cos q – sin q cos q + xcos2q = 8 3 2 2 –x – x + x(sin q + cos q) = 8 –x3 – x + x = 8 ⇒ x3 + 8 = 0 (x + 2) (x2 – 2x + 4) = 0 ⇒ x + 2 = 0 [Q x2 – 2x + 4 > 0 "x] x = –2
1 1 1 −1 1 + tan −1 + tan −1 + tan −1 5. L.H.S. = tan 3 5 7 8
1 1 1 1 + + −1 7 8 3 5 + tan = tan 1 1 1 1 1− × 1− × 7 8 3 5 15 8 −1 15 −1 56 = tan + tan 14 55 56 15 4 3 = tan −1 + tan −1 11 7 65 4 3 + −1 7 11 −1 77 = tan = tan 65 1 − 4 × 3 77 7 11 p = tan −1 1 = = R.H.S. 4 −1
6. We have, A2 – 5A + 4I 2 0 1 2 0 1 1 0 0 2 0 1 = 2 1 3 2 1 3 −5 2 1 3 + 4 0 1 0 1 −1 0 1 −1 0 0 0 1 1 −1 0
5 −1 2 = 9 −2 5 − 0 −1 −2
dS = 8pr dr dS \ DS = Dr = 8pr Dr dr = 8p × 9 × 0.03 = 2.16 p cm2
10 0 5 4 0 0 10 5 15 + 0 4 0 5 −5 0 0 0 4
⇒
−1 −1 −3 = −1 −3 −10 −5 4 2
This is the approximate error in calculating surface area.
(1 + 1) 2 2 7. We have, f (1) = = = 1 and f (2) = = 1 2 2 2 Thus, f(1) = f(2) while 1 ≠ 2 \ f is many-one. In order to show that f is onto, consider an arbitrary element n ∈N. If n is odd, then (2n – 1) is odd (2n − 1 + 1) 2n \ f (2n − 1) = = =n 2 2 If n is even, then 2n is even 2n \ f (2n) = = n 2 Thus, for each n ∈ N (whether even or odd) there exists its pre-image in N. \ f is onto. Hence, f is many-one onto function.
11. We have, a * b = (2a – b)2 \ 3 * 5 = (2 × 3 – 5)2 = (6 – 5)2 = 1 5 * 3 = (2 × 5 – 3)2 = (10 – 3)2 = 49 Thus, 3 * 5 ≠ 5 * 3
12. Given, xy + y2 = tan x + y Differentiating w.r.t. x, we get dy dy dy x + y + 2y = sec2 x + dx dx dx 2 dy sec x − y dy ⇒ (x + 2 y − 1) = sec2 x − y ⇒ = dx x + 2 y − 1 dx 13. The given curves are y = x2 ... (i) and y = x ...(ii)
8. We have, lim f (x ) = lim k(h2 + 2) = 2k x →0−
h →0
The point of intersection of (i) and (ii) are A (1, 1) and O(0, 0). \ Required area = Area of shaded region
lim f (x ) = lim (3h + 1) = 1 and f(0) = 2k
x →0+
h →0
As f(x) is continuous at x = 0 \
x → 0−
x → 0+
⇒ 2k = 1 ⇒ k =
1 2
9. Here, y = a cos (x + b) Differentiating (i) w.r.t. x, we get dy = −a sin ( x + b ) dx Again differentiating w.r.t. x, we get d2 y dx 2
⇒
1
lim f ( x ) = lim f ( x ) = f (0)
d2 y dx 2
...(i)
= −a cos ( x + b ) = −y ⇒
d2 y dx 2
+ y = 0.
10. Let r be the radius of sphere and Dr be the error in measuring the radius. Then, r = 9 cm, Dr = 0.03 cm Now, surface area (S) of the sphere is given by 4pr2
1
3 2 1 1 1 = ∫ (x − x )dx = x − x = − = sq.unit. 2 3 0 2 3 6 0 dy 14. We have, cosec x log y + x 2 y 2 = 0 dx log y x2 dy + dx = 0 cosec x y2 2
Integrating both sides, we get log y 2 ∫ y 2 dy + ∫ x sin x dx = 0 1 Put log y = t ⇒ dy = dt and y = et y
Solution Sender of Maths Musing set-169 1. N. Jayanthi Hyderabad 2. ravinder Gajula Karimnagar mathematics today | february ‘17
79
⇒
∫ t. e
⇒ t.
