clep calculus

7/30/2019 CLEP Calculus

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CLEPCalculus

1

Copyright 2006Peterson's

CLEP is a registered trademark of the College Entrance Examination Board,which was not involved in the production of and does not endorse this product.

Time60 Minutes45 Questions For each question below, choose the best answer from thechoices given.

1. 5

7lim

5 +x xis

(A) 7

(B) 0

(C)7

10 (D) 7

(E) Nonexistent

2. Iff(x) = 133x , thenf(x) =

(A) x

(B)43

9

4x

(C)13

x

(D)23

x

(E)233

x

3. Which of the following is an equation of the line tangent to the graph off(x) = x3 +x4 at the pointx = 2?

(A) 20x +y = 32

(B) 20x +y = 48

(C) 44x +y = 36

(D) 8x y = 4

(E) x y = 2


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4. ( )+ + + e xx e x e dx =

(A) 11 + + +e xex e C

(B)2

2+ + + +

e xx

e x e C

(C)2 2 1 1

2 2 1 1

+ +

+ + + ++ +

e xx e x eC

e x

(D)2 1

2 1

+

+ + + ++

exx xex e C

e

(E)2

2+ + + +

e xx ex x e C

5.

The graph of ,f the second derivative of the functionf, is shown above. On which of the following

intervals is fconcave upward?

(A) (,2)

(B) (1, 1)

(C) (,2) and (0, 1)

(D) (,2) and (2, )

(E) (2, 1) and (1, 2)


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6. ( )2 x x x dx =

(A) 0 + C

(B)5

2+

x xC

(C)3 3 32 2

3 3 3

+

x x xC

(D)3 22

7 2 +

x x xC

(E)5 22

7 2 +

x xC

7.

At which of the five points on the graph in the figure above isdy

dxnegative and

2

2

d y

dxpositive?

(A) A

(B) B

(C) C

(D) D

(E) E

8. The acceleration, at time t, of a particle moving along the x-axis is given by 3( ) 4 1= +a t t . At time t= 0,

the velocity is 0 and the position of the particle is 12. What is the position of the particle when t= 1?

(A) 0.7

(B) 1.0(C) 5.0

(D) 12.0

(E) 12.7


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9. What is5 3

5 3

3 2 5lim

5 3 7

+ +

+ +x

x x

x x?

(A)3

5

(B)2

3

(C)5

7

(D) 1

(E) The limit does not exist

10.23

1

+ x dx =

(A) 1.5

(B) 1.0

(C) 1.0

(D) 1.5

(E) 2.0

11. Letfand g be the functions defined by3 21 1( )

3 2= +f x x x and g(x) =x

2 + 6x. For which of the following

positive values ofa is the tangent line to fat x = a parallel to the tangent line to g atx = a?

(A)2

3

(B)3

2

(C) 2

(D) 3

(E) 6

12. The function fis given byf(x) = 8x3 + 2x +1. What is the average value offover the closed interval [1,

2]?

(A) 3

(B) 9(C) 12

(D) 20

(E) 24


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13. Iff(x) =tan

,3

x

xthen ( ) =F x

(A)21 sec

3x

(B)21 csc

3x

(C)( )

cos

3 sin cos +

x

x x x

(D)2

2

sec tan

3

x x x

x

(E)2

2

tan sec

3

x x x

x

14. For selected values ofx, the functions fand g and their derivatives,f(x) and g

(x) are evaluated and

presented in the table above. Ifh(x) = g(f(x)), what is h(5)

x f(x) f(x) g(x) g(x)

5 10 10 15 10

10 20 5 5 15

15 30 10 10 5

-(A) 5

(B) 15

(C) 150

(D) 250

(E) 750

15.24

1

1

=

+dx

x

(A) 4

(B) ln 3(C) ln 2 ln 4

(D) ln 3 ln 5

(E)2

3


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16.0

tan tan3 3

lim

+

=

h

h

h

(A) 0

(B)3

4

(C) 1

(D)1

2

(E) 4

17. The functionfis differentiable and has a relative maximum value of 5 atx = 1. If 3( )

( ) ,=f x

h xx

then

( )1 h =

(A) 15

(B) 5

(C) 0

(D) 15

(E) 45

18.

