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Sample Tests for The Advanced Placement Examinations
The Advanced Placement Calculus Examinations are three hours and 15 minutes long and evaluate how well students have mastered the concepts and techniques in either AB or BC Calculus course.
Each examination consists of (1) a multiple-choice section that tests proficiency over a broad spectrum of topics and (2) a problem section that requires students to demonstrate their ability in solving
problems requiring a more extensive chain of reasoning. Students taking the BC test will receive both
an AB and a BC grade to better assist colleges in giving credit for scores.
The examination will be based on the Advanced Placement Course Description.
MULTIPLE CHOICE (Section I)
Part A
Part B
28 questions
17 questions
55 minutes
50 minutes
No calculator allowed
Graphing calculator required (students will
have to decide whether a calculator is appropriate since not all problems need a calculator)
45 minutes
45 minutes
Graphing calculator required
No calculator allowed
FREE RESPONSE (Section II)
Part A
Part B
3 questions
3 questions
The problem section (6 questions completed in 90 minutes) is designed to utilize the graphing calculator for some problems. During the timed portion for Part B, students may work on Part A questions without the use of a calculator.
Both the multiple-choice and free-response sections are given equal weight in determining the
grade for the examination.
The following tests are intended to give students practice with the type of exam they will take. You
may wish to simulate the conditions of the exam, providing the exact amount of time for each section
and using the multiple choice answer sheet that is provided on page 137.
100
Sample AB Test
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Advanced Placement Calculus AB test
Section I—Part A (55 minutes)
Choose the best answer for each question. Your score is determined by subtracting one-fourth of the
number of wrong answers from the number of correct answers. Calculators are not permitted.
1.
dy
d2y
6 0 and 2 , 6 0?
For the graph shown, at which point is it true that
dx
dx
(A) A
(B) B
(C) C
(D) D
(E) E
2. Find the area of the region bounded by the x-axis and the graph of y = 1 x + 1 2 1 x - 2 2 2.
(A)
5
4
(B) 2
3
4
(C) 5
1
4
(D) 6
1
4
(E) 6
3
4
3. Which of the following is an antiderivative of x2 sec2 x3?
(A) 2x sec2 x3 П© 6x4 sec2 x3 tan x3
(B) 2x sec2 x3 П© 6x3 sec x3
(C)
1
tan x3 - 5
3
(D) 3 tan x3 П© вђІ
1
(E) - cot x 3 + 4
3
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Sample AB Test
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4. Line L is tangent to the curve defined by 2xy2 ПЄ 3y П­ 18 at the point (3, 2). The slope of line L is
(A)
21
8
(B)
32
3
(C) -
10
21
(D)
8
21
(E) -
8
21
5.
A bicyclist rides along a straight road starting from home at t П­ 0. The graph above shows the
bicyclist’s velocity as a function of t. How far from home is the bicyclist after 2 hours?
(A) 13 miles
(B) 16.5 miles
(C) 17.5 miles
6. Find the value of x at which the graph of y =
(B) 4 2>3
(A) 2
7. Find lim
xSq
(A)
(D) 18 miles
(E) 20 miles
1
+ 1x has a point of inflection.
x
(C) 4
(D) 6
(C) 1
(D) -
(E) 8
2x - 4x3
8x3 + 4x2 - 3x
2
3
102 Sample AB Test
(B)
3
2
1
2
(E) -
3
4
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8. Which of the following is a slope field for the differential equation dy>dx = - 2x + y?
(A)
(B)
y
(E)
y
y
x
x
x
(D)
(C)
y
y
x
x
1
9. Let f (x) П­ cos (3вђІx2). Find fВї a b .
3
(A) - 13вђІ
(B) 13вђІ
(C) 0
(D) -
13p
2
(E) ПЄвђІ
10. Assume that f (x) is a one-to-one function. The area of the
shaded region is equal to which of the following definite integrals?
I.
II.
III.
4
3 f1x2 - 3 4 dx
0
3 f1x2 - 3 4 dx
3
f -1 1y2 dy
30
34
31
(A) I only
(B) II only
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(C) III only
(D) I and III
(E) II and III
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11.
The graph of a function y П­ f(x) is shown above. Which of the following are true for the function f?
I. fВї 1 22 is defined.
II. lim+ f1x 2 = lim- f1x2
xS2
xS2
III. fВї 1x 2 6 0 for all x in the open interval (ПЄ1, 2).
(A) I only
(B) II only
12. Let f 1x 2 = sin-1x. Find fВї a
(A)
p
4
13. Evaluate
(B)
3
12
2
12
2
(C) III only
(D) II and III
(E) I, II and III
(D) 12
(E) Undefined
b.
(C)
1
2
1cos - e2x 2 dx.
(A) -sinx -
1 2x
e + C
2
1
(B) sinx - e2x + C
2
(C) - sinx - 2e2x + C
(D) sinx - 2e2x + C
(E) - cosx -
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Sample AB Test
1 2x
e + C
2
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3
14. Let f(x) П­ ex
-2x2 -4x +5
(A) ПЄ2
. Then f has a local minimum at x П­
(B) -
2
3
(C)
2
3
(D) 1
(E) 2
15. The acceleration of a particle moving along the x-axis is a(t) П­ 12t ПЄ 10.
At t П­ 0, the velocity is 4.