⇒
−t
dt + ∫ x 2 sin x dx = 0
e −t e −t − ∫ 1. dt + x 2 (− cos x ) − ∫ 2x ( − cos x ) dx = C −1 −1
–t e–t –e–t –x2 cos x + 2x sin x −2 ∫ 1 ⋅ sin x dx = C
⇒ −
1 + log y − x 2 cos x + 2 x sin x + 2 cos x = C y
is the required solution. 15. The total amount spent by the party in two cities X and Y is represented in the matrix equation by matrix C as, C = BA 140 X 1000 500 5000 ⇒ = 200 Y 3000 1000 10000 150 ⇒
\
X 1000 × 140 + 500 × 200 + 5000 × 150 = Y 3000 × 140 + 1000 × 200 + 10000 × 150 990000 = 2120000 X = ` 990000 and Y = ` 2120000
Thus, amount spent by the party in city X and Y is ` 990000 and ` 2120000 respectively. One should consider about the social activity before casting his/her vote. OR 1 2 2 Given, A = 2 1 2 2 2 1 1 2 2 1 Now, A2 = 2 1 2 2 2 2 1 2 2 \ A – 4A – 5I 1 9 8 8 = 8 9 8 − 4 2 2 8 8 9
2 2 9 8 8 1 2 = 8 9 8 2 1 8 8 9
1 2 2 1 2 − 5 0 2 1 0 9 8 8 4 8 8 5 0 = 8 9 8 − 8 4 8 − 0 5 8 8 9 8 8 4 0 0 0 0 0 = 0 0 0 = O 0 0 0
80
mathematics today | february ‘17
0 0 1 0 0 1 0 0 5
Now, A2 – 4 A – 5 I = O Pre-multiplying by A–1 both sides, we get (A–1 A)A – 4 (A–1A) – 5(A–1 I) = O ⇒ IA – 4I = 5A–1 −3 2 2 1 −1 1 ⇒ A = (A − 4 I ) = 2 −3 2 5 2 2 2 −3 16. Given, a = 3, b = 4, c = 5 ...(i) ( ) ( ) and a ⋅ b + c = 0 , b ⋅ (c + a ) = 0 , c ⋅ a + b = 0 \ a ⋅b + a ⋅c + b ⋅c + b ⋅a + c ⋅a + c ⋅b = 0 ...(ii) ⇒ 2(a ⋅ b + b ⋅ c + c ⋅ a ) = 0 2 Now, a + b + c 2 2 2 = a + b + c + 2 ( a ⋅ b ) + 2 (b ⋅ c ) + 2 ( c ⋅ a ) = 32 + 42+ 52 + 0 = 50 [Using (i) and (ii)] \ a +b + c = 5 2. 2 2 = 17. Here, p = P (success of the experiment) = 2 +1 3 2 1 \ q = 1− p = 1− = 3 3 Also n = 6 Let X denote the number of successes. \ Required probability P(X ≥ 4) = P(X = 4) + P(X = 5) + P(X = 6) 2
4
1
5
0
1 2 1 2 1 2 = 6C4 + 6C5 + 6C6 3 3 3 3 3 3 16 32 64 = 15 × 6 + 6 × 6 + 6 = 240 + 192 + 64 = 496 3 3 3 729 36 18. We have, y = 3 cos (log x) + 4 sin (log x) Differentiating w.r.t. x, we get 1 1 dy = −3 sin(log x ) × + 4 cos(log x ) × x x dx dy ⇒ x = −3 sin(log x ) + 4 cos(log x ) dx Again differentiating w.r.t. x, we get x ⇒ x2
⇒ x2
d2 y dx
2
d2 y dx 2
d2 y dx 2
+
dy 1 1 = − 3 cos(log x ) × − 4 sin(log x ) × dx x x
+x
dy = −[3 cos(log x ) + 4 sin(log x )] dx
+x
2 dy dy 2d y + x + y =0 =− y ⇒x 2 dx dx dx
6
OR
dt −dt + = − log 1 − t + log 2 − t + C 1−t ∫ 2 −t 2−t 2 − sin x = log + C = log +C 1−t 1 − sin x
I=∫
\
l(x 2 + 2), if x ≤ 0 Here, f (x ) = if x > 0 4 x + 6,
At x = 0, f(0) = l(02 + 2) = 2l L.H. L. = lim f (x ) = lim l[(−h)2 + 2] = 2l x →0
−
OR p
h →0
Let I = ∫
R.H.L. = lim f (x ) = lim[4(h) + 6] = 6 x →0
+
h →0
For f to be continuous at x = 0, 2l = 6 ⇒ l = 3. Hence, the function becomes 3(x 2 + 2), if x ≤ 0 f (x ) = if x > 0 4 x + 6, \
f ′(0− ) = lim
h →0
f (0 − h) − f (0) 0−h
f (0 + h) − f (0) 0+h
h →0
⇒ \
–
I=∫ 0
0
0
= lim
0
p
4h + 6 − 6
h →0
h
0 p
=4
f ′(0 ) ≠ f ′(0 ) f is not differentiable at x = 0.