The shaded region in the figure above is the region bounded (enclosed) by the graphs off(x) = x + 3 and

g(x) =x3 6x + 7. Find the area of the shaded region.

(A) 41.75

(B) 38.25

(C) 8.5(D) 38.25

(E) 41.75


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19. Ifx5+y

5+xy = x

4, then =

dy

dx

(A)

4 4

4 5 3

5

5 4

+

x x y

x y y x

(B)

3 4 5

5 4

4 5

5

+

x x y y

x y x

(C)

4

5 45 +

x

x y x

(D)

4

5 4+

x

x y x

(E)

4 4

3 4 5

5

4 5

+

+

x y x

y x x y

20.

The graph of the function fis shown in the figure above. What is1

lim ( )x

f x ?

(A) 1

(B) 2

(C) 2.5

(D) 3

(E) The limit does not exist


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21.ln

lim

=xx

x

e

(A) 0

(B)

1

e

(C) 1

(D) e

(E)

22. If the function fis continuous, nonpositive, and strictly increasing on the closed interval [a, b], which of

the following statements is NOT true?

(A) ( ) ( )( )< ba f x dx f b b a

(B) ( ) ( )( )> ba f x dx f a b a

(C) ( ) ( )( )= baf x dx f c b a for some number c contained in (a, b)

(D) ( ) ( ) ( ),= ba f x dx F b F a where ( ) ( ) =F x f x on the open interval (a, b)

(E) ( ) 0>ba f x dx

23.

4

3 8 +

= xx e dx

(A)4

48

4

++

xx e C

(B)42 83 + +xx e C

(C)4 8+

+xe C

(D) ( )414 8++

xe C

(E)4 81

4

++

xe C


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24.

The functionfis shown in the figure above. At which of the following points could the derivative offbe

equal to the average rate of change offover the closed interval [1, 1]?

(A) A

(B) B

(C) C

(D) D

(E) E

25. For which of the following functions does4 2

4 2?=

d y d y

dx dx

I.y = ex

II.y = cosx

III. y = sin x

(A) I only

(B) II only

(C) III only(D) I and III only

(E) II and III only

26. The height, in feet, of a ball thrown vertically upward from a building 50 feet high is given by s(t) = 16t2

+ 256t+ 50, where tis measured in seconds. What is the height of the ball, in feet, when its velocity is

zero?

(A) 178

(B) 306

(C) 434

(D) 562(E) 1074


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27. Iffis a function such that( ) ( )

2

2lim 8,

2

=

+x

f x f

xthen which of the following statements must be true?

(A) ( )2 8 = f

(B) ( )2 8

=f (C) ( )2 8 =f

(D) fis continuous at x = 2

(E) f(x) = 2x2

28. ( )( )tan =xd

edx

(A) ex sec2(e

x)

(B)cos

sin

x

x

e

e

(C) sec2(e

x)

(D) ex

sec2x + (tan x)e

x

(E) ex

tan(ex)

29. Which of the following statements about the function f(x) = 3x4

4x3is true?

(A) fhas no relative extrema and no inflection points

(B) fhas one relative maximum value, one relative minimum value, and one inflection point

(C) fhas one relative maximum value and two inflection points

(D) fhas one relative minimum value and two inflection points

(E) fhas three inflection points

30.2 2cot csc = x xdx

(A)3

cot

3 +

xC

(B)3csc

3 +

xC

(C)3cot

3+

xC

(D)3csc

3+

x C

(E)2cot

2+

xC


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31. Let s(t) be a differentiable function that is positive and increasing with respect to time t. For what value,

s(t), is the rate of increase of the volume of a cube with edge s equal to 48 times the rate of increase in s?(The volume of a cube with edge s is V= s

3.)

