AP Calculus AB Practice Exam

CALCULUS AB
SECTION I, Part A
Time—55 minutes
Number of questions—25
A CALCULATOR MAY NOT BE USED ON THIS PART OF THE
EXAMINATION.
Directions: Solve each of the following problems. After examining the
choices, select the choice that best answers the question. No credit will
be given for anything written in the test book.
In this test: Unless otherwise specified, the domain of a function f is as-
sumed to be the set of all real numbers x for which f(x) is a real number.
7

1. # 1
^ x - 3h2
dx =
4

1
A. 27

3
B. 4

C. 91

D. ln 4 - ln 3

E. ln 16 - ln 9
2 AP CALCULUS AB PRACTICE EXAM AP CALCULUS AB PRACTICE EXAM 3

5. Find the average value of y = x 3 x 4 - 1 on the interval [1,3].

x

2. # cos tdt =
r
2
A. 216 5
A. cos x
B. 2163 5
B. - cos x
C. sin x - 1 C. 80 5
D. sin x
D. 803 5
E. 1 - sin x
E. 640
27
5

e2

cx + 1m
3
3. # x dx = x 1 3 5
1
e

6 f  (x) 4 k 3
A. e -2 1
6e
12 6 6. Given that f is a continuous function on the interval [1,5] and that
B. e + 26e - 1
f takes values shown in the table. The function f will have two zeros
in the interval [1,5] if k =
9 3
C. e + e3 - 1
e
9 3 A. –1
D. e + 9e3 - 1
3e B. 0
9
E. e -3 1 + ln 2 C. 1
e
D. 2
E. 3
4. A particle moves along the y-axis so that its position at time
0 # t # 20 is given by y ^ t h = 5t - 3 . At what time does the particle
t2
dy
7. If dt = ky and k ! 0 , which of the following could be the equation
change direction?
of y?
A. 5 seconds
A. y = kx - 7
B. 7.5 seconds
B. y = 95e kt
C. 10 seconds
C. y = 5 + ln k
D. 15 seconds
D. y = ^ x - k h2
E. 18 seconds
E. y = k x
4 AP CALCULUS AB PRACTICE EXAM AP CALCULUS AB PRACTICE EXAM 5

x
12. If f m ^ x h = x ^ x - 2 h^ x + 1 h2 , then the graph of f has points of inflec-
8. If F ^ x h = # t - t dt , then F l ^3h =
2

1
tion when x =
A. 6
A. –2 and 1
B. 5
B. 2 and –1
C. 6
C. 2 and 0
D. 5
D. –2 and 0
E. 3
E. 0, 2 and –1

9. If f ^ x h = tan ^ e 2x h , then f m ^ x h = dy
13. If 3x 2 - 4xy = 1 , then when x = 1, dx =
A. 2e 2x sec 2 ^ e 2x h 3
A. 2
B. 8e 2x tan ^ e 2x h
C. 4e 2x sec 2 ^ e 2x h
B. 1

D. 8e 4x sec 2 ^ e 2x h tan ^e 2x h C. 21
E. 4e 2x sec 2 ^ e 2x h62e 2x tan ^ e 2x h + 1@ D. 0
E. - 21
10. If f ^ x h = sec ^3x h , then f l a 3r k
4 =
A. - 3 2
14. If a curve is defined by f ^ x h = 1 - x cos x , an equation of the normal
B. -
3 2 to the curve at a r k
2 , 1 is
2
3
C. 2 A. y = r2x + r2
D. 3 2 2 B. y = r2x + a r k
2

2
E. 3 2 C. y = 2rx - 2

D. y =- 2rx + 2
for 0 # x 1 1
11. If f ^ x h = * e x
x3 e x
, then lim f ^ x h is
for 1 1 x # 3 x"1
E. y =- 2rx - 2
x3
A. 0
B. 1
C. e
D. e 3
E. nonexistent
6 AP CALCULUS AB PRACTICE EXAM AP CALCULUS AB PRACTICE EXAM 7

15. If a function is given by f ^ x h = x2 + 3 , what is the instantaneous rate 18. Let f and g be differentiable functions such that f ^ x h ! 0 and
x -1 g^ xh - g^ xhf l ^ xh
g ^ 2 h = 3 . If h ^ x h = and hl ^ x h = , then g ^ x h =
6 f ^ x h@2
of change of the function at x = 3 ?
f^ xh
A. f ^3 h
7
A. 16

B. - 167 B. 0
6 f ^ x h@2
f l ^ xh
11 C. -
C. 16
D. 3
11
D. - 16
E. f l ^2 h x + 3
E. 61

19. Over the interval 0 # t # 5 , the position of a particle is given by

16. If f ^ x h is a twice differentiable function on the interval a 1 x 1 b , s ^ t h = t 4 - t 3 - t + 1 . What is the minimum velocity of the particle

and f l ^ x h = c for all a 1 x 1 b , then f m ^ x h dx =

b

# on the interval 0 # t # 5 ?
A. t = 21
a

A. b - a
B. c ^ b - a h B. t = 41
C. 0
C. t = 1
D. c ^ a - b h
D. t =- 1
E. a - b
E. t = 2

17. If f ^ x h is continuous on the closed interval [a,b] and differentiable

on the open interval (a,b), which of the following must be true? 20. The function f is given by f ^ x h = 2x 4 - 3x 2 + 1 . On which of the fol-
lowing intervals is f decreasing?
I. If f ^a h = f ^b h , then for some value c between a and b, f l ^ c h = 0
A. c 23 , 3 m
II. If k is any number between f ^a h and f ^b h, there is a value c d^ a, b h
such that f ^ c h = k . B. c - 23 , 23 m
f ^bh - f^ah
III. There is a value c d^ a, b h such that f l ^ c h = C. c - 3, - 23 m
b-a .
A. I only D. c 0, 23 m
B. II only
C. III only E. c - 3, - 23 m and on c 0, 23 m