= p∫ ⇒ I=
26 = 26l2 ⇒ l2 = 1 ⇒ l = ±1 Equation of the required plane is r ⋅ (4i + 2 j − 4k ) + 6 = 0 or r ⋅ (−2i + 4 j + 4k ) + 6 = 0
⇒ \
21. L.H.S. =
0 2
cos x dx (1 − sin x )(2 − sin x ) Put sinx = t ⇒ cosxdx = dt dt \ I=∫ (1 − t )(2 − t ) 1 A B We write, = + (1 − t )(2 − t ) 1 − t 2 − t
...(1)
...(2)
p
tan x dx = p ∫ sin2 x dx sec x cos ec x 0
p
p sin 2 x p2 1 − cos 2 x = dx = x − 2 2 2 0 2
p 4
9 p 9 −1 1 9 p 1 − sin = − sin −1 3 8 4 4 2 3 9 1 = cos −1 3 4
p −1 −1 Qsin x + cos x = 2
Q cos −1 x = sin −1 1 − x 2 for 0 ≤ x ≤ 1 8 9 −1 2 2 = sin = R.H.S. 3 9 4
9 1 = sin −1 1 − 4 9 9 = sin −1 4
20. Let I = ∫
p
( p − x )tan(p − x ) ( p − x )tan x dx = ∫ dx sec( p − x )cosec (p − x ) sec x cos ec x
+
1 = A (2 – t) + B(1 – t) Putting t = 1 in (1), we get 1 = A(2 – 1) ⇒ A = 1 Putting t = 2 in (1), we get 1 = B(1 – 2) ⇒ B = –1
a
2I = p∫
19. Equation of plane passing through intersection of given planes is x + 3y + 6 + l(3x – y – 4z) = 0 ⇒ r ⋅[(1 + 3 l) i + (3 − l)j − 4 lk ] + 6 = 0 Q Distance of plane from origin is unity. 6 \ =1 2 (1 + 3l) + (3 − l)2 + 16 l2
⇒
p
...(1)
∫ f (x)dx = ∫ f (a − x)dx, we get
Adding (1) and (2), we get
3(h2 + 2) − 6 = lim (−3h) = 0 = lim −h h →0 h →0 and f ′(0+ ) = lim
Using
0 a
x tan x dx sec x cosec x
22. The given function is f(x) = x4 – 8x3 + 22x2 – 24x + 21 ⇒ f ′(x) = 4x3 – 24x2 + 44x – 24 = 4(x3 – 6x2 + 11x – 6) = 4(x – 1)(x2 – 5x + 6) = 4(x – 1)(x – 2)(x – 3) Thus f ′(x) = 0 ⇒ x = 1, 2, 3. Hence, possible disjoint intervals are (–∞, 1), (1, 2), (2, 3) and (3, ∞). In the interval (–∞, 1), f ′ (x) < 0 In the interval (1, 2), f ′(x) > 0 In the interval (2, 3), f ′(x) < 0 In the interval (3, ∞), f ′(x) > 0 \ f is increasing in (1, 2) ∪ (3, ∞) and f is decreasing in (–∞, 1) ∪ (2, 3). mathematics today | february ‘17
81
x+y x x 23. L.H.S. = 5x + 4 y 4 x 2 x . 10 x + 8 y 8 x 3x
Since each element in the first column of determinant is the sum of two elements, therefore, determinant can be expressed as the sum of two determinants given by x x x y x x 5x 4 x 2 x + 4 y 4 x 2 x 10 x 8 x 3x 8 y 8 x 3x Taking x common from R1, R2, R3 in first determinant and x common from C2, C3, y common from C1 in second determinant, we get 1 1 1 1 1 1 1 1 1 3 2 3 x 5 4 2 + yx 4 4 2 = x 5 4 2 + yx 2 ⋅ 0 10 8 3 8 8 3 10 8 3 (Q C1 and C2 are identical in the second determinant) Applying C1 → C1 – C3 and C2 → C2 – C3, we get 0 0 1 3 3 x 3 3 2 2 = x ⋅1⋅(15 – 14) = x = R.H.S. 24. The equation of the plane passing through the line of intersection of the planes 2x + y – z = 3 and 5x – 3y + 4z + 9 = 0 is (2x + y – z – 3) + l(5x – 3y + 4z + 9) = 0 ⇒ x(2 + 5l) + y(1 – 3l) + z(4l – 1) + 9l – 3 = 0 ...