(A) 4

(B) 4 3(C) 16

(D) 16 3

(E) 48

32. If 0( ) sin ,= x

F x tdt then ( )6

=F

(A)3

2

(B)

1

2

(C)3

12

+

(D)1

2

(E)3

2

33. ( )

2 64for 8

Let be defined by .816 for 8

= + =

xx

f f x x

x

Which of the following statements aboutfare true?

I.8

lim ( )x

f x exists.

II.f(8) exists.

III. fis continuous atx = 8.

(A) None of the above

(B) I only

(C) II only(D) I and II only

(E) I, II, and III


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34. The functionfis continuous on the closed interval [0, 2] and has values that are given in the table above.

If two subintervals of equal length are used, what is the right endpoint Riemann sum approximation of20 ( ) ? f x dx

x 0 0.5 1.0 1.5 2.0

f(x) 3 4 6 5 7

(A) 5.2

(B) 9.0

(C) 10.0

(D) 12.5

(E) 13.0

35. What is the average rate of change of the functionfdefined by f(x) = cosx on the closed interval [0, ]?

(A)2

(B)2

(C)

(D)2

(E)2

36. Letfbe continuous atx = a. Which of the following statements is NOT necessarily true?

(A) ( )lim ( ) lim

=x a x a

f x f x

(B) f(a) exists

(C) lim ( )x a

f x exists

(D)( )

0lim

+

h

f a h

hexists

(E) ( )lim ( ) ( ) 0

=

x a

f x f a


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37. A rectangular field is to be adjacent to a river having a straight-line bank. It is to have fencing on three

sides, the side on the river bank requiring no fencing. If 100 yards of fencing is available, find the lengthof the side of the field parallel to the river that will yield maximum area.

(A) 25 yards

(B) 30 yards(C) 40 yards

(D) 50 yards

(E) 60 yards

38. Iff(x) = tan2(cot x), then ( )f x =

(A) 2 tan(cotx)

(B) 2 tan(cot x)(csc2x)

(C) 2 tan(cotx)(csc2x)

(D) 2tan(cot x)(sec2(cotx))csc

2x

(E) 2 tan(cot x)(sec2(cotx))csc2x

39. Letfbe continuous at all points of the open interval (a, b), and let c be a point in (a, b) such that

I. f exists at all points of (a, b), except possibly at (C).

II. ( )f x < 0 for all values ofx in some open interval having c as its right endpoint.

III. ( )f x > 0 for all values ofx in some open interval having c as its left endpoint.

Which of the following statements must be true?

(A) ( )

f c = 0(B) ( )f c does not exist

(C) fhas a point of inflection at c

(D) fhas a relative minimum value at c

(E) fhas a relative maximum value at c


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40.

A rectangle with one side on the x-axis and one side on the line x = 0.5 has its upper right vertex on the

graph of2

1,=y

xas indicated in the figure above. For what value ofx does the area of the rectangle

attain its maximum value?

(A) 0.5

(B) 0.8

(C) 1.0

(D) 2.0

(E)

41. Let f(x) = x3+ 2x

2+ 10x. Ifh is the inverse function off, then ( )9 h =

(A) 9

(B) 1

9

(C) 1

(D)1

9

(E) 9


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42. Iffis continuous for all x, which of the following integrals necessarily have the same value?