D. I and III
E. I, II, and III
8 AP CALCULUS AB PRACTICE EXAM AP CALCULUS AB PRACTICE EXAM 9

21. The function f shown in the graph above has horizontal tangents 22. The graph above shows the rate, in hundreds of passengers per
at (–2,1) and (1,–2) and vertical tangents at (–1,0) and (3,0). hour, at which commuters passed through a subway station during
For how many values of x in the interval (–5,5) is the function a 12-hour period. Which of the following is the best estimate of
not differentiable? the number of commuters who passed through the station in that
A. 0 12-hour period?
B. 1 A. 840
C. 2 B. 900
D. 3 C. 1300
E. 4 D. 8400
E. 9600
10 AP CALCULUS AB PRACTICE EXAM AP CALCULUS AB PRACTICE EXAM 11

23. The graph of f l , the derivative of f , is shown in the figure above. 24. The graph of a piecewise linear function f ^ x h for - 2 # x # 3 is

f ^ x h dx ?
Which of the following could be the graph of f ?
3

shown above. What is the value of #

-2
D. A. 13
A. 2
B. 41
6
C. 8
D. 9

B. E. 37
E. 6

C.
12 AP CALCULUS AB PRACTICE EXAM AP CALCULUS AB PRACTICE EXAM 13

27. What is the area of the region enclosed by the graphs of y = 2x 2 + 1

and y = x 3 - 1 and the vertical lines x =- 1 and x = 2 ?
A. 9
99
B. 12
C. 7
D. 155
12
E. 14

28. Which of the following is the x-coordinate of a point of inflection

25. The graph of a twice-differentiable function f is shown in the figure
above. Which of the following is true? on the graph of y = 41 x 4 - 21 x 2 + 3 ?

A. f l ^0 h 2 f m ^0 h 2 f ^0 h A. 33
B. f ^0 h 2 f l ^ 0 h 2 f m ^0 h
C. f ^0 h 2 f m ^0 h 2 f l ^0 h B. - 33
D. f m ^0 h 2 f ^0 h 2 f l ^0 h
C. ! 33
E. f l ^ 0 h 2 f ^ 0 h 2 f m ^ 0 h
D. ! 1
E. - 1
# ^x - 2 h dx = 2 ?
k
k2
26. What are all the values of k for which 2

-1
3
A. k =- 2

B. k = 6

C. k = ! 6
3
D. k = ! 6 or k =- 2
3
E. k = 2
14 AP CALCULUS AB PRACTICE EXAM AP CALCULUS AB PRACTICE EXAM 15

SECTION I, Part B 30. The side of a square is increasing at a constant rate of 0.2 centi­
Time—50 Minutes meters per second. In terms of the perimeter, P, of the square, what
Number of Questions—17 is the rate of change of the area of the square in square centimeters
A GRAPHING CALCULATOR IS REQUIRED FOR SOME QUES- per second?
TIONS ON THIS PART OF THE EXAMINATION A. 0.8P
Directions: Solve each of the following problems. After examining the B. 0.2P
choices, select the best answer. No credit will be given for anything writ- C. 0.1P
ten in the test book. D. 0.01P
In this test: E. 0.04P
1. The exact numerical value of the correct answer does not always ap-
pear among the answer choices given. When this happens, select the 31. The population of bacteria in a culture grows at a rate that can be
dy
answer that best approximates the exact numerical value. described by the equation dt = ky , where y is the population and
2. Unless otherwise specified, the domain of a function f is assumed to be t is the time, measured in hours. If the population doubles every 3
the set of all real numbers x for which f(x) is a real number. hours, k =

29. Which of the following is an equation of a tangent line to the graph A. ln a 2

3
k
of 2xy - x 2 = y at the point where y = 1 ? B. ln 3
A. y = 2x - 1 C. ln 2
B. y = 2x - 3
D. ln32
C. y = 1
D. y = x E. 1
E. y = x - 1
16 AP CALCULUS AB PRACTICE EXAM AP CALCULUS AB PRACTICE EXAM 17

33. The graph of a function f is shown above. Which of the following

statements about f is true?
A. x = 1 is not in the domain of f.
B. f has a relative minimum at x = 1
C. lim
x"1
f ^ x h = lim f ^ x h
- +
x"1

D. lim
x"1
f^ xh = 1

E. f is continuous at x = 1

32. The graphs of the derivatives of functions f, g, and h are shown

x 1 2 3 4 5
above. Which of the functions have a relative minimum on the
f(x) 4 3 7 1 3
interval - 3 1 x 1 3 ?
A. g only 34. The function f is continuous on the closed interval [1,5] and values
B. h only of the function are shown in the table above. If the values in the
C. f and g table are used to calculate a trapezoidal sum, the approximate value

f ^ x h dx is
5

D. g and h of #
E. f and h 1

A. 14
B. 14.5
C. 15
D. 29
E. 50
18 AP CALCULUS AB PRACTICE EXAM AP CALCULUS AB PRACTICE EXAM 19

35. The first derivative of a function f is given by 38. Let f be the function given by f ^ x h = ^ x - 4 h2 and let g be the func-
cos x ^ 2x sin x - cos x h tion given by g ^ x h = e 3x . At what value of x do the graphs have tan-
f l ^ xh = . On the interval 0 1 x 1 8 , how many
x2
gent lines that are perpendicular?
relative maxima does the function f have?
A. x c 1.143
A. 0
B. x c- 1.143
B. 1
C. x c 0.512
C. 2
D. x c- 0.512
D. 3
E. x c- 0.750
E. 4

39. Let f be the function defined by f ^ x h = x 2 - 13 . Which of the fol-

36. The base of a solid is a region in the first quadrant bounded by the x
x-axis, the y-axis, and the curve x 2 + 2y = 4 . If the cross sections of lowing statements about f are true?
the solid perpendicular to the x-axis are squares, the volume of the
I. f is differentiable at x =- 1 .
solid is
II. f is continuous at x =- 1 .
A. 72
5
III. f has an absolute maximum at x =- 1 .
B. 128
5
A. I only
C. 192
5 B. I and II
C. III only
D. 56
15 D. II and III
E. 64
15 E. I, II and III