(i) Since, plane (i) is parallel to the line x −1 y − 3 z − 5 = = 2 4 5 \ 2(2 + 5l) + 4(1 – 3l) + 5(4l – 1) = 0 1 ⇒ 18l + 3 = 0 ⇒ l = − 6 Putting the value of l in (i), we obtain 7x + 9y – 10z – 27 = 0 This is the equation of the required plane. Now, distance of the plane 7x + 9y – 10z – 27 = 0 from the origin is, −27 27 unit = 72 + 92 + 102 230 OR The line joining the given points P(1, 4, 2) and Q(2, 3, 5) has direction ratios <1 –2, 4 –3, 2 – 5> i.e., <– 1, 1, –3> The plane through (1, 2, 1) and perpendicular to the line PQ is –1(x – 1) + 1(y – 2) – 3(z – 1) = 0 mathematics today | february ‘17
x – y + 3z – 2 = 0
x +3 y −5 z −7 Now, direction ratios of line = = 2 −1 −1 are 2, –1, –1. Now, 2(1) + (–1) (–1) + (3) (–1) = 2 + 1 – 3 = 0 \ Line is parallel to the plane. Since, (–3, 5, 7) lies on the given line. \ Distance of the point (–3, 5, 7) from plane is −3 − 5 + 3(7) − 2 11 d= ⇒ d= = 11 units. 1+1+ 9 11 dy 25. We have, x + y − x + xy cot x = 0,(x ≠ 0) dx dy ⇒ x + (1 + x cot x ) ⋅ y = x dx dy 1 + x cot x .y = 1 ⇒ + dx x dy This is linear D.E. of the form + Py = Q dx 1 + x cot x 1 where P = = + cot x , Q = 1 x x Pdx log (x sin x) ∫ Now, I.F. = e =e = x sin x \
7 5 3
82
⇒
...(i)
The solution of (i) is
y ⋅ x sin x = ∫ 1⋅ x sin x dx + C
= x(− cos x ) + ∫ 1 ⋅ cos x dx + C
...(ii) p Putting x = , y = 0 in (ii), we get 2 p p p 0 = − cos + sin + C ⇒ C = −1 2 2 2 \ xy sinx = sinx – xcos x – 1 is the required particular solution. ⇒ x y sin x = − x cos x + sin x + C
1 −1 0 2 2 −4 26. Here, A = 2 3 4 ; B = −4 2 −4 0 1 2 2 −1 5 1 −1 0 2 2 −4 6 0 \ AB = 2 3 4 ⋅ −4 2 −4 = 0 6 0 1 2 2 −1 5 0 0 ⇒
0 0 = 6I 6
1 1 A ⋅ B = I ⇒ A is invertible and A−1 = B 6 6
Now, the given system of equations is x–y=3 2x + 3y + 4z = 17 y + 2z = 7 The system of equations can be written as AX = P
mathematics today | february ‘17
83
1 −1 0 x 3 where, A = 2 3 4 ; X = y ; P = 17 0 1 2 z 7
Since A–1 exists, so system of equations has a unique solution given by X = A–1 P x 2 2 −4 3 1 1 ⇒ y = BP = −4 2 −4 17 6 6 z 2 −1 5 7 12 2 1 = −6 = −1 6 24 4 ⇒ x = 2, y = – 1, z = 4.