I. ( )aaf x dx

II. 2 ( )ao f x dx

III. 20 ( ) a f x dx

(A) I and II only

(B) I and III only

(C) II and III only

(D) I, II, and III

(E) No two necessarily have the same value

43. Let F(t) be the amount of an investment, in dollars, at time t, in years. Suppose that Fis growing at a rate

given by ( ) ( ) 0.251000 , = tF t e where tis time in years. IfF(0) = 4000, then F(t) =

(A) (2000)e4t+ (2000)e

4t

(B) (4000)0.25t

(C) (1000)e0.25t

+ 4000

(D) (4000)e0.25t

(E) (1000)e0.25t

44. The volume of a cube is increasing at a rate of 30 cubic feet per hour3ft

.h

At what rate is s, the edge of

the cube, changing when the length of the edge is 5 feet?

(A)ft

0.17h

(B)ft

0.40h

(C)ft

2.00h

(D)ft

3.00h

(E)ft

6.00 h


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45. The Riemann sum

3100

1

1 1

4 100 100=

i

ion the closed interval [0, 1] is an approximation for which of the

following definite integrals?

(A)100 3

1 x dx

(B)2 30 x dx

(C)3

10 4

x

dx

(D) 1 30 x dx

(E)3

1001

100

xdx


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ANSWER KEYCLEP Calculus

ANSWER KEY - Page 17

1. The correct answer is E. From the definition for the limit you know that5

7lim

5 + xxcan exist only if

5

7

+x approaches a single finite value asx approaches 5 from both the left and right. Asx approaches 5

from the left, the number5

7

+xis decreasing without bound; symbolically, you indicate this behavior by

writing5

7lim

5 +

xx= . Asx approaches 5 from the right, the number

5

7

+xis increasing without bound;

symbolically, you indicate this behavior by writing5

7lim

5 ++

xx= . Since

5

7

+xdoes not approach a single

finite value asx approaches 5 from both the left and right of5,5

7lim

5 + xxis nonexistent.

2. The correct answer is D. If13( ) 3 ,f x x= then

13

23

11( ) 3( )

3

.

=

=

f x x

x

3. The correct answer is A. They-value whenx is 2 is given by

( ) ( ) ( )2 2 3 2 4

8.

= +

=

f

Sincef(x) =x3

+x4, then ( ) 2 23 4 . = +f x x x Thus, the slope of the tangent line at the point (2, 8) is

( ) ( ) ( )2 3

2 3 2 4 2

20.

= +

=

f

4. The correct answer is D.

( )

2 1

1

.2 1

+

+ + + = + + +

= + + +

= + + + ++

e x e x

e x

ex

x e x e dx xdx edx x dx e dx

xdx e dx x dx e dx

x x

ex e C e

5. The correct answer is E. The graph offis concave upward for values ofx for which ( ) 0. >f x Looking

at the graph of ,f you can see that ( )f x is positive on (2, 1) and (1, 2). Therefore, the functionfis

concave upward on (2, 1) and (1, 2).


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6. The correct answer is D.

( ) ( )

( )

1 12 2

52

52

72

2 2

2

72

3 2

2

2.

7 2

=

=

=

= +

= +

x x x dx x x x dx

x x dx

x dx x dx

x xC

x x xC

7. The correct answer is A. The slope of the tangent line at pointA is negative, indicating thatdy

dx

is

negative atA. In a neighborhood about the pointA, the slope of the tangent line is increasing and, thus,dy

dxis

increasing, indicating the curve is concave upward about the pointA. Since2

2

d y

dxis positive when the curve is

concave upward,dy

dxis negative, and

2

2

d y

dxis positive at pointA. This condition does not occur at any of the

other four points.

8. The correct answer is E. Theds

dtis the velocity, v(t), of the particle; and

dv

dtis the acceleration, a(t), of

the particle. Thus,

( )3

41

( )

(4 1)

.

=

= +

= + +

v t a t dt

t dt

t t C

At t= 0, v(t) = 0.

Substituting and solving for C1, you have

( )4

1

1

0

0 0 0

0.

=

+ + =

=

v o

C

C

Now sinceds

dtis v(t),


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( )4

5 2

2

( )

( )

.5 2

=

= +

= + +

s t v t dt

t t dt

t t C

At t= 0, s(t) = 12.