At t П­ 1, the position is x П­ 8.
Find the position at t П­ 2.
(A) 5
(B) 4
(C) 10
(D) 11
(E) 7
16. Let f be differentiable for all real numbers. Which of the following must be true for any real
numbers a and b?
a
I.
32
b
II.
b
f1x2 dx =
3a
32
a
f1x2 dx +
3a
f1 x 2 dx
¢ 3 f1x2 4 2 + f¿ 1x2 ≤ dx = 3 f1 b 2 4 2 - 3f 1 a 2 4 2
b
III.
3b
b
3 f1x2 dx = 3
(A) I only
3a
f1x 2 dx
(B) II only
(C) I and II
(D) I and III
3x
at x П­ 3.
x -6
(A) 5x П© y П­ 18 (B) 5x ПЄ y П­ 12 (C) 5x П© 3y П­ 24 (D) x ПЄ 5y П­ ПЄ12
17. Find an equation of the line normal to the graph of y =
В© 2007 Pearson Education, Inc.
(E) I, II, and III
2
(E) x П© y П­ 6
Sample AB Test
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1x + h2 2 - x2
. For what value of x does g(x) П­ 2?
hS0
h
18. Let g 1x2 = lim
(A) x П­ 1
(B) x П­ 2
(C) x П­ 3
(D) x П­ 4
(E) x П­ 5
19. Let f be a differentiable function of x that satisfies f(1) П­ 7 and f(4) П­ 3. Which of the following
conditions would guarantee that the tangent line at x П­ c is parallel to the secant line joining
(1, f(1)) to (4, f(4))?
(A) f1c 2 =
3
2
(B) f1c 2 = 5
(C) fВї 1c2 = -
3
4
(D) fВї 1 c2 = -
4
3
(E) fВї 1 c 2 = -
4
3
20. Let f(x) П­ x3 ПЄ 12x. Which statement about this function is false?
(A) The function has a relative minimum at x П­ 2.
(B) The function is increasing for values of x between ПЄ2 and 2.
(C) The function has two relative extrema.
(D) The function is concave upward for x > 0.
(E) The function has one inflection point.
3
21.
32
(A)
8x 1x2 - 52 dx
74
3
22. Let f 1x2 =
(B) 30
(C) 90
370
3
(D) 112
(E)
(D) 4
(E) 5
x
d
1t2 + 16 dt. What is f1 -3 2 ?
dx 30
(A) ПЄ5
(B) ПЄ4
(C) 3
dy
1
= xy2 and y = - when y П­ 2, what is y when x П­ 4?
dx
3
1
1
1
1
(A) (B) (C) (D)
3
5
9
3
23. If
106
Sample AB Test
(E)
1
9
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24. Use the Trapezoidal Rule with n П­ 3 to approximate the area between the curve y П­ x2 and the
x-axis for 1 Х… x Х… 4.
(A) 14
(B) 21
(C) 21.5
(D) 29
(E) 30
25. Let f(x) be a continuous function that is defined for all real numbers x.
x2 - x - 6
If f1x2 = 2
when x2 - 5x + 6 Z 0, what is f (3)?
x - 5x + 6
(A) 5
(B) 4
(C) 2
(D) 1
(E) 0
26. Find the derivative of cos3 2x.
(A) ПЄsin3 2x
(B) ПЄ6 cos2 2x.
(C) 6 cos2 2x sin 2x
(D) ПЄ3 cos2 2x sin 2x
(E) ПЄ6 cos2 2x sin 2x
27. Let f be a twice-differentiable function whose derivative f Вї (x) is increasing for all x. Which of the
following must be true of all x?
I. f(x) Пѕ 0
II. f Вї (x) Пѕ 0
III. f – (x) Ͼ 0
(A) I only
(B) II only
(C) III only
(D) I and II
(E) II and III
28. The function f (x) П­ x3 ПЄ 6x2 П© 9x ПЄ 4 has a local maximum at
(A) x П­ 0
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(B) x П­ 1
(C) x П­ 2
(D) x П­ 3
(E) x П­ 4
Sample AB Test
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Section I—Part B (50 minutes)
Choose the best answer for each question. (If the exact answer does not appear among the choices,
choose the best approximation for the exact answer.) Your score is determined by subtracting one-fourth
of the number of wrong answers from the number of correct answers. You may use a graphing calculator.
29. Which of the following functions has the fastest rate of growth as x S q ?
(A) y П­ x18 ПЄ 5x
(B) y П­ 5x2
(C) y П­ ln x2
(D) y П­ (ln x)2
(E) y П­ e0.01x
30. The velocity of a particle moving along a straight line is given by v(t) П­ 3x2 ПЄ 4x. Find an
expression for the acceleration of the particle.
(A) x3 ПЄ 4
(B) x3 ПЄ 2x2
(C) 3x2 ПЄ 4
(D) 3x ПЄ 4
(E) 6x ПЄ 4
31. Find the average value of the function y П­ x3 ПЄ 4x on the closed interval [0, 4].
(A) 8
(B) 12
(C) 24
(D) 32
(E) 48
32. A region is enclosed by the x-axis and the graph of the parabola y П­ 9 ПЄ x2. Find the volume
of the solid generated when this region is revolved about the x-axis.