37. Let f and g be differentiable functions on the interval ^3, 3 h such

that gl ^ x h = f ^ x h ln ^ x - 3 h , and f ^ x h 2 0 for all x 2 3 . Which of the
following must be true?
A. g ^ x h has a relative minimum at x = 4 .
B. g ^ x h has a relative maximum at x = 3 .
C. g ^ x h has a relative minimum at x = 4 and a relative maximum at
x = 3.
D. g ^ x h has no relative minimum or maximum.
E. There is not enough information to determine the relative ex-
trema of g ^ x h .
20 AP CALCULUS AB PRACTICE EXAM AP CALCULUS AB PRACTICE EXAM 21

40. Let f be a function that is differentiable on the open interval ^0, 5h . 43. If F ^ x h is an antiderivative of f ^ x h = 4x + 1 and F ^ 0 h = 1 , then
If f ^1 h = 2 , f ^3 h =- 1 , and f ^4 h = 5 , which of the following must F^2h =
be true?
A. 67
3
I. For some value 1 1 c 1 4 , f l ^ c h = 1 .
B. 64
3
II. The function f has at least three zeros on the interval ^0, 5h .
C. 53
3
III. For some value 1 1 c 1 4 , f ^ c h = 4 .
D. 50
3
A. I only
B. I and II E. 16
3
C. I and III
D. II and III
44. If r r
2 1 k 1 r and the area under the curve y = 2 sin x from x = 2 to
E. I, II and III
x = k is equal to 1.228, then k =

A. 2.617
41. If the base of a triangle is increasing at a rate of 2 centimeters per min-
B. 1.984
ute, and its area remains constant, at what rate is the height changing?
C. 1.797
A. b - 4h D. 0.882
E. 0.117
B. - 4hb

C. - 2bh
45. F ^ x h and f ^ x h are continuous functions such that F l ^ x h = f ^ x h for
D. 4bh
f ^ 3x h dx =
5

all x. #
E. 4bh 2

A. F ^ 5h - F ^ 2h

B. F ^ 15h - F ^ 6 h
3 3
x -c
42. If c ! 0 , then lim
x " c x2 - c2
is
A. 0 C. 31 6F ^5h - F ^2h@
B. 32c
D. 3 6F ^ 5h - F ^ 2h@
C. 2c
D. 2c 2 E. 31 6F ^15 h - F ^6 h@
E. nonexistent
22 AP CALCULUS AB PRACTICE EXAM AP CALCULUS AB PRACTICE EXAM 23

SECTION II SECTION II, PART A

GENERAL INSTRUCTIONS Time—45 minutes
You may wish to look over the problems before you begin, since you Number of problems—2
may not complete all parts of all problems. All problems are given equal A GRAPHING CALCULATOR IS REQUIRED FOR SOME PROB-
weight, but the parts of a particular problem are not necessarily given LEMS OR PARTS OF PROBLEMS.
equal weight. During the timed portion for Part A, you may work only on the prob-
A GRAPHING CALCULATOR IS REQUIRED FOR SOME PROB- lems in Part A.
LEMS OR PARTS OF PROBLEMS ON THIS SECTION OF THE EX- On Part A, you may use your calculator to solve an equation, find the
AMINATION. derivative of a function at, or calculate the value of a definite integral.
• You should write all work for each part of each problem in the However, you must clearly indicate the setup of your problem. If you
space provided for that part in the booklet. Be sure to write clearly use other built-in features, you must show the steps necessary to pro-
and legibly. If you make an error, you may save time by crossing it duce your results.
out rather than trying to erase it. Erased or crossed-out work will 3 2
1. Let R be the region bounded by y = x - 14x2x+ 53x - 40
and the
not be graded. +1
horizontal line y = 3 , and let S be the region bounded by the graph
• Show all your work. You will be graded on the correctness and 3 2

completeness of your methods as well as your answers. Correct of y = x - 14x2x+ 53x - 40

+1 and the horizontal lines y = 1 and y = 3 .
answers without supporting work may not receive credit.
a. Find the area of R.
• You may use your calculator to solve an equation, find the derivative
of a function at a point, or calculate the value of a definite integral. b. Find the area of S.
However, you must clearly indicate the setup of your problem,
namely the equation, function, or integral you are using. If you c. Set up, but do not evaluate, an integral that could be used to find
use other built-in features or programs, you must show the the volume of the solid generated when R is rotated about the hori-
mathematical steps. zontal line y = 1 .
• Unless otherwise specified, answers (numeric or algebraic) do not need
to be simplified. If your answer is given as a decimal approximation, it
should correct to three places after the decimal point.

• Unless otherwise specified, the domain of a function f is assumed to

be the set of all real numbers x for which f(x) is a real number.
24 AP CALCULUS AB PRACTICE EXAM AP CALCULUS AB PRACTICE EXAM 25

PART B
Time—60 minutes
Number of problems—4
NO CALCULATOR IS ALLOWED FOR THESE PROBLEMS.
During the timed portion for Part B, you may continue to work on
the problems in Part A without the use of any calculator.

2. A particle moves along the x-axis so that its velocity v at time t $ 0

is given by v ^ t h = 3t cos ^ t h . The graph of v is shown above for
0 # t # 3r . The position of the particle at time t is x ^ t h and its posi-
tion at time t = 0 is x ^ 0 h = 5 .

a. Find the acceleration of the particle at time t = r .