Substituting and solving for C2, you have

( )

( ) ( )5 2

2

2

2

0 12

0 012

5 2

0 0 12

12.

=

+ + =

+ + =

=

s

C

C

C

Therefore, the position function of the moving particleis

5 2

( ) 12.5 2= + +t t

s t

When t= 1,

( ) ( )5 2

1 1(1) 12.

5 2

12.7.

= + +

=

s

9. The correct answer is A. First, divide every term by the highest power ofx, and then evaluate the limit.

Thus,5 3

5 3 5 5 5

5 3 5 3

5 5 5

2 5

2 5

3 2 5

3 2 5lim lim

5 3 7 5 3 7

2 53

3 75

3.

5

+ ++ +

=

+ ++ +

+ +

=

+ +

=

x x

x x

x x x x x

x x x x

x x x

x x

x x

10. The correct answer is D. Note that

( )

1 if 11 .

1 if 1

+ + =

+ <

x xx

x x

Therefore, since the upper limit of the integral, 2, is less than 1,


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( )

( )

( )( )

( )( )

( ) ( )( )

2

3

2 23 3

23

2

2 2

1 1

1

2

2 32 3

2 2

0 1.5 1.5.

+ = +

= +

= +

= + +

= =

x dx x dx

x dx

xx

11. The correct answer is D. Atx= a, the tangent line tofis parallel to the tangent line to g when

( ) ( ) , =f a g a that is, when a2 + a = 2a + 6. Solving a2 + a = 2a + 6 for a, you obtain

( )( )

2

2

2 6

6 0

2 3 0

2 (reject)

or

3.

+ = +

=

+ =

=

=

a a a

a a

a a

a

a

12. The correct answer is C. The average value of a continuous functionfover a closed interval [a, b] is

given by1

( ) .

baf x dx

b a

Thus, the average value offover the closed interval [1, 2] is

( )( ) ( )

( ) ( ) ( ) ( )( )( )

( )

2

1

2 3 4 21

4 24 2

1 18 2 1 2

2 1 3

12 2 2 2 2 1 1 1

3

138 2

3

36

3

12.

+ + = + +

= + + + +

=

=

=

x x dx x x x

13. The correct answer is D. Applying the rule for the derivative of the quotient of two differentiable

functions, you have


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( ) ( )( )

( )

( )

2

2

2

2

2

2

3 sec tan 3( )

3

3 sec tan

9

sec tan.

3

=

=

=

x x xf x

x

x x x

x

x x x

x

14. The correct answer is C. Using the chain rule for the derivative of a composite function, you have

( )

( )( )

(5) (5) (5)

(10) (5)

15 10150.

=

=

=

=

h g f f

g f

15. The correct answer is B.

2

4

24

1ln 1

1

ln 2 1 ln 4 1

ln1 ln 3

ln3.

= +

+

= + +

=

=

dx xx

16. The correct answer is E. Using( ) ( )

0( ) lim

+ =

h

f a h f af a

hand selectingf(x) = tanx and ,

3

=a you

have


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( )

( )

( )

2

2

2

212

0

3

tan tan3 3

lim3

sec3

1

cos

1

cos 60

1

4.

+

=

=

=

=

=

=

h

h

fh

17. The correct answer is A. Sincefis differentiable and has a relative maximum value of 5 atx = 1, you

know thatf(1) = 5 and that ( )1 0. =f Using the quotient rule for differentiation, you have

( ) ( )

( )

3 2

23

4

( ) ( ) 3( )

( ) 3 ( ).

=

=

x f x f x xh x

x

xf x f x

x

Thus,

( )

( )

( ) ( )

( )

4

4

1 ( 1) 3 ( 1)( 1)

1

1 (0) 3 5

1

0 15

1

15.