(A) 36вђІ
(B) 40.5вђІ
(C) 129.6вђІ
(D) 194.4вђІ
(E) 259.2вђІ
33. Which of the following is an antiderivative of x 1x2 + 3?
1
1
1
2
(A) x3>2
(B) x3
(C) 1 x2 + 3 2 3>2 (D) 1 x2 + 3 2 3>2
3
3
3
3
108 Sample AB Test
(E) (x2 П© 3)3/2
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34.
x
f(x)
3.3
3.69
3.4
3.96
3.5
4.25
3.6
4.56
3.7
4.89
Let f be a differentiable function that is defined for all real numbers x. Use the table above to
estimate fВї 13.62 .
(A) 0.3
(B) 1.8
(C) 2.7
(D) 3.0
(E) 3.2
35. The weight in pounds of a certain bear cub t months after birth is given by w(t). If w(2) П­ 36,
dw
w(7) П­ 84, and
was proportional to the cub’s weight for the first 15 months of his life, how
dt
much did the cub weigh when he was 11 months old?
(A) 125 pounds
36. Let f1x2 = b
(B) 135 pounds
(C) 145 pounds
(D) 155 pounds
(E) 165 pounds
3x2 - 4, for x Х… 1
.
6x - 5, for x 7 1
Which of the following are true statements about this function?
I. lim f1x2 exists.
xS1
II. fВї 112 exists.
III. lim fВї 1x2 exists.
xS1
(A) None
(B) II only
(E) III only
(D) II and III
(E) I, II, and III
37. Two particles are moving along the x-axis. Their positions are given by x1(t) П­ 2t2 ПЄ 5t П© 7 and
x2(t) П­ sin 2t, respectively. If a1(t) and a2(t) represent the acceleration functions of the particles,
find the numbers of values of t in the closed interval [0, 5] for which a1(t) П­ a2(t).
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4 or more
(D) 3
(E) 4 or more
38. The function f(x) П­ ex ПЄ x3 has how many critical points?
(A) 0
(B) 1
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(C) 2
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39. A dog heading due north at a constant speed of 2 meters per second trots past a fire hydrant at
t П­ 0 sec. Another dog heading due east at a constant speed of 3 meters per second trots by the
hydrant at t П­ 1 sec. At t П­ 9 sec, the rate of change of the distance between the two dogs is
(A) 3.2 m/sec
(B) 3.6 m/sec
40. Let f (x) П­ x5 П© x. Find the value of
(A) -
1
6
(B)
1
6
(C) 4.0 m/sec
(D) 4.4 m/sec
(E) 4.8 m/sec
(D) 6
(E) 81
d ПЄ1
f (x) at x П­ 2.
dx
(C)
1
81
1 ln t 2 2 3
ft >sec for t Х† 1 sec. If the
t
volume of the balloon is 1.3 ft3 at t П­ 1 sec, what is the volume of the balloon at t П­ 5 sec?
41. Suppose air is pumped into a balloon at a rate given by r1 t 2 =
(A) 2.7 ft3
(B) 3.0 ft3
(C) 3.3 ft3
(D) 3.6 ft3
(E) 3.9 ft3
42. Find the approximate value of x where f1 x 2 = x2 - 3 1x + 2 has its absolute minimum.
(A) ПЄ4.5
(B) ПЄ2
(C) 0
(D) 0.5
(E) 2.5
43.
The graph of y = fВї 1 x2 is shown. Which of the following statements about the function f(x) are true?
I. f(x) is decreasing for all x between a and c.
II. The graph of f is concave up for all x between a and c.
III. f(x) has a relative minimum at x П­ a.
(A) I only
110
Sample AB Test
(B) II only
(C) III only
(D) I and III
(E) I, II, and III
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44. Suppose f and g are even functions that are continuous for all x, and let a be a real number. Which
of the following expressions must have the same value?
a
I.
3-a
3 f1x2 + g1x 2 4 dx
a
II. 2
30
3 f1x2 + g1x 2 4 dx
a
III.
3-a
a
f1x2 dx +
(A) I and II only
3-a
g1 x2 dx
(B) I and III only
(C) II and III only
(D) I, II, and III
(E) None
45. Let f(x) П­ g(h(x)), where h(2) П­ 3, hВї 1 2 2 = 4, g(3) П­ 2, and gВї 1 32 = 5. Find fВї 1 2 2 .
(A) 6
(B) 8
(C) 15
(D) 20
(E) More information is needed to find fВї 1 2 2 .
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Sample AB Test
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Advanced Placement Calculus AB test
Section II (90 minutes)
Show your work. In order to receive full credit, you must show enough detail to demonstrate a clear understanding of the concepts involved. You may use a graphing calculator. Where appropriate, you may
give numerical answers in exact form or as decimal approximations correct to three decimal places.
For Problems 1–3, a graphing calculator may be used. (45 minutes)
1. For t Х† 0, a particle moves along the x-axis with a velocity given by v(t) П­ 2t ПЄ 5 sin вђІt.
At t П­ 0, the particle is located at x П­ 0
(a) Write an expression for the acceleration a (t) of the particle.