3. Jerry runs on a straight track starting at time t = 0 seconds and end-
b. Find the total distance traveled by the particle from time t = 0 to ing at time t = 10 seconds. During the time interval 0 # t # 10 , his
t = 3r . velocity v ^ t h in meters per second is modeled by the piecewise-
linear function whose graph is shown above.
c. Find the position of the particle at time t = 3r
2 .
a. Find Jerry’s acceleration at time t = 4.5 seconds. Indicate units of
d. For 0 # t # 3r , find the time t at which the particle is farthest to the measure.
left. Explain your answer.
v ^ t h dt and how (if at
10

b. Using correct units, explain the meaning of #

0

v ^ t h dt .
10

all) it differs from #

0

c. At what time (if any) did Jerry change direction? Explain your
reasoning.

d. Jeff runs on the same track, starting from the same starting line,
with a velocity given by f ^ t h = 5 - 25
x2
. At time t = 10 seconds, who
is closer to the starting line: Jerry or Jeff?
26 AP CALCULUS AB PRACTICE EXAM AP CALCULUS AB PRACTICE EXAM 27

d2 y
b. Find in terms of x and y. Describe the region in the xy-plane
dx 2
in which all solution curves to the differential equation are concave
4. Let f be a function defined on the closed interval - 3 # x # 3 with down.
f ^ - 1 h =- 2.8 and f ^ 1 h = 2.8 . The graph of f l , the derivative of f ,
consists of two line segments and a semicircle, as shown above. c. Let y = f ^ x h be a particular solution to the differential equation with
the initial condition f ^0 h = 1 . Does f have a critical point at x = 1 ?
a. For - 3 # x # 3 , find all values x at which f has a relative minimum. If so, describe the behavior of f at that point.
Justify your answer.
6. Let f be the function given by f ^ x h = x e-x x for all x.
2

b. For - 3 # x # 3 , find all values x at which the graph of f has a point

of inflection. Justify your answer. a. Write an equation for the tangent to the graph of f at x = 1 .

c. Find all intervals on which the graph of f is increasing and concave b. Find any critical points of f, and determine whether each is a relative
down. Explain your reasoning. maximum, relative minimum, or neither.

d. Find the absolute maximum value of f ^ x h over the closed interval c. Find the x-coordinate of each point of inflection.
- 3 # x # 3 . Explain your reasoning.

dy
5. Consider the differential equation dx = xy - y

a. On the axes provided, sketch a slope field for the given differential
equation at the points indicated.
28 AP CALCULUS AB PRACTICE EXAM AP CALCULUS AB PRACTICE EXAM 29

Solutions: AP Calculus AB Practice Test rem guarantees that f will be equal to zero at least once in the interval
[1,3] and at least once in the interval [3,5].

Multiple Choice 7. B.
dy
dt = ky and k ! 0 should signal that this is a differential equation
Section I Part A that leads to exponential growth, but if you don’t immediately recognize
dy
that, separate the variables and take the antiderivative. If y = kdt , then
dx = # ^ x - 3 h-2 dx = - ^ x - 3 h-1 4 = 7 - 3 - 4 - 3 =- 4 + 1 = 4
7 7

1. B. -1 -1 1 3
# 1
^ x - 3h2
7
ln y = kt + c and y = Ce kt .
4 4

8. C. If F ^ x h =
x

#
x

2. C. # cos tdt = sin t x

r
r
= sin x - sin 2 = sin x - 1 t 2 - t dt , then by the Fundamental Theorem of Cal-

culus, F l ^ x h = x 2 - x and F l ^ 3 h = 9 - 3 = 6 .
2
1
r
2

e2

cx + 1m
e2

a x2 +
x dx = ; 3 + ln xE =
3
1k x3
3. D. # #
9. E. If f ^ x h = tan ^ e 2x h , then f l ^ x h = 2e 2x sec 2 ^e 2x h and
x dx =
1 1
e e

^e 2 h3 ^1 h3 f m ^ x h = 2e 2x $ 2 sec ^ e 2x h sec ^e 2x h tan ^e 2x h $ 2e 2x + 4e 2x sec 2 ^e 2x h

c m - c e + ln 1 m = e + 2 - 1 3 + 1 = e -3 1 + 3 = e + 9e3 - 1
6 9 9 3
2
3 + ln e 3 e 3
= 8e 4x sec 2 ^ e 2x h tan ^ e 2x h + 4e 2x sec 2 ^ e 2x h = 4e 2x sec 2 ^ e 2x h62e 2x tan ^ e 2x h + 1@ .
3e 3e 3e

4. B. At the moment the particle changes direction, its velocity will be

zero. Velocity is the derivative of position, so v ^ t h = 5 - 23t . Set veloc- 10. A. If f ^ x h = sec ^3x h , then f l ^ x h = 3 sec ^3x h tan ^3x h , and
f l a 4 k = 3 sec 4 tan 4 = 3 ^ - 2 h^ 1 h =- 3 2 .
ity to zero, multiply through by 3, and solve 2t = 15 . Velocity is zero at r 3r 3r

t = 7.5 seconds, so the particle changes direction at 7.5 seconds.

x 3 e x for 0 # x 1 1
11. C. If f ^ x h = * e x , then lim f ^ x h = 1 3 $ e 1 = e and
^ 3 4 h
3

1 # x x - 1 dx . Use u-substitution
1 for 1 1 x # 3
-

5. D. The average value is 3 -

x"1
3
x
lim f ^ x h = 3 = e , so lim f^ xh = e .
1
e1
with u = x 4 - 1 and du = 4x 3 dx . Adjust the limits of integration. If x = 1 , x"1+
1 x"1

u = 0 , and if x = 3 , u = 80 . Then the average value is

12. C. If f m ^ x h = x ^ x - 2 h^ x + 1 h2 , then f m ^ x h = 0 when x = 0 , x = 2 ,
9 u C = 9 ^ 80 h C
80 80
1
# du 1
# 1 2 1 2
80

4^3 - 1h 0 4^3 - 1h 3
and x =- 1 , but check for a change in concavity before deciding that all
3 3
= 3-1 u 4 = u du = 2

8 3
2

0
0
three values represent points of inflection.
= 12 ^ 4 5 h =
1 64 $ 5 5 80 5
= 3 .
3

12
x x 1- 2 - 1 - 1 1 x 1 0 0 0 1 x 1 2 2 x22

6. A. If f is continuous on [1,5] and has two zeros in the interval, then f m ^ xh + 0 + 0 – 0 +

either f changes sign twice in the interval [1,5] or f has a relative max or
The graph of f has points of inflection when x = 0 and when x = 2 .
relative min with a y-value of 0. If k = 0 , the Intermediate Value Theo-
30 AP CALCULUS AB PRACTICE EXAM AP CALCULUS AB PRACTICE EXAM 31

f ^ x h ! 0 , it must be that gl ^ x h = 0 and therefore g is a constant function.

