Exit Exam First Quarter-key (1)

MATH23X EXIT EXAMINATION 1st Quarter SY 2011–2012

DIRECTIONS: Solve each problem and choose the best answer from the choices given. Detailed solution should be shown and all final answers should be encircled. Write out your solution as shown below:

Sample Item: 1. What is the x-intercept of the line passing through A(1,4) and B(4,1)? a. 4 b. 4.5 c. 5 d. 6 In your answer sheet write: 1. C Solution:

A (1, 4) x 1 = 1 y1 = 4 B(4, 1) x 2 = 4 y 2 = 1 using two point form :  y - y1   y 2 - y1    x - x1   x 2 - x1 

 y - y1    y 2 - y1   x - x1   x 2 - x1   y - 4  1- 4  x -1  4 -1  y - 4    x -1 x y 5 Solving for x - intercept, let y = 0 x5

EXAMINATION STARTS HERE! 1. Find x in the equation shown: 4x2 + 48x + 144 = 400 a. 16, 4 16 Answer:

b. -16, 4

b

Solution: 4x2 + 48x + 144 = 400 2 4x + 48x – 256 = 0 x2 + 12x – 64 = 0

c. -32, 8

d. -4,

(x + 16)(x – 4) = 0,

2.

x = -16, x = 4

The age of a crocodile at the Manila Zoo 13 years ago was 1/3 of its age 7 years from now. How old is the crocodile at present? a. 15 b. 21 c. 23 d. 27

Answer:

c

Solution: Let x = present age, x – 13 = age 13 years ago, x+ 7 = age 7 years from now

1 ( x+7 )=x−13 3 x + 7 = 3x – 39 -2x = -46 x = 23 So, the crocodile is 23 years old at present. 3. An airplane flying with the wind can cover a certain distance in 2 hours. The return trip against the wind takes 2.5 hours. How fast is the plane and what is the speed of the air, if the one-way distance is 600 miles? (Let x-speed of the plane, y-speed of the wind) a. x = 40, y = 260 b. x = 30, y = 270 c. x = 270, y = 30 d. x = 230, y = 70 Answer:

c

Solution:

600 =2 x+y

,

600 5 = x− y 2

2x + 2y = 600 ---------------- >

x + y = 300

5x – 5y = 1200 -------------- > x – y = 240 ----------------------2x = 540 x = 270 y = 300 – 270 = 30 speed of the plane: x = 270 , speed of the wind : y = 30

4. How many gallons of 20% alcohol solution and 50% alcohol solution must be mixed to get 9 gallons of 30% alcohol solution? a.

3 gallons of 20% solution and 6 gallons of 50% solution

b.

6 gallons of 20% solution and 6 gallons of 50% solution

c.

5 gallons of 20% solution and 3 gallons of 50% solution

d.

6 gallons of 20% solution and 3 gallons of 50% solution

Answer:

d

Solution: (0.20)(x) + (0.50)(9-x) = (0.30)(9) 2x +45 – 5x = 27 -3x = -18 x = 6 gallons -------- > amount of 20% alcohol solution y = 9 – x = 3 gallons --------- > amount of 50% alcohol solution

5. How many different second order determinant can be formed from four different numbers? a. 4 b. 8 c. 16 d. 24 Answer:

d

Solution: 4! = 4 x 3 x 2 x 1 = 24

6. The angle of elevation of the top of an incomplete vertical pillar at a horizontal distance of 100 meters from its base is 45 degrees. If the angle of elevation of the top of the complete pillar at the same point is to be 60 degrees, then the height of the incomplete pillar is to be increased by how much? a.