(b) Write an expression for the position x(t) of the particle.
(c) For what values of t (t Х† 0) is the particle moving to the left?
(d) For t Пѕ 1, find the position of the particle the first time the velocity of the particle is zero.
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2. A function y П­ f(x) is defined by 5 1x + xy + y3 + 11.
(a) Find an expression for f' 1 x2 =
dy
in terms of x and y.
dx
(b) Find the equation of the line that is tangent to the graph of y П­ f (x) at the point (0.25, 2).
(c) Use the tangent line from part (b) to estimate f (0.6).
(d) Write an equation whose solution is the exact value of f(0.6).
To the nearest thousandth, what is f(0.6)?
(e) Would it be appropriate to use the tangent line from part (b) to estimate f(ПЄ0.1)? Explain.
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3.
The figure above shows a pump connected by a flexible tube to a spherical balloon. The pump
consists of a cylindrical container of radius 8 inches, with a piston that moves up and down accord24
+ ln 1 t + 1 2 for 0 Х… t Х… 100, where t is measured in seconds and
ing to the equation h1 t2 =
t +1
h (t) is measured in inches. As the piston moves up and down, the total volume of air enclosed in
the pump and the balloon remains constant, and r (t) П­ 0 at t П­ 0. Вў The volume of a sphere with
4
radius r is ␲r 3. ≤
3
(a) Write an expression in terms of h (t) and r(t) for the total volume of the air enclosed in the
pump and the balloon. (Do not include the air in the flexible tube.)
(b) Find the minimum volume of air in the pump and when it occurs.
(c) Find the rate of change of the volume of the air enclosed in the pump at t П­ 3 sec.
(d) At t П­ 3 sec, find the radius of the balloon and the rate of change of the radius of the balloon.
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No calculator may be used for Problems 4, 5, & 6. Students may continue working on Problems 1–3, but
may not use a calculator. (45 minutes)
4.
The shaded region is enclosed by the graphs of y П­ x3 and y П­ 4 14x.
(a) Find the coordinates of the point in the first quadrant where the two curves intersect.
3
(b) Use an integral with respect to x to find the area of the shaded region.
(c) Set up an integral with respect to y that could be used to find the area of the shaded region.
(d) Without using absolute values, write an integral expression that gives the volume of the solid
generated by revolving the shaded region about the line x П­ ПЄ1. Do not evaluate.
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Sample AB Test
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5.
The graph of a differentiable function g is shown. The area of the shaded region is 8 square units.
Let f be a differentiable function such that f(0) П­ ПЄ3 and fВї 1 x 2 = g 1 x2 for ПЄ1 Х… x Х… 5.
(a) Find f(4)
(b) For what values of x is the graph of y П­ f(x) concave upward? Explain.
(c) Write an expression for f(x). Your answer should involve a definite integral and should be
expressed in terms of the function g.
(d) Sketch a possible graph for y П­ f(x).
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dy
П­ (y П© 5) (x П© 2).
dx
(a) On the axes provided below, sketch a slope field for the given differential equation at the
nine points indicated.
6. Consider the differential equation given by
y
3
2
1
–3
–2
–1
x
(b) Find the particular solution y П­ f(x) to the given differential equation with the initial
condition f(0) П­ 2.
(c) Find the domain and range of your particular function.
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Advanced Placement Calculus BC test
Section 1—Part A (55 minutes)
Choose the best answer for each question. Your score is determined by subtracting one-fourth of the
number of wrong answers from the number of correct answers. Calculators are not permitted.
dy
= -x with initial condition y(0) П­ 1
dx
(A) is always concave up
1. The solution to
(B) is always concave down
(C) is undefined at x П­ 0
(D) is always decreasing
(E) is always increasing
2. Which of the following is a term in the Taylor series about x П­ 0 for the function f(x) П­ cos 2x?
1
(A) - x2
2
3. Evaluate
(A)
(B)
(C)
(D)
(E)
118
3
4
(B) - x3
3
(C)
2 4
x
3
(D)
1 5
x
60
(E)
4 6
x
45
x cos 2x dx.
1
1
x cos 2x - sin 2x + C
2
4
1
1
x sin 2x - cos 2x + C
2
4
1
1
x sin 2x - sin 2x + C
2
4
1
1
x cos 2x + sin 2x + C
2
4
1
1
x sin 2x + cos 2x + C
2
4
Sample BC Test
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dy
= 1x + 3 2e -2y, then which of the following is a possible expression for y?
dx
1
(A) ln 1 x2 + 6x + 52
2
(B) ln 1x2 + 6x - 4 2
1
(C) ln 1 x2 + 6x2 - 3
2
3
1
1
(D) ln a x2 + x b
2
4
2
1
(E) ln 1 x2 + 3x2
2
4. If
5. Let f1x2 = b
2x - 5,
1x + 1,
for x Х… 3
.
for x 7 3
8
Find
30
f1x2dx.
(A) 24
(B)
45
2
(C)
52
3
(D)
20
3
(E)
32
12 - 2 13
3
6. The line tangent to the graph of y П­ x3 ПЄ 3x2 ПЄ 2x П© 1 at x П­ ПЄ1 will also intersect the curve at
which of the following values of x?