84
mathematics today | february ‘17
3 2 1 A = 4 1 3 = −5 ≠ 0 1 1 1
A is invertible and system of equations has a unique solution given by X = A–1 B Now, A11 = –2, A12 = –1, A13 = 3, A21 = –1, A22 = 2, A23 = –1, A31 = 5, A32 = –5, A33 = –5 \
OR
According to question, we have, 3x + 2y + z = 1000
4x + y + 3z = 1500 ...(ii) x + y + z = 600 ...(iii) The given system of equations can be written as AX = B 3 2 1 x 1000 where, A = 4 1 3 , X = y and B = 1500 1 1 1 z 6000
...(i)
\
−2 −1 5 adj A = −1 2 −5 3 −1 −5
\ A −1
−2 −1 5 adj( A) −1 = = −1 2 −5 | A| 5 3 − 1 −5
Now, X = A–1B x −2 −1 5 1000 −500 −1 −1 ⇒ y = −1000 −1 2 −5 1500 = 5 5 z 3 −1 −5 600 −1500 x 100 ⇒ y = 200 z 300 ⇒ x = 100, y = 200, z = 300
Hence, the money awarded for discipline, politeness and punctuality are ` 100, ` 200 and ` 300 respectively. 27. Suppose, number of electronic sewing machines purchased = x and number of manually operated sewing machines purchased = y Mathematical formulation of given problem is : Maximize Z = 22x + 18y Subject to constraints : x + y ≤ 20 ...(i) 360x + 240y ≤ 5760 or 3x + 2y ≤ 48 ... (ii) x ≥ 0, y ≥ 0 Solving equations x + y = 20 and 3x + 2y = 48, we get x = 8, y = 12 Let P ≡ (8, 12) Now, we plot the graph of system of inequalities.
The shaded portion OCPB represents the feasible region which is bounded.
The corner points of feasible region are : O(0, 0), C(16, 0), P(8, 12) and B(0, 20). Now, we calculate Z at each corner point. Corner points Value of Z = 22x + 18y O(0, 0) 22(0) + 18(0) = 0 C(16, 0) 22(16) + 18(0) = 352 P(8, 12) 22(8) + 18(12) = 392 (Maximum) B(0, 20) 22(0) + 18(20) = 360 \ Maximum value of Z is 392 which occurs at the point P(8, 12). Hence, the maximum profit is ` 392 when 8 electronic machines and 12 manually operated sewing machines are purchased. Manually operated machine should be promoted, to save energy and increase employment for rural people. 28. Let ABCD be the given trapezium. Then AD = DC = CB = 10 cm In DAPD and DBQC, DP = CQ = h AD = BC = 10 cm ∠DPA = ∠CQB = 90° \ DAPD @ DBQC (by R.H.S. congruency) ⇒ AP = QB = x cm (say) \ AB = AP + PQ + QB = x + 10 + x = (2x + 10) cm Also from DAPD, AP2 + PD2 = AD2 ⇒ x2 + h2 = 102 ⇒ h = 100 − x 2
...(i)
Now, area S of this trapezium is given by 1 1 S = ( AB + DC ) ⋅ h = (2 x + 10 + 10)h 2 2 = (x + 10) ⋅ 100 − x 2
...(ii)
[Using (i)]
Differentiating (i) w.r.t. x, we get 1 dS = 1 ⋅ 100 − x 2 + (x + 10) ⋅ ⋅ (−2 x ) dx 2 100 − x 2 100 − x 2 − x 2 − 10 x −2(x 2 + 5x − 50) = = 100 − x 2 100 − x 2 −2(x + 10)(x − 5) = 100 − x 2 dS For max. or min. value of S, =0 dx ⇒ (x + 10)(x – 5) = 0 ⇒ x = 5 (Reject x = – 10 as x
85
A will win the best of 3 games match if he wins in 2 or 3 games E2 = the event that A wins ‘a best of 5 games’ match. A will win a best of 5 games match if he wins in 3 or 4 or 5 games. Now P(E1) = P(X = 2 or X = 3) = P(X = 2) + P(X = 3) = 3C2 p2q + 3C3p3= 3(0.4)2(0.6) + (0.4)3 = (0.4)2[1.8 + 0.4] = (0.4)2(2.2) = 0.352 P(E2) = P(X = 3 or X = 4 or X = 5) = P(X = 3) + P(X = 4) + P(X = 5) =5C3 p3q2 + 5C4 p4q + 5C5 p5 = (10p3q2 + 5p4q + p5) = p3(10q2 + 5pq + p2) = (0.4)3 × [10 × (0.6)2 + 5 × (0.4) × (0.6) + (0.4)2] = (0.064) × (3.6 + 1.2 + 0.16) = 0.064 × 4.96 = 0.317 Since P(E1) > P(E2), hence A should choose the first option.
dS <0 For this value of x, dx \ S is maximum at x = 5.