13. B. If 3x 2 4xy = 1 , differentiate implicitly. 6x - c 4x dx + 4y m = 0 or
dy

dy dy 6x - 4y 3x - 2y Since g ^2 h = 3 , g ^ x h = 3 .
6x - 4y = 4x dx . Solve for dx = 4x = 2x . Then when x = 1 ,
1 dy
3 - 4y = 1 so 2 = 4y and y = 2 . Evaluate dx =
3 $ 1 - 2 $ 21 2
= 2 = 1. 19. A. If the position of a particle is given by s ^ t h = t 4 - t 3 - t + 1 , then
2$1
the velocity v ^ t h = 4t 3 - 3t 2 - 1 and the acceleration a ^ t h = 12t 2 - 6t .
14. D. If a curve is defined by f ^ x h = 1 - x cos x , f l ^ x h = x sin x - cos x The minimum velocity of the particle on the interval 0 # t # 5 will be at-
tained when the acceleration is zero. 6t ^2t - 1 h = 0 when t = 0 or t = 21 .
and f l a r k
r r r r r
2 = 2 sin 2 - cos 2 = 2 . The slope of the tangent is 2 , so the Check the sign of a ^ t h .
2
slope of the normal line is m =- r . Use point-slope form to show that

an equation of the normal to the curve at a r k

2a rk 1 1 1
01t1 2 t2 2
2 , 1 is y - 1 =- r x - 2
t t=0 t= 2

or y =- 2rx + 2 . a^ t h 0 – 0 +

The velocity is a minimum when acceleration changes from negative to

15. B. If a function is given by f ^ x h = x2 + 3 , the instantaneous rate
x -1 positive, at t = 21 .
^ x 2 - 1 h - 2x ^ x + 3 h
of change of the function at x = 3 is f l ^ x h =
^ x 2 - 1 h2
=
20. E. Find the critical values of f ^ x h = 2x 4 - 3x 2 + 1 by finding
f l ^ x h = 8x 3 - 6x and solving f l ^ x h = 8x 3 - 6x = 0 . 2x ^ 4x 2 - 3 h = 0 so
x 2 - 1 - 2x 2 - 6x x 2 + 6x + 1
^ x - 1h ^ x 2 - 1h2
2 2 =- , evaluate at x = 3 .
3
f l ^ 3 h =-
9 + 18 + 1 28 7 x = 0 and x = ! 2 . Test the sign on the derivative between critical values.
^9 - 1 h2
=- 64 =- 16 .

16. C. If f l ^ x h = c for all a 1 x 1 b , f m ^ x h = 0 for all a 1 x 1 b . Then

3 3 3 3
x 1- 2 - 2 1x10 01x1 2 x2 2

f m ^ x h dx = f l ^ xh
b b

# # 0dx = 0x + C b
a
= 0. – + – +
a a

17. E. Rolle’s Theorem guarantees that if f ^a h = f ^b h , then for some The function f is decreasing on c - 3, - 23 m and on c 0, 23 m because the
value c between a and b, f l ^ c h = 0 . If k is any number between f ^a h and derivative is negative.
f ^ b h , the Intermediate Value Theorem assures that there is a value such
that c d^ a, b h. According to the Mean Value Theorem, there is a value 21. D. The function is not differentiable at the points where the tangent
f ^bh - f^ah
c ! ^ a, b h such that f l ^ c h = b-a .
becomes vertical, x =- 1 and x = 3 , at the jump discontinuity, x = 3 ,
and at the cusp, approximately x =- 2.7 .
g^ xh f ^ x h gl ^ x h - g ^ x h f l ^ x h
18. D. The derivative of h ^ x h = should be hl ^ x h =
f^ xh 6 f ^ x h@2
- g^ xhf l ^ xh
but you are told hl ^ x h = , so f ^ x h gl ^ x h = 0 . Given that
6 f ^ x h@2
32 AP CALCULUS AB PRACTICE EXAM AP CALCULUS AB PRACTICE EXAM 33

25. E. f ^0 h = 0 . At x = 0 , f is increasing so f l ^0 h 2 0 , and the graph of

f is concave down, so f m ^0 h 1 0 . Therefore, f l ^0 h 2 0 2 f m ^0 h .

# ^x - 2 h dx = 2 , find the antiderivative.

k
k2
26. D. If 2

-3

= c 3 - 2k m - ^ - 9 + 6 h = 3 - 2k + 3 . Then 3 - 2k + 3 = 2
x3 k k3 k3 k3 k2
3 - 2x -3

or 2k 3 - 12k + 18 = 3k 2 . Solve 2k 3 - 3k 2 - 12k + 18 = 0 by factoring

k 2 ^ 2k - 3 h - 6 ^ 2k - 3 h = 0 , to find that k = ! 6 or k = 2 .
3

27. B. The area of the region enclosed by the graphs of y = 2x 2 + 1

22. D. The total number of commuters who passed through the sta- and y = x 3 - 1 and the vertical lines x =- 1 and x = 2 is
tion in the 12-hour period can be approximated by the area of the
# ^2x + 1 h - ^ x 3 - 1 h dx = # ^- x + 2x 2 + 2 h dx =- 4 x 4 + 3 x 3 + 2x
2 2
2 3 1 2 2

two trapezoids, as shown, using the points ^0, 9 h , ^6, 3 h , and ^12, 13 h .
-1
=
-1 -1

A = 2 $ 6 ^ 9 + 3 h + 2 $ 6 ^ 3 + 13 h = 36 + 48 = 84 . The best estimate of the a- 4 + k a k

1 1 16 1 2 99
3 + 4 - - 4 - 3 - 2 = 12 .
number of commuters who passed through the station in that 12-hour
period is 8400 commuters. 28. C. Find the second derivative of y = 41 x 4 - 21 x 2 + 3 . yl = x 3 - x and
ym = 3x 2 - 1 . Set the second derivative equal to zero and solve to find
23. A. The graph of f l has zeros at x = 1 and x =- 1 , so the graph of f 1 3
x =! = ! 3 . Test for change in concavity.
3
should have critical points at ! 1 . At x =- 1 , the derivative changes from
positive to negative, indicating a maximum, and at x = 1 , the change x 3
x 1- 3 - 3 1 x 1 3 x2 3
3
from negative to positive indicates a minimum. This eliminates (C), (D), 3 3

and (E). The end behavior of the graph of the derivative indicates that f m ^ xh + – +
the graph of f increases to the right and decreases to the left at steeper
and steeper rates. The graph of (B) shows a slowing in the rates of in-
The function has points of inflection at both ! 33 .
crease and decrease. Graph (A) is most likely to be the graph of f.
Section I Part B
f ^ x h dx can be determined geometrically by calculating the
3