(A) x П­ 4
7. lim
tan a
hS0
(B) x П­ 5
(C) x П­ 6
(D) x П­ 7
(E) x П­ 8
p
p
+ h b - tan
3
3
=
h
(B) 13
(A) 4
(C)
1
13
(D)
13
2
(E)
1
2
8. A curve in the xy-plane is defined by the parametric equations x П­ t3 ПЄ 2 and y П­ t2 П© 4t. Find the
slope of the line tangent to the curve at the point where x П­ 6.
(A) -
3
2
(B)
2
3
(C) -
2
3
(D)
1
2
(E) 2
9. Assume that gВї 1x2 П­ h(x) and f(x) П­ x2. Which of the following expressions is equal to
(A) 2x g (x)
(B) 2x h(x)
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(C) 2 g(x) h(x)
(D) fВї 1 x2 g 1 x2 h1 x2
d
f 1 g 1x 2 2 ?
dx
(E) x2 h(x) П© 2x g(x)
Sample BC Test
119
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10. Let f 1x 2 = b
for x 6 1
2x,
.
2x - 3, for x 6 1
Let g (x) П­ ln [(x ПЄ 1)2].
Which of the following functions are continuous at x П­ 1?
I. g (x)
II. fВї 1x 2
x
III.
30
f1t 2 dt
(A) I only
(B) II only
(C) III only
(D) I and II
(E) I and III
q 1x - 22n
11. Find the values of x for which the series a
n converges.
n =1 n1 -3 2
(A) x П­ 2 only
(B) ПЄ1 Х… x ПЅ 5
(C) ПЄ1 ПЅ x Х… 5
(D) ПЄ1 ПЅ x ПЅ 5
(E) All real numbers
12. The slope field for a certain differential equation is shown below. Which of the following could be
a specific solution to that differential equation?
y
5
O
5
x
(A) y = 1x
(B) y = x2
(C) y = 1>x
(D) y = ln x
(E) y = ex
120 Sample BC Test
В© 2007 Pearson Education, Inc.
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13. A particle is moving along the x-axis according to the equation x (t) П­ 4t2 ПЄ sin 3t where x is given
p
in feet and t is given in seconds. Find the acceleration at t П­ .
2
(A) ПЄ1 ft/sec2
(B) 5 ft/sec2
(C) 11 ft/sec2
(D) 17 ft/sec2
(E) 2вђІ ft/sec2
14. If the derivative of f is fВї П­ x (x ПЄ 1)2 (x ПЄ 2)3 (x ПЄ 3)4, find the number of points where f has a
local maximum.
(A) None
(B) One
(C) Two
(D) Three
(E) Four
1 -4
e
2
1
(D) - e-4
2
(E) q
q
15. Evaluate
2
32
xe -x dx.
1
(A) e -2
2
1
(B) - e-2
2
(C)
16. Let f and g be functions that are differentiable for all real numbers, with lim f1 x 2 = 3 and lim g 1x 2 = 5.
xS0
xS0
f1 x2
f1 x 2
? Вў You may assume that lim
exists. ≤
xS0 g 1 x2
xS0 g 1 x 2
Which of the following must be equal to lim
I.
II.
3
5
f102
g102
fВї 102
xS0 gВї 102
III. lim
(A) None
В© 2007 Pearson Education, Inc.
(B) I and II
(C) I and III
(D) II and III
(E) I, II, and III
Sample BC Test
121
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q
2p
b.
17. Let f (x) П­ a 1cosx2 3n. Evaluate f a
3
n =1
1
7
1
9
1
7
8
9
The series diverges.
(A) (B)
(C)
(D)
(E)
18. Let f 1x 2 =
2
(A) ex
x2
2
30
et
+t
dt. Find fВї 1x2.
(B) 2x ex 1x
2
+2x
2
+12
4
(C) ex
+x2
(D) 2e1x +x2
2
2
(E) 2x ex
+2x
19. A particle is moving along the graph of the curve y П­ ln (3x П© 5). At the instant when the particle
crosses the y-axis, the y-coordinate of its location is changing at the rate of 15 units per second.
Find the rate of change of the x-coordinate of the particle’s location.
(A)
(B)
(C)
(D)
(E)
5 ln 3 units per second
9 units per second
25 units per second
45 units per second
3 ln 5 units per second
20. Find lim a
xSq
2x + 1 3x
b
2x
(A) 1.5
122 Sample BC Test
(B) 6
(C) e1.5
(D) e6
(E) q
В© 2007 Pearson Education, Inc.
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21. Use implicit differentiation to find
1
(A) - exy
4
(B)
y
x + 4e-xy
3
22. Which of following is equal to
3
(A)
1
u3 du
4 31
dy
for the equation 4y ПЄ exy П­ 7.
dx
31
(C) -
y
(D) - exy + 7
4
(E)
7 - yexy
4 + xexy
1 2x2 - 5 2 3 x dx?