From (ii), the max. value of S = (5 + 10) ⋅ 100 − 52 = 15 75 = 75 3 sq.cm.
29. Let Ei(i = 1, 2, 3) be the event of drawing a queen in the ith draw. Let X denote the discrete random variable “Number of Queens” in 3 draws one by one with replacement. Here X = 0, 1, 2, 3 4 Now P(Ei ) = ; P(Ei ) = 48 (i = 1, 2, 3) 52 52 \ P(X = 0) = P(E1E2 E3 ) = P(E1)P(E2 )P(E3 ) 3
3
48 12 1728 = = = 52 13 2197 P(X = 1) = P(E1E2 E3 or E1E2 E3 or E1E2 E3 ) = P(E1E2 E3 ) + P(E1E2 E3) + P(E1E2 E3) = P(E1)P(E2 )P(E3 ) + P(E1)P(E2 )P(E3 ) = 3×
+ P(E1)P(E2 )P(E3 )
2
2
4 48 1 12 432 × = 3× × = 52 52 13 13 2197
P(X = 2) = P(E1E2 E3 or E1E2 E3 or E1E2 E3)
= P(E1E2 E3) + P (E1E2 E3 ) + P ( E1E2 E3 )
= P(E1)P(E2 )P(E3 ) + P (E1)P(E2 )P(E3 ) + P(E1)P(E2 )P(E3 ) 2
2
36 1 12 4 48 = 3× × = 3× × = 13 13 2197 52 52 P(X = 3) =P(E1E2E3) = P(E1)P(E2)P(E3) 3
\
1 4 = = 52 2197 The required probability distribution of X is X
0
1
2
3
P(X)
1728 2197
432 2197
36 2197
1 2197
OR Let E = the event that A wins a game against B. Let occurrence of the event E be called a success and X denote the number of successes. Let E1 = the event that A wins ‘a best of 3 games’ match. 86
mathematics today | february ‘17
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8
T
Class XII
his specially designed column enables students to self analyse their extent of understanding of specified chapters. Give yourself four marks for correct answer and deduct one mark for wrong answer. Self check table given at the end will help you to check your readiness.
VECTORS & THREE DIMENSIONAL GEOMETRY Total Marks : 80
Time Taken : 60 Min.
Only One Option Correct Type 1. If the volume of parallelopiped with a × b , b × c , c × a as co-terminous edges is 9 cu. units, then the volume of the parallelopiped with (a × b ) × (b × c ), (b × c ) × (c × a ), (c × a ) × (a × b ) as co-terminous edges is (a) 9 (b) 729 (c) 81 (d) 27 2. If the plane 2ax – 3ay + 4az + 6 = 0 passes through the midpoint of the line joining the centres of the spheres, x2 + y2 + z2 + 6x – 8y – 2z =13 and x2 + y2 + z2 – 10x + 4y – 2z = 8, then a equals (a) 1 (b) – 1 (c) 2 (d) – 2 3. If p × q = r and q × r = p, then (a) r = 1, p = q (b) p = 1, q = 1 (c) r = 2p, q = 2 (d) q = 1, p = r x − 2 y −1 z + 2 lie in the plane = = 3 −5 2 x + 3y – az + b = 0. Then (a, b) equals (a) (–6, 7) (b) (5, –15) (c) (–5, 5) (d) (6, –17) 5. Let a = a1 i + a2 j + a3 k , b = b1i + b2 j + b3 k and c = c1 i + c2 j + c3 k be three non-zero vectors such that c is a unit vector perpendicular to both the vectors a and b and if the angles between a and 4. Let the line
2
a1 a2 a3 p b is , then b1 b2 b3 is equal to 6 c1 c2 c3 (a) 1 1 (b) (a12 + a22 + a32 ) (b12 + b22 + b32 ) 4
3 2 2 2 2 2 2 2 2 2 (a + a2 + a3 ) (b1 + b2 + b3 ) (c1 + c2 + c3 ) 4 1 (d) none of these y −1 z − 3 6. If the angle between the line x = and = 2 l 5 , then l the plane x + 2 y + 3z = 4 is cos −1 14 equals (a) 2/5 (b) 5/3 (c) 2/3 (d) 3/2 One or More Than One Option(s) Correct Type (c)
7. If a vector r satisfies the equation r × (i + 2 j + k ) = i − k , then r may be equal to (a) i + 3j + k (b) 3i + 7 j + 3k (c) j + t (i + 2 j + k ), where t is any scalar (d) i + (t + 3)j + k , where t is any scalar