24. A. # 29. C. At the point where y = 1 , 2y - x 2 = y becomes 2x - x 2 = 1 or

-2

area under the graph. The key is to determine the x-intercept between
x 2 - 2x + 1 = 0 , so x = 1 . Find the derivative. 2xyl + 2y - 2x = yl so
–2 and –1. The line segment connects ^- 2, - 1 h to ^- 1, 2h and so has the 2^ x - yh
2xyl - yl = 2x - 2y and yl ^ 2x - 1 h = 2 ^ x - y h . Isolating yl = 2x - 1 .
equation y - 2 = 3 ^ x + 1h . Substituting 0 for y will give an x-intercept of
Evaluate at x = 1 , y = 1 to find the slope of the tangent line is
f ^ x h dx = 2 $ 3 ^ - 1 h + 2 $ 2 a 3 + 3 k =- 6 + 3 = 6 = 2 . 2^1 - 1h
3
5
x =- 3 . # 1 1 1 14 6 1 20 39 13
m = 2 - 1 = 0 . The equation of a tangent line to the graph of
-2

2xy - x 2 = y at ^ 1, 1 h is y - 1 = 0 ^ x - 1 h or y = 1 .
34 AP CALCULUS AB PRACTICE EXAM AP CALCULUS AB PRACTICE EXAM 35

cos x ^2x sin x - cos x h

30. C. You know that ds dt = 0.2 . The area of the square is tive is zero, solve = 0 or cos x ^ 2x sin x - cos x h = 0 .
x2
dA ds P
A = s so dt = 2s dt . The perimeter, P, is P = 4s , so s = 4 and Set each factor equal to zero. cos x = 0 at x = r 3r 5r
2

2 , 2 , 2 and
cos x ^ 2x sin x - cos x h = 0 by using your calculator, when x c 0.653 ,
dt = 2 $ 4 dt = 2 ^ 0.2 h = 0.1P .
dA P ds P

dy dy x c 3.292 , and x c 6.362 . For relative maxima, you’re looking for a

31. D. If dt = ky , y = kdt and ln y = kt + C . This becomes the expo-
change in the sign of the derivative from positive to negative, repre-
nential growth equation y = Ce kt . If the population doubles every 3 hours, senting the function changing from increasing to decreasing. From the
ln 2
2 = le 3k so ln 2 = 3k and k = 3 . graphs, you can see that this occurs at x = r 5r
2 , x = 2 , and x c 3.292 .
32. C. The graph will have a relative minimum when the graph changes On the interval 0 1 x 1 8 , the function f has 3 relative maxima.
from decreasing to increasing, which is when the derivative changes from 36. E. The base of a solid is a region in the first quadrant bounded by the
negative to positive. f l has one change of sign, from negative to positive, in- x2
x-axis, the y-axis, and x 2 + 2y = 4 or y = 4 - 2 . Each square cross sec-
dicating that f has a relative minimum. gl has two changes of sign, indicating
c4-x m .
2 2
x2
both a relative maximum and a relative minimum. hl has one sign change, tion has a side equal to s = 4 -2 and therefore an area of A = 2
Integrate to find the volume. V = # c 4 - dx = 4 # ^ 16 - 8x 2 + x 4 h dx
x m
2 2
2 2
but the change from positive to negative indicates a relative maximum, 1
2
rather than a minimum. 0 0

= 4 916x - 3 x + 5 x C = 4 932 - 3 + 5 C = 15 .
1 8 3 1 52 1 64 32 64
33. C. The graph shows that the function is defined for x = 1 , because 0

f ^ 1 h = 1 , but the graph does not change from decreasing to increas- 37. A. The critical values of g occur when the derivative is zero.
gl ^ x h = f ^ x h ln ^ x - 3 h = 0 when f ^ x h = 0 or ln ^ x - 3 h = 0 . Since f ^ x h 2 0 ,
ing at x = 1 , so there is no relative minimum at x = 1 . lim f ^ x h = 5 and
the relative extremum of g occurs when ln ^ x - 3 h = 0 , which is when
-
x"1

lim f ^ x h = 5 so C is true, but D is not. Since lim f ^ x h = 5 but f ^ 1 h = 1 , the

x " 1+ x"1
x - 3 = 1 , or x = 4 . For values 3 1 x 1 4 , gl ^ x h 1 0 and when x 2 4 ,
function is not continuous at x = 1 . gl ^ x h 2 0 , which means that at x = 4 , the graph of g ^ x h changes from

x 1 2 3 4 5 decreasing to increasing. g ^ x h has a relative minimum at x = 4 .

f  (x) 4 3 7 1 3 38. B. Find the derivatives f l ^ x h = 2 ^ x - 4 h and gl ^ x h = 3e 3x . If the tan-

34. B. The function f is continuous on the closed interval 61, 5@ and gent lines are perpendicular, the slopes are negative reciprocals, so solve
-1
2^ x - 4h
values of the function are shown in the table above. If the values in 3e 3x = by calculator to find x c- 1.143 .

the table are used to calculate a trapezoidal sum, the approximate

39. B. If f ^ x h = x 2 - 13 , then f l ^ x h = 2x + 34 . The function is differen-
f ^ x h dx = 2 ^ 4 + 3 h + 2 ^ 3 + 7 h + 2 ^ 7 + 1 h + 2 ^ 1 + 3 h
5
x x
value of # 1 1 1 1
tiable and continuous at x =- 1 . To determine where the relative extre-
1