13
(B)
yexy
xexy - 4
1
u3 du
4 3-3
13
(C)
3-3
3
u3 du
(D) 4
31
13
u3 du
(E) 4
3-3
23. Find the area of the region above the x-axis and beneath one arch of the graph of y П­
(A)
2p
+ 13
3
(B)
2p
+1
3
(C) 13 -
p
3
(D) 13 +
4p
3
(E)
u3 du
1
П© sin x.
2
7p
13
+
+1
12
2
24. A curve is defined parametrically by x П­ t3 ПЄ 5 and y П­ e2t for 0 Х… t Х… 4.
Which of the following is equal to the length of the curve?
4
(A)
19t4 + 4e4t dt
30
4
(B)
16t2e2t + 1 dt
30
4
(C) 2
4
(D)
1t4 + e4t dt
30
30
1 1t3 - 52 2 + e4t dt
4
(E) 2p
30
1t3 - 52 19t4 + 4e4t dt
В© 2007 Pearson Education, Inc.
Sample BC Test
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25. Find the values of x for which the graph of y П­ xex is concave upward.
(A) x ПЅ ПЄ2
(B) x Пѕ ПЄ2
(C) x ПЅ ПЄ1
26. Find the sum of the geometric series
(A)
3
5
(B)
5
8
(D) x Пѕ ПЄ1
(E) x ПЅ 0
3
1
1
9
- + - + ...
8
4
2
3
(C)
13
24
(D)
27
8
(E)
27
40
27. The graph of f(x) П­ x3 П© x2 has a point of inflection at
(A) x =
1
3
(B) x = -
1
3
(C) x = -
2
3
(D) x =
2
27
(E) x = 0
5
28. Use partial fractions to evaluate
(A) ln 3 П© ln 5
124
Sample BC Test
4x - 9
dx
33 2x - 9x + 10
(B) 2 ln 3 П© ln 5
2
(C) ln 3 П© 2 ln 5 (D) ln 5 ПЄ ln 3
(E) 2 ln 5 ПЄ ln 3
В© 2007 Pearson Education, Inc.
5223_Test_100-137 04/10/06 11:24 AM Page 125
Section I—Part B (50 minutes)
Choose the best answer for each question. (If the exact answer does not appear among the choices,
choose the best approximation for the exact answer.) Your score is determined by subtracting one-fourth
of the number of wrong answers from the number of correct answers. You may use a graphing calculator.
29. Find the average value of the function y = x 1cos x on the closed interval [5, 7].
(A) 4.4
(B) 5.4
30. The series x + x3 +
(A) x ln (1 П© x2)
(C) 6.4
(D) 7.4
(E) 10.8
x7
x2n +1
x5
+
+p +
+ p is the Maclaurin series for
2!
3!
n!
(B) x ln (1 ПЄ x2)
(C) x2ex
2
(D) xex
2
(E) ex
31. Find the area, in terms of k, for the region enclosed by the graphs of y П­ x4 and y П­ k. (Assume
k Пѕ 0.)
(A) 1 2 + k2 1k
4
(B) 2k a k -
k2
b
5
4
(C) 2 1 1 + k2 1k
(D) 1.6k5>4
(E) 1.8k5>4
(D) 12.5вђІ
(E) 25вђІ
32. The area enclosed by the graph of r П­ 5 cos 4вђЄ is
(A) 5
(B) 10
(C) 6.25вђІ
33. A region is enclosed by the graphs of the line y П­ 2 and the parabola y П­ 6 ПЄ x2. Find the volume
of the solid generated when this region is revolved about the x-axis.
(A) 76.8
В© 2007 Pearson Education, Inc.
(B) 107.2
(C) 167.6
(D) 183.3
(E) 241.3
Sample BC Test
125
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34. Let f(x) be a differentiable function whose domain is the closed interval [0, 5], and let
x
F(x) =
30
f1t2 dt. If F(5) П­ 10, which of the following must be true?
I. F(x) П­ 2 for some value of x in [0, 5].
II. f(x) П­ 2 for some value of x in [0, 5].
III. f Вї (x) П­ 2 for some value of x in [0, 5].
(A) I only
35. Let g 1x2 =
(B) II only
x
(C) III only
(D) I and II
(E) I, II and III
1t + 22 1 t - 32e-t dt.
30
For what values of x is g decreasing?
(A) x ПЅ ПЄ1.49
(B) x Пѕ 0.37
(C) –2 Ͻ x Ͻ 3
(D) x ПЅ ПЄ2.72, x Пѕ 0
(E) Nowhere
36.
Use the Trapezoidal Rule with the indicated subintervals to estimate the area of the shaded region.
(A) 48
126
Sample BC Test
(B) 50
(C) 51
(D) 52
(E) 54
В© 2007 Pearson Education, Inc.
5223_Test_100-137 04/10/06 11:24 AM Page 127
37. The velocity of a particle moving along the x-axis is given by v(t) ПЄ t sin t2. Find the total distance
traveled from t П­ 0 to t П­ 3.
(A) 1.0
(B) 1.5
(C) 2.0
(D) 2.5
(E) 3.0
38.