=

=

=

f fh

18. The correct answer is B. From the figure, you can see that the graphs offand g intersect at (1, 2) and

(4, 1). Sincef(x) = x + 3 lies above g(x) =x3 6x + 7 on the interval fromx = 1 tox = 4, the area of the

shaded region between the two graphs is given by


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( ) ( )( ) ( )

( ) ( ) ( ) ( )( )

4

1

4 43 31 1

4 2

3 2 4 2

6 7 5 4

5

44 2

4 5 4 1 5 14(4) 4 1

4 2 4 2

740

4

38.25.

+ + = +

= +

= + +

= +

=

x x x dx x x dx

x xx

19. The correct answer is B. The equationx5

+y5

+xy =x4

is not easily solved fory in terms ofx, so you

should use the technique of implicit differentiation to find .dy

dx

On the assumption that there is a differential

functionfdefined implicitly by the equationx5

+y5

+xy =x4, you have

( ) ( )

( )

( )

5 5 4

5 4 5 4 3

5 4 4 5 3

5 4 3 4 5

3 4 5

5 4

5 5 1 4

5 5 4

5 4 5

4 5.

5

+ =

+ + + =

+ + + =

+ =

=

+

d dx y xy x

dx dx

dy dyx y y x x y x

dx dx

dy dyx y x y x y x

dx dx

dyx y x x x y y

dxdy x x y y

dx x y x

20. The correct answer is E. From the definition for the limit you know that1

lim ( )x

f x can exist only iff(x)

approaches a single finite value asxapproaches 1 from both the left and right. Asxapproaches 1 from the

left, the graph offapproaches the point (1, 3), so1

lim ( ) 3;

=

xf x but asxapproaches 1 from the right, the

graph offapproaches the point (1, 2), so1

lim ( ) 2.+

=x

f x Becausef(x) approaches two different values asx

approaches 1,

1

lim ( )

x

f x does not exist.

21. The correct answer is A. Since lim ln

= x

x and lim ,

= x

xe you can use LHpitals rule to obtain


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( )

( )

lnln

lim lim

1

lim

1lim

0.

=

=

=

=

x

xx

xx

x x x

dx

x dxde edx

x

e

xe

22. The correct answer is E. Sincef(x) is nonpositive and strictly increasing on the closed interval [a, b],

then when a


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(1) ( 1) (1) 0

(1) ( 1) 2

1

(1),2

=

=

f f f

f

which you can estimate, by looking at the graph, as about

1(1.5) 0.75

2

0.

=

>

Geometrically, the average rate of change offover [1, 1] is the slope of the line connecting the points (1,

0) and (1,f(1)). Since the derivative, ( ) ,f x offat a pointx equals the slope of the tangent line at the pointx,

you can work this problem by comparing the slope of the tangent line at each of the indicated points to the

slope of the line connecting the points (1, 0) and (1,f(1)).

You can eliminate point Cfrom consideration because the slope of the tangent line at Cis zero, so the

derivative offat Cwould not equal the average rate of change offover [1, 1].

You can also eliminate pointsA andB from consideration because the slope of the tangent line at each of

these points is negative, so the derivative offevaluated at either of these points would not equal the average

rate of change offover [1, 1].

Looking at points D and Eand considering their tangent lines, you can see that the tangent line at point D

would more likely be parallel to the line connecting the points (1, 0) and (1,f(1)) than the tangent line atpoint E, which is too steep in comparison, indicating that the derivative at D could be equal to the average

rate of change offover [1, 1].

25. The correct answer is E. This question is an example of a multiple response set question. An efficient

approach to working this type of question is to check Roman numeral options and eliminate answer choices,

if possible, as you go along.

Checking Roman I, you have


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( )

( )

( )

( )

2

2

3

3

4

4.

=

= =

= =

= =

= =

x

x

x

x

x

x

x

x

x

y e

d edye

dx dx

d ed ye

dxdx

d ed ye

dxdx

d ed ye

dxdx

Thus,4 2

4 2,

d y d y

dx dx

so eliminate choices A and D because these answer choices contain Roman I.