8. If OABC is a tetrahedron such that OA2 + BC2 = OB2 + CA2 = OC2 + AB2, then (a) OA ⊥ BC (b) OB ⊥ CA (c) OC ⊥ AB (d) AB ⊥ BC 9. If r is equally inclined to the co-ordinate axes and magnitude of r = 6 then r equals 2(i + j + k ) (a) 3(k + i + j) (b) 3 6(i + j + k ) (c) (d) − 2 3(i + j + k ) 3 10. If the unit vectors a and b are inclined at an angle 2q and | a − b |< 1, then if 0 ≤ q ≤ p, q lies in the interval (a) [0, p/6) (b) (5p/6, p] (c) [p/6, p/2] (d) [p/2, 5p/6] mathematics today | february‘17
87
11. Let the unit vectors A and B be perpendicular and the q to both at an angle unitvector C be inclined A and B . If C = a A + b B + g( A × B ) then (a) a = b (b) g2 = 1 – 2a2 1 + cos 2q (c) g2 = – cos 2q (d) b2 = 2 12. The equation of a sphere which passes through (1, 0, 0), (0, 1, 0) and (0, 0, 1) and whose centre lies on the curve 4xy = 1 is (a) x2 + y2 + z2 – x – y – z = 0 (b) x2 + y2 + z2 + x + y + z – 2 = 0 (c) x2 + y2 + z2 + x + y + z = 0 (d) x2 + y2 + z2 – x – y – z – 2 = 0 x −1 y −1 z + 3 = = 13. The points in which the line 1 −1 1 cuts the surface x2 + y2 + z2 – 20 = 0 are (a) (0, 2, 4) (b) (0, 2, – 4) (c) (4, 2, 0) (d) (0, – 2, – 4) Comprehension Type If a, b, c are three given non-coplanar vectors and r be any arbitrary vector in space, where r ⋅a b⋅a c⋅a a⋅ a r ⋅ a c ⋅ a ∆1 = r ⋅ b b ⋅ b c ⋅ b , ∆ 2 = a ⋅ b r ⋅ b c ⋅ b , a⋅c r ⋅c c ⋅c r ⋅c b⋅c c⋅c a⋅ a b⋅ a r ⋅ a a⋅ a b⋅ a c ⋅ a ∆ 3 = a ⋅ b b ⋅ b r ⋅ b and ∆ = a ⋅ b b ⋅ b c ⋅ b , then a⋅ c b⋅ c r ⋅ c a⋅c b⋅c c ⋅c 14. If vector r is expressible as, r = xa + yb + zc, then a⋅b a⋅a (a) a = (b × c ) + (c × a) [a b c] [a b c] c⋅a + (a × b) [a b c] (b) a = a ⋅ a(b ×c ) + a ⋅ b(c × a) + c ⋅ a(a × b) (c) a = [ a b c ](b × c) + [ a b c ](c × a ) + [ a b c ](a × b) (d) None of these. a b c 15. The value for a ⋅ p b ⋅ p c ⋅ p , is a⋅q b⋅q c ⋅q
(a) ( p × q)[ a × b b × c c × a] (b) 2( p × q)[ a × b b × c c × a] (c) 4( p × q)[ a × b b × c c × a] (d) ( p × q) [ a × b b × c c × a]. Matrix Match Type
16. Match the following: Column I
Column II
P. A line is perpendicular to x + 2y +
1.
5 3
Q. A plane passes through (1, –2, 1)
2.
2 2
Point (a, b, g) lies on the plane x + y + z = 2. Let a = ai + b j + g k , k × ( k × a ) = 0 , then g =
3.
2
2z = 0 and passes through (0, 1, 0). The perpendicular distance of this line from the origin is and is perpendicular to two planes 2x – 2y + z = 0 and x – y + 2z = 4. The distance of the plane from the point (1, 2, 2) is
R.
(a) (b) (c) (d)
P 2 3 3 1
Q R 1 3 4 2 2 4 2 3 Integer Answer Type
17. The line passes through the points (5, 1, a) and 17 13 (3, b, 1) crosses the yz-plane at the point 0, , − . 2 2 then a – b = 18. If [a × b b × c c × a] = l[abc ]2 , then l is equal to 19. The plane x + 2y – z = 4 cuts the sphere x2 + y2 + z2 – x + z – 2 = 0 in a circle of radius 20. Let u = i + j , v = i − j , w = i + 2 j + 3k . If n is a unit vector such that u . n = 0 and v . n = 0 then w . n equals Keys are published in this issue. Search now! J
Check your score! If your score is > 90%
excellent work ! You are well prepared to take the challenge of final exam.