= 2 ^ 4 + 2 $ 3 + 2 $ 7 + 2 $ 1 + 3 h = 2 ^ 4 + 6 + 14 + 2 + 3 h = 14.5 . ma fall, solve f l ^ x h = 2x + 34 = 0 . 2x =- 34 and 2x 5 =- 3 so x = 5 - 2

1 1 3
.
x x
35. D. The first derivative of a function f is given by The extrema do not occur at x =- 1 .
cos x ^ 2x sin x - cos x h
f l ^ xh = . To determine where the first deriva-
x2
36 AP CALCULUS AB PRACTICE EXAM AP CALCULUS AB PRACTICE EXAM 37

40. C. If f is differentiable (and therefore continuous) on the open in- Section II Part A
terval ^0, 5h , and f ^ 1 h = 2 , f ^3 h =- 1 , and f ^4 h = 5 ,then the Mean Val- 1.
ue Theorem guarantees that there is at least one value 1 1 c 1 4 such a. Use the calculator to find the intersection points of
f^4h - f^1h 5 - 2
that f l ^ c h =
3 2
= 4 - 1 = 1 , and the Intermediate Value Theo- x - 14x + 53x - 40
4-1 y= 2x + 1 and y = 3 , and then integrate. The area of R is
rem guarantees that for any value f ^1 h 1 k 1 f ^4 h there is some value
# c x - 14x2x++53 - 3 m dx . 0.546 .
2.894
3 2
x - 40
1 1 c 1 4 such that f ^ c h = k . Therefore I is true and taking k = 1 , III
1
1.556

is true. The Intermediate Value Theorem will also assure that there is b. Use the calculator to find the intersection points of
a zero between x = 1 and x = 3 , and a second zero between x = 3 and x 3 - 14x 2 + 53x - 40
y= and y = 1 . Find the area bounded by
x = 4 . There is no information to allow you to determine if there is a 2x + 1
third zero in the interval. x 3 - 14x 2 + 53x - 40
y= 2x + 1 and y = 1 .
41. C. The area of the triangle is A = 21 bh so the area is changing at
c x - 14x + 53x - 40 - 1 m dx . 5.075 . Then subtract the area of
4.224
3 2
# 2x + 1
dt = 2 9b dt + h dt C . Since the area remains constant,
a rate equal to dA 1 dh db 1.122

region R. 5.075 - 0.546 = 4.529

dt = 0 . Substitute 0 = 2 9b dt + 2hC and solve for dt . 2 b dt =- 2h
dA 1 dh dh 1 dh
3 2
c. The outer radius is the distance from y = 1 to y = x - 14x2x+ 53x - 40
,
and dh 2h
dt =- b .
+1

x -c 3 3
^ x - c h^ x 2 + cx + c 2 h x 2 + cx + c 2
and the inner radius is the distance from y = 1 to y = 3 . The volume of
42. B. If c ! 0 , then lim
^ x - c h^ x + c h
x " c x2 - c2
= lim
x"c
= lim
x"c x+c
3c 2
3c the solid generated when R is rotated about the horizontal line y = 1 is
= 2c = 2 .
;c 1 - x - 14x2x+ - ^1 - 3 h2E dx .
53x - 40 m2
2.894
3 2
given by r # +1
f ^ x h dx = F ^ 2 h - F ^ 0 h.
2

43. E. By the Fundamental Theorem of Calculus, # 1.556

0 2.
4x + 1 dx = 4 $ 3 ^ 4x + 1 h = 6 69 - 1 @ = 6 627 - 1@ = 3 and
2

# a. The acceleration of the particle is given by a ^ t h = vl ^ t h = 3 cos ^ t h - 3t sin ^ t h.

1 2 1 1 13
Find
3
2
2 3
2
3
2
0

At time t = r , a ^r h = 3 cos ^r h - 3r sin ^r h =- 3 .

0

substitute F ^ 2h - F ^ 0 h = 13
3 or F ^ 2 h = 3 + 1 = 3 .
13 16

b. The total distance traveled by the particle from time t = 0 to

# ^2 - sin xhdx = 2x + cos x
k

44. A. = 2k + cos k - r - 0 . You’re given

k r 3r
3r 2 2 3r

# # # #
r

t = 3r is
2
r
2 3t cos t dt = 3t cos tdt - 3t cos tdt - 3t cos tdt
that the area under the curve from x = r
2 to x = k is equal to 1.228, so
0 0 r 5r

= 1.712 - ^ - 18.850 h + 37.699 - ^ - 26.562 h = 84.823 .

2 2

2k + cos k - r = 1.228 . Solve by calculator and k . 2.617 .

c. The position of the particle at time t = 3r
2 is
45. E. If u = 3x , du = 3dx or dx = du 3 . When x = 2 , u = 6 and when
3r

x^0h +
2

# 3t cos tdt = 5 - 17.137 =- 12.137 .

x = 5 , u = 15 . # f ^ 3x h dx = 3 # f ^ u h du = 3 6F ^ 15 h - F ^ 6 h@ .
5 15
1 1 0

2 6
38 AP CALCULUS AB PRACTICE EXAM AP CALCULUS AB PRACTICE EXAM 39

d. The time t at which the particle is farthest to the left is the abso- 4.

lute minimum value of x ^ t h . First consider the critical points. From a. The function f will have a relative minimum when the derivative
changes from negative to positive, which occurs at x =- 2 .
the graph of v ^ t h = 3t cos ^ t h , you can see that velocity is zero at t = r
2, b. Point of inflection occur when concavity changes. We look for the
3r 5r r
t = 2 , and t = 2 . At t = 2 , the change in the sign of the deriva- points at which the second derivative changes sign, or when the de-
tive is from positive to negative, indicating a relative maximum, and the rivative changes from increasing to decreasing or vice versa. The rela-
tive maximum of the derivative, at x = 0 , represents such a change and
a
3r k
same is true at t = 5r
2 . From part c, you know that x 2 =- 12.137 therefore a point of inflection.
and the change of sign of the derivative from negative to positive in- c. The graph of f is increasing when the derivative is positive, and con-
dicates a relative minimum. Check the ends of the interval. x ^ 0 h = 5 cave down when the second derivative is negative or the first derivative