Answer:

73.21

b. 100

c. 173.21

d. 26.79

a

Solution:

7. The function f is defined for 0 ≤ x ≤ 360, f(t) = 3sin(2t-1). What is the period of the waveform? a. π/4 b. 2 c. π d. π/2 Answer:

c

Solution:

Compare f(t) = a sin(bt – c) + d with f(t) = 3 sin(2t – 1), then b = 2

P=

2π 2π = =π b 2

8. If an equilateral triangle is circumscribed about a circle of radius 12 cm, determine the side of the triangle.

a. 34.64

Answer:

b. 41.57

c. 83.14

d. 20.78

c. 4

d. 25

b

Solution:

9. Solve for x: 2 log5(x-4) = 4 a. 29 b. 21 Answer:

a

Solution: 2 log5(x – 4) = 4 log5(x – 4)2 = 4 (x – 4)2 = 54 x2 – 8x + 16 = 625 x2 – 8x – 609 = 0 (x – 29)(x + 21) = 0

--------- > x = 29 , x = -21 (extraneous)

10.The two vertices of a triangle are A(2,4) and B(-2,3). Find the locus of the third vertex of the triangle if its area is 2 square units. a. 4x-y = 12 b. 4x + 4y = 10 c. x + 4y = 12 d. x – 4y = -10 Answer:

d

Solution:

11.What are the coordinates of the center of the curve x 2 + y2 -2x -4y -31 = 0? a. (-1, -1) b. (-2, -2) c. (1, 2) d. (2, 1) Answer:

c

Solution:

x 2  y 2  2 x  4 y  31  0

x x

2 2

 2 x    y 2  4 y   31

 2 x  1   y 2  4 y  4   31  5

 x 1  2   y  2  2 Center   1, 2 

 6 2  circle

12.Find the eccentricity of the curve 9x2 – 4y2 -36x +8y = 4. a. 1.76 b. 1.80 c. 1.86 1.92 Answer:

d.

b

Solution: 9x2  4y2  36 x  8y  4









9 x2  4x  4  4 y2  2y  1  4  9  4   4 1 9 x  2  4 y  1  9  4  2

 x  2 2 4 2 a 4



2

 y  1 2 9

 1  hyperbola

b2  9 c2  a2  b2  13; c  3.605 e

c 3.605   1.80 a 2

13.If the edge of a cube is increased by 30%, by how much is the surface area of one side of the cube is increased?

a. 30% 69%

b. 33%

c. 60%

d.

Answer: d Solution: x = initial edge y = increased edge A x = initial area of the cube A y = area of the cube when the side is increased by 30% By ratio and proportion , Ay Ax



y2 ; y  1.30 x x2

Ay

 1.30 x    Ax  x 

2

Ay  1.69 Ax

Therefore, the surface area on one side increased by 69%. 14.A conical vessel has a height of 24 cm and a base diameter of 12 cm. It holds water to a depth of 18 cm above its vertex. Find the volume (in cu cm) of its content. a. 188.40 c. 298.40 c. 381.70 d. 412.60 Answer: c Solution:

12 cm = diameter of cone , 24 cm = height of cone , 18 cm  depth of water By ratio and proportion : x 6  ,      x  4.5 18 24 1 1 2 V   r 2 h    4.5 18        V  381.70 3 3 15.A 10m-diameter hemispherical rubber ball is filled with liquid to a depth of 4m. Determine the surface area of the ball which is not in contact with the liquid. a. 5π b. 10π c. 20π d. 120π Answer:

b

Solution: Z = 2πRh = 2π(5)(1) = 10π

16.Find the value of k such that the function defined by continuous at every real number. a. 3 b. 4

c. 5

{

f ( x )= 3 x+7 if x ≤ 4 kx −1if x > 4 d. 7

is

Answer:

c

Solution: −¿

lim f (x ) x→4

must exist, so

x→4 f (x) lim ¿ ¿

+¿

=

x→4 f (x) ) lim ¿ ¿

3(4) + 7 = k(4) – 1 4k – 1 = 19 4k = 20 k=5 17.If f’(2) = g’(2) = f(2) = g(2) = 2, then what is the value of a. 8 20 Answer:

b. 12

( fg )' (2) ?

c. 16

d.