No. of questions attempted
……
90-75%
Good work !
You can score good in the final exam.
No. of questions correct
……
74-60%
satisfactory !
You need to score more next time.
Marks scored in percentage ……
< 60%
88
mathematics today | february‘17
not satisfactory! Revise thoroughly and strengthen your concepts.
E 4
60° F 60° soLUtioN set-169
120°
1. (c) : z1, z2, z3 are roots ⇒ Sz1 = –p, Sz1z2 = q
60° 60° A 1 B
= 32 ⋅ 112 ⋅ 612 The number of all pairs is the number of divisors of 32 ⋅ 112 ⋅ 612, which is 33 = 27. 27 − 1 = 13 Number of pairs (a, b), a < b is 2 3. (a) : x2 – ix – 1 = 0. Solving, 5π π x = cis and x = cis 6 6 π π 2013 x = cis (2013) and x2013 = cis (10065) 6 6 2013 2013 –2013 ⇒ x = –i, x +x = –i + i = 0 4. (c) : =
100
100
0 9
0
∫ { x }dx = ∫
xdx −
9
(r +1)2
∑ ∫
r =0 r2
rdx
2000 2000 2 − ∑ (2r 2 + r ) = − 615 = 51 3 3 3 r =1
5. (b) : The parabola is open upwards with vertex (2, – 1). The focus is (2, –1 + 1) = (2, 0). The reflected ray passes through the focus (2, 0). 6. (a, d) : The equation represents a pair of planes. 3 1 a 2 2 3 ⇒ 1 1 = 0 ⇒ 4ab – 4a – 9b + 5 = 0 …(i) 2 1 1 b 2 For perpendicular planes a + 1 + b = 0 …(ii) Eliminating b from (i) and (ii), we get 7 4a2 – a – 14 = 0 ⇒ a = 2, − 4 7. (c) : ABEF and BCDE are trapeziums. BE = 2 + 2 · 4 cos60° = 6 AF = BE – 2cos60° = 6 – 1 = 5 Perimeter = 1 + 4 + 2 + 4 + 1 + 5 = 17
2
120° C
p2 = Sz21 + 2q For equilateral triangle, Sz21 = Sz1z2 \ p2 = q + 2q = 3q 1 1 1 2. (b) : + = a b 2013 ⇒ (a – 2013)(b – 2013) = 20132
D
4
8. (d) : Area = Area of BCDE + Area of ABEF 1 1 = (6 + 2)4 sin 60° + (6 + 5)sin 60° 2 2 11 43 =8 3+ 3= 3 4 4 9. (4) : N is the coefficient of x30 in (1 + x + x2 + … + x10)3 (1 + x + x2 + … + x20) (1 – x11)3 (1 – x21)(1 – x)–4 4 5 = (1 – 3x11 + 3x22 – x21 …) 1 + x + x 2 + ... 2 1 33 22 11 12 which is − 3 + 3 − 30 19 8 9 33 22 11 12 = − 3 + 3 − = 1111, 3 3 3 3 with digit sum 4. 10. (a)-(s, t); (b)-(p), (c)-(r), (d)-(r) f ( x ) − f (0) 1 = lim x a −1 sin = 0 x x x →0 x →0
(a) f ′(0) = lim
if a > 1 1 1 x ≠ 0 ⇒ f ′(x ) = ax a −1 sin − x a − 2 cos x x lim f ′( x ) = 0 if a > 2 \ a = 3, 4. x →0
(b) y(1) = 2 ⇒ 2 = a + b + c, y(0) = 0, y′(0) = 1 ⇒ c = 0, b = 1, a = 1 \ y = x2 + x, y(–1) = 0 (c) f(x) = sinx(1 + a), where a = \
f ( x ) = 2 sin x ,
π /2
∫
π /2
∫
cos tdt = 1
0
f ( x )dx = 2.
0
x2 y2 − y 2 = 1 and − x 2 = 1 have 2 2 common tangents with slopes ±1. The four tangents are x + y ± 1 = 0, x – y ± 1 = 0. They form a square of area 2 × 2 = 2
(d) The hyperbolas
mathematics today | February‘17
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mathematics today | February‘17
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