A 15-foot ladder is leaning against a building as shown, so that the top of the ladder is at (0, y) and the
dy
= -12 feet per
bottom is at (x, 0).The ladder is falling because the ground is slippery; assume that
dt
dx
second at the instant when x П­ 9 feet. Find
at this instant.
dt
(A) 6 feet per second
(B) 9 feet per second
(C) 12 feet per second
(D) 16 feet per second
(E) 20 feet per second
5
in the first quadrant is revolved about the x-axis
x +1
to generate a solid. The volume of this solid is
39. The infinite region beneath the curve y =
(A) 5
В© 2007 Pearson Education, Inc.
(B) 5вђІ
(C) 25
(D) 25вђІ
(E) q
Sample BC Test
127
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40. Let f (t) П­ sin t ПЄ 2 cos t2, where 0 Х… t Х… 4. For what value of t is f(t) increasing most rapidly?
(A) 1.76
(B) 2.81
(C) 3.32
(D) 3.56
(E) 3.77
41.
2
A rectangle is inscribed under the curve y = e -x as shown above. Find the maximum possible area of
the rectangle.
(A) 0.43
(B) 0.61
(C) 0.71
(D) 0.86
(E) 1.77
42. Let fn(x) denote the nth-order Taylor polynomial at x П­ 0 for cos x (that is, the sum of the terms
up to and including the xn term). For what values of n is
fn(0.8) ПЅ cos x?
(A) 0, 2, 4, 6, 8, 10, …
(B) 1, 3, 5, 7, 9, 11, …
(C) 1, 2, 5, 6, 9, 10, …
(D) 2, 3, 6, 7, 10, 11, …
(E) 3, 4, 7, 8, 11, 12, …
43. Find the average rate of change of y with respect to x on the closed interval [0, 3] if
(A)
1
ln 10
6
128 Sample BC Test
(B)
1
ln 3
6
(C)
1
ln 10
2
(D)
1
10
dy
x
= 2
.
dx
x +1
(E)
3
10
В© 2007 Pearson Education, Inc.
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44. The position vector of a particle moving in the xy-plane is given by r 1 t 2 = 8sin-1 t, 1t + 42 2 9 9 for
ПЄ1 Х… t Х… 1. The velocity vector at t П­ 0.6 is
(A) 8sin -1 0.6, 21.169
(B) 81. 25, 9.29
5
(C) h , 1.2 i
3
5
(D) h , 9.2 i
3
(E) h
75
, 2i
64
45. The base of a solid is the region in the xy-plane beneath the curve y П­ sin kx and above the x-axis for
p
0 Յ x Յ . Each of the solid’s cross-sections perpendicular to the x-axis has the shape of a rectangle
2k
with height cos2 kx. If the volume of the solid is 1 cubic unit, find the value of k. (Assume k Пѕ 0.)
(A) 3
(B) 3вђІ
В© 2007 Pearson Education, Inc.
(C)
1
3p
(D)
p
3
(E)
1
3
Sample BC Test
129
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Advanced Placement Calculus BC test
Section II
Show your work. In order to receive full credit, you must show enough detail to demonstrate a clear understanding of the concepts involved. You may use a graphing calculator. Where appropriate, you may
give numerical answers in exact form or as decimal approximations correct to three decimal places.
For Problems 1–3, a graphing calculator may be used. (45 minutes)
1.
The figure above shows a pump connected by a flexible tube to a spherical balloon. The pump
consists of a cylindrical container of radius 8 inches, with a piston that moves up and according to
24
+ ln 1t + 12 for 0 Х… t Х… 100, where t is measured in seconds and h(t) is
the equation h1t 2 =
t +1
measured in inches. As the piston moves up and down, the total volume of air enclosed in the pump
and the balloon remains constant, and r(t) = 0 at t = 0 a The volume of a sphere with radius r
4
is pr3. b
3
(a) Write an expression in terms of h(t) for the total volume of the air enclosed in the pump and
the balloon. (Do not include the flexible tube.)
(b) Find the rate of change of the volume of the air enclosed in the pump at t П­ 3 sec.
(c) Find the rate of change of the radius of the ballon at t П­ 3 sec.
(d) Find the maximum volume of the balloon and when it occurs.
130 Sample BC Test
В© 2007 Pearson Education, Inc.
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2. Let f be a function that has derivatives of all orders on the interval (ПЄ1, 1).
Assume that f102 = 6, f¿ 1 0 2 = 8, f– 1 0 2 = 30, f– 1 02 = 48, and 0 f142 1 x2 0 Յ 75 for all x in the
interval (0, 1).
(a) Find the third-order Taylor series about x П­ 0 for f(x).
(b) Use your answer to part (a) to estimate the value of f(0.2).
What is the maximum possible error in making this estimate?
(c) Let g(x) = x f(x2). Find the Maclaurin series for g(x). (Write as many nonzero terms as possible.)
(d) Let h(x) be a function that has the properties h(0) П­ 5 and hВї (x) П­ f(x).
Find the Maclaurin series for h(x). (Write as many terms as possible.)
В© 2007 Pearson Education, Inc.
Sample BC Test
131
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3. Consider the family of polar curves defined by r П­ 2 П© cos ku, where k is a positive integer.
(a) Show that the area of the region enclosed by the curve does not depend on the value of k.
What is the area?