a

Solution: Using product rule, (fg)’(2) = f(2)g’(2) + f’(2)g(2) = (2)(2) + (2)(2) = 4 +4=8 18.Find the equation of the tangent line to the curve y = x 3 at the point (2,8). a. 12x – y – 16 = 0b. x + 12y – 98 = 0 c. 12x + y -98 = 0 d. x – 12y + 16 = 0 Answer:

a

Solution: f(x) = x3 -------- > f’(x) = 3x2 mT = f’(2) = 3(22) = 12 using point slope of a line: y = mT(x – x1) + y1 y = 12(x – 2) + 8 y = 12x – 24 + 8 12x – y – 16 = 0

dy dx

19.Find

a.

dy y −sec 2 x = dx sec2 y −x

d.

Answer:

in the equation tan x +tan y =xy

b.

dy sec2 y −x = dx x−sec 2 x

a

Solution:

tan x +tan y =xy sec 2 x+ sec 2 y

dy dy =x + y dx dx

dy x−sec 2 y = dx sec 2 x− y

.

dy sec 2 x− y c. dx = y −sec 2 y

sec 2 y

dy dy −x = y −sec 2 x dx dx

dy ( sec2 y−x ) = y−sec 2 x dx dy y −sec2 x = dx sec 2 y −x

3

∫

20.Evaluate:

1

(

3 x2 +

a. 68/3 86/3 Answer:

4 dx x2

)

b. 70/3

c. 92/3

d.

d

Solution: 3

∫ 1

21.Evaluate:

3

3

] (

4 4 4 86 dx=∫ ( 3 x 2 + 4 x−2 ) dx= x 3− = 27− −( 1−4 )= 2 x 3 3 x 1 1

)

3

c. 4

)

3

∫ x 4 √ 3−5 x 5 dx

3 5 4 /3 (3−5 x ) +C 100

a.

Answer:

(

3 x2 +

b.

5 4 /3

(3−5 x ) +C

d.

−3 5 4 /3 (3−5 x ) +C 100

−3 5 4 /3 (3−5 x ) +C 4

b

Solution: Let u = 3 – 5x5, du = -25x4dx 3

∫ x 4 √ 3−5 x 5 dx=

[ ] 4 /3

4

−1 −1 u −1 3 ∫ u1 /3 du= 25 4 / 3 +C= 25 4 ( 3−5 x 5 ) 3 +C 25

=

()

−3 5 4 /3 (3−5 x ) +C 100

22.The region bounded by the lines y = 3 – 2x, y = 2 and the coordinate axes is revolved about the y-axis . Find the volume of the solid formed. a. 6π/5 b. 5π/6 c. 13π/6 d. 7π/6

Answer:

c

Solution: 2

V=

π∫ 0

3− y 2 dy 2

( )

=

13 π 6

23. Find the area of the region bounded by the curve y 2 = 4x, the line y = 2 and the y-axis. a. 1/3 b. 2/3 c. 1/6 d. 5/6 Answer:

b

Solution: 1

A=

∫ ( 2−√ 4 x ) dx= 23 0

24.Find a vector orthogonal to both of the vectors u = <2, -1, 3> and v = <-7, 2, -1>. a. -5i – 19j – 3k b. 5i + 19j + 3k c. 5i – 19j + 3k d. -5i + 19j – 3k Answer:

a

Solution: uxv=

|

|

i j k 2 −1 3 −7 2 −1

= -5i – 19j – 3k

3 4

25.Evaluate a. 11/3

Answer:

b

Solution:

∫∫ ( x 2− y ) dy dx 1 1

b. 11

. c. 5/12

d. 12

1 2 2 ( 4 x −8 ) −(¿)dx ¿ ¿ 3 2 y 4 2 x y− dx=∫ ¿ 2 1 1 x 2−

(

|)

3 4

3

∫∫ ( x 2− y ) dy dx=∫ ¿ 1 1

3

(

¿∫ 3 x 2− 1

1

|

15 15 x 3 dx= x 3− =11 2 2 1

)