Checking Roman II, you have

( )

( )

( )

( )

2

2

3

3

4

4

cos

cossin

sincos

cossin

sincos .

=

= =

= =

= =

= =

y x

d xdyx

dx dx

d xd yx

dxdx

d xd yx

dxdxd xd y

xdxdx

Thus,4 2

4 2,=

d y d y

dx dxso eliminate choice C because this answer choice does not contain Roman II.

Checking Roman III, you have


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( )

( )

( )

( )

2

2

3

3

4

4

sin

sincos

cossin

sincos

cossin .

=

= =

= =

= =

= =

y x

ddyx

dx dx

dd yx

dxdx

d xd yx

dxdx

d xd yx

dxdx

Thus,4 2

4 2,=

d y d y

dx dxso eliminate choice B because this answer choice does not contain Roman III.

Choice E is correct because it includes every Roman numeral option that is correct and no incorrect Roman

numeral options.

26. The correct answer is E. Since ( )s t is the velocity, v(t), you have

( ) ( )

32 256.

=

= +

s t v t

t

Setting v(t) equal to zero and solving for tyou have

32 256 0

32 256

8

+ =

=

=

t

t

t

seconds when the velocity is zero.

Evaluating s(t) at t= 8 seconds, you obtain

( ) ( ) ( )2

8 16 8 256 8 50

1024 2048 50

1074

= + +

= + +

=

s

feet, which is the height of the ball when the velocity is zero.


28/42



27. The correct answer is C. The functionfis differentiable atx = 2 provided

( ) ( )

( )

( ) ( )2

2 2lim

2 2

=

+x

f x f f x f

x xexists. Since ( )2 8, =f the statement given in Cmust be true.

The statement given in A is false.

The question stem fails to provide sufficient information for determining whether the statements given in the

other answer choices are true or false.

28. The correct answer is A. Using the chain rule for the derivative of a composite function and lettingf(u)

= tan u and u = ex, you have

( )( ) ( )( )

( ) ( )

( )( )( )

( )

2

2

tan

sec

sec .

=

=

=

=

x

x x

x x

d f u d f u du

dx du dx

d ed u

du dx

e e

e e

29. The correct answer is D. To investigate relative extrema forf, first identify potential critical points for

the function by solving ( ) 0. =f x

Since ( ) 3 212 12 , = f x x x you have:

( )

3 2

2

12 12 0

12 1 0,

=

=

x x

x x

andx = 0 and x = 1 are critical numbers forf.

The second derivative offis given by ( )f x = 36x2 24x. Since

( ) ( ) ( )2

1 36 1 24 1

12 0,

=

= >

f

fhas a minimum atx = 1.


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Notice that you can eliminate choice A at this point. This is a smart test-taking strategy that you should use

when taking the actual test.

Continuing with the analysis, evaluate ( )f x atx = 0 to obtain

( ) ( )2

(0) 36 0 24 0

0,

=

=

f

and the second-derivative test fails atx = 0.

Since the second-derivative test fails atx = 0, using the first derivative test,

( ) ( )3 2

(0.5) 12 0.5 12 0.51.5

=

=

f

and

( ) ( )3 2

( 0.5) 12 0.5 12 0.5

4.5,

=

=

f

andf(0)is not a relative extrema forf. Eliminate choices B and C.

Now, investigateffor points of inflection. Points of inflection forfcan occur only at numbers where

36x2

24x = 12x(3x2) = 0; therefore 0=x and2

.3

=x

Thus, (0,f(0)) and ( )2 23 3( , )f are potential points of inflection.

Testing for concavity to the left of 0 and between 0 and2

,3

( ) ( )2

( 1) 36 1 24 1

60 0

=

= >

f

and

( ) ( )2

1 1 12 2 2

( ) 36 24

3 0.

=

=

clep calculus

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