and x ^3r h = x ^ 0 h +
3r
is decreasing. The graph of the derivative is positive but decreasing from
# 3t cos tdt = 5 - 6 =- 1 . The relative minimum at
0 x = 0 to x = 2 , so f is increasing and concave down on 0 1 x 1 2 .
t = 2 is the absolute minimum of x ^ t h . The particle is farthest to the
3r
d. From x =- 3 to x =- 2 , f is decreasing. From x =- 2 to x = 2 , f is
left at t = 3r
2 .
increasing, and from x = 2 to x = 3 , it is decreasing. The absolute maxi-
mum is occurring either at x =- 3 or x = 2 . It is possible to determine
Section II, Part B the value of f ^- 3h and f ^2h by integrating.
3.
x 2 + 4x + c 1 - 3 # x #-1
- 1 1 x 1 1 so f ^ x h = * # ^ 2 + 1 - x 2 h dx
2x + 4 - 3 # x #-1
a. Acceleration is the derivative of velocity, so his acceleration at time f l ^ xh = * 2 + 1 - x2 -1 1 x 1 1 .
t = 4.5 seconds is the slope of the line segment connecting ^ 4, 5 h and - 2x + 4 11x13 - x 2 + 4x + c 2 1#x#3
^7, - 1 h . m = 7 - 4 =- 3 =- 2 meters per second.
-1 - 5 6 x 2 + 4x + 0.2 - 3 # x #-1
Since f ^ - 1 h =- 2.8 and f ^1 h = 2.8 , f ^ x h = * # ^2 + 1 - x h dx -1 1 x 1 1 .
2

v ^ t h dt represents Jerry’s displacement, or distance from the starting

10

#
2
- x + 4x - 0.2 1#x#3
b.
0 Then f ^- 3h = ^ - 3 h2 - 12 + 0.2 =- 2.8 and f ^2 h =- 4 + 8 - 0.2 = 3.8 .
v ^ t h dt is the total distance
10

line, in meters, after 10 seconds, while # The absolute maximum occurs at ^2, 3.8 h .
0

Jerry travels, in meters, over the course of the ten seconds.

c. Jerry changes direction at 6.5 seconds, indicated by the velocity chang-
ing from positive to negative.
f ^ t h = 5 - 25 , his displacement is
x2
d. If Jeff’s velocity is

# c 5 - 25
x2 m
10 10
x3 1000 40 100 2
dx = 5x - 75 = 50 - 75 = 50 - 3 = 3 = 36 3 meters.
0 0

Jerry’s displacement can be found by the area under the curve. Using
area calculations from geometry, 26.75 - 7.25 = 19.5 meters. At time
t = 10 seconds, Jerry is closer to the starting line.
40 AP CALCULUS AB PRACTICE EXAM AP CALCULUS AB PRACTICE EXAM 41

5. f^ xh = e and f l ^ x h = e ^ x - 1h .
x2 x2
ln 1 = c and c = 0 . Therefore 2
-x 2
-x

dy
a. Calculate the value of dx = xy - y at each point and plot.
f l ^ 0 h =- 1 and f l ^ 2 h = 1 so x = 1 is a relative minimum.
y –2 –1 0 1 2
6. Let f be the function given by f ^ x h = x e-x x for all x.
2

x
e x ^ 2x - 1h - ^ x 2 - x h e x e x ^ - x 2 + 3x - 1 h
–2 6 3 0 –3 –6 a. Find the derivative f l ^ x h =
^ e x h2
=
e 2x
–1 4 2 0 –2 –4
- x 2 + 3x - 1
. Evaluate f ^1 h = 1 -1 1 = 0 and f ^1 h = - 1 +13 - 1 = 1e .
2 2

0 2 1 0 –1 –2 =
ex e e
1 0 0 0 0 0 The equation for the tangent to the graph of f at x = 1 is y - 0 = 1e ^ x - 1h
2 –2 –1 0 1 2
or y = x - 1
e .
b. To find any critical points of f, set f l ^ x h = - x +e3x x - 1 = 0 and solve.
2

Since e x 2 0 for all x, solve - x 2 + 3x - 1 = 0 to find that the critical

- 3 ! 9 - 4 ^ - 1 h^ - 1 h -3 ! 5 3! 5
2^- 1h
numbers are x = = -2 = 2 . Do a first
derivative test for determine the behavior of f at each point.

x 0 3- 5 1 3+ 5 3
2 2
f l ^ xh – 0 + 0 –

= x dx + dx = x ^ xy - y h + y - ^ xy - y h = x 2 y - xy + y - xy + y
d2 y dy dy
b. The function f has a relative minimum at x = 3 -2 5 and a relative maxi-
dx 2
= x 2 y - 2xy + 2y . Solution curves to the differential equation are con- mum at x = 3 +2 5 .
= x 2 y - 2xy + 2y 1 0 or y ^ x 2 - 2x + 2 h 1 0 . Since
d2 y
cave down when c. Find the x-coordinate of each point of inflection, find the second
dx 2
e x ^ - 2x + 3 h - e x ^ - x 2 + 3x - 1 h e x ^ x 2 - 5x + 4 h
x 2 - 2x + 2 2 0 for all x, the second derivative is negative when y 1 0 . derivative. f m ^ x h =
^e x h2
=
^ x - 1h^ x - 4 h
e 2x
dy
c. At x = 1 , dx = xy - y = y - y = 0 so x = 1 is a critical point. Check the = . The second derivative will be zero at x = 1 and x = 4 .
ex
dy
value of the derivative above and below x = 1 . At x = 0 , dx =- y and at Examine the sign of the second derivative to verify that there is a
dy
x = 2 , dx = 2y - y = y . This is adequate to tell you that there is a change change in the concavity of the graph of f.
of sign, so the point is a relative of extremum of some kind, but further
x 0 1 2 4 5
investigation is necessary to determine if it is a maximum or minimum. If
f m ^ xh
dx = xy - y , then y = ^ x - 1 h dx and ln y = 2 x - x + c . Since f ^ 0 h = 1 ,
dy dy 1 2 + 0 – 0 +

The graph of f has inflection points at x = 1 and x = 4 .