(b) Write an expression in terms of k and u for the slope
(c) Find the value of
132 Sample BC Test
dy
of the curve.
dx
dy
p
at u = , if k is a multiple of 4.
dx
4
В© 2007 Pearson Education, Inc.
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No calculator may be used for Problems 4–6. Students may continue working on Problems 1–3, but
may not use a calculator. (45 minutes)
4. A particle travels in the xy-plane according to the equations x(t) П­ t3 П© 5 and y(t) П­ 4t2 ПЄ 3 for t Х† 0.
(a) For t П­ 5, find the velocity vector and its magnitude.
(b) Find the total distance traveled (i.e., the length of the path traced) by the particle during the
interval 0 Х… t Х… 5.
(c) Find
dy
as a function of t.
dx
(d) Find
d2y
as a function of t.
dx2
В© 2007 Pearson Education, Inc.
Sample BC Test
133
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5.
3
The shaded region is enclosed by the graphs of y П­ x3 and y = 4 14x.
(a) Find the coordinates of the point in the first quadrant where the two curves intersect.
(b) Use an integral with respect to x to find the area of the shaded region.
(c) Write an integral with respect to y that could be used to confirm your answer to part (b).
(d) Without using absolute values, write an integral expression that gives the volume of the solid
generated by revolving the shaded region about the line x П­ ПЄ1. Do not evaluate.
134 Sample BC Test
В© 2007 Pearson Education, Inc.
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6.
The graph of a differentiable function g is shown. Assume that the area of the shaded region is 8
x>3 +2
square units. Let f1x2 =
32
g 1 t 2 dt.
(a) Find f(12).
(b) Find f Вї (6).
(c) Write an expression for f– 1 x 2 in terms of the functions g.
В© 2007 Pearson Education, Inc.
Sample BC Test
135
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(d) For what values of x is the graph of y П­ f(x) concave downward? Explain.
(e) Sketch a possible graph for y П­ f(x).
136 Sample BC Test
В© 2007 Pearson Education, Inc.
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DATE
NAME
SAMPLE AB TEST
SAMPLE BC TEST
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В© 2007 Pearson Education, Inc.
Sample BC Test
137
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Answers
Concepts Worksheets
(g)
Sections 1.2–1.6
1. (a)
(h)
(b)
(i)
(c)
(j)
(d)
(k)
(e)
(l)
(f)
138 Answers
В© 2007 Pearson Education, Inc.
5223_ans_138-156 04/10/06 11:18 AM Page 139
Symmetry
with
respect
to y-axis
or origin
Even or
Odd
Function—
f(ПЄx) П­ f(x)
or
f(ПЄx) П­ ПЄf(x)
Is the
function
periodic?
If so,
state
the
period.
Is f(x) a
one-toone
function?
(For each
f(x) only
one x
exists)
Function
Domain
Range
y П­ f(x)
Zeros
(Find x
when
f(x) П­ 0)
(a) f(x) П­ x2
R
yՆ0
xП­0
y-axis
Even
No
No
(b) f(x) П­ x3
R
R
xП­0
Origin
Odd
No
Yes
(c) f(x) П­ 0 x 0
R
yՆ0
xП­0
y-axis
Even
No
No
(d) f(x) П­ sin x
R
0y0 Х… 1
k
Origin
Odd
Yes, 2вђІ
No
(e) f(x) П­ cos x
R
0y0 Х… 1
p
x = 12k + 12 ,
2 y-axis
k HJ
Even
Yes, 2вђІ
No
(f) f(x) П­ tan x
x H R,
p
x Z 1 2k + 1 2 ,
2
k H J,
R
Origin
Odd
Yes, вђІ
No
(g) f(x) П­ sec x
x H R,
p
x Z 1 2k + 1 2 ,
2
k H J,
0y0 Х† 1
None
y-axis
Even
Yes, 2вђІ
No
(h) f(x) П­ 2x
R
yПѕ0
None
Neither
Neither
No
Yes
(i) f(x) П­ log2 x
xПѕ0
R
xП­1
Neither
Neither
No
Yes
x
x
x
y
None
Origin
Odd
No
Yes
(j) f(x) П­
1
x
H R,
0
x П­ kвђІ,
HJ
x П­ kвђІ,
k
H R,
0
HJ
(k) f(x) П­ 1x
xՆ0
yՆ0
xП­0
Neither
Neither
No
Yes
(l) f(x) П­ 1a2 - x2
0 x 0 Х…a
0Х…yХ…a
x П­ П®a
y-axis
Even
No
No
В© 2007 Pearson Education, Inc.
Answers
139
5223_ans_138-156 04/10/06 11:18 AM Page 140
3. Yes. Even functions have y-axis symmetry, while
odd functions are symmetric about the origin.
4. (a)
(b)
9.
10. x П­ g(t), y П­ f(t); 3
11. x П­ e2t, y П­ et
[–␲, ␲] by [–1.5, 1.5]
(c)
[–4, 4] by [–3, 3]
5. (a)
(b)
12. (a) One possible answer:
x П­ t2, y П­ ПЄt
(b) One possible answer:
1 1
x = 2,
t 0t0
[–␲, ␲] by [–1.5, 1.5]
(c)
Sections 2.2–2.3
[–4, 4] by [–3, 3]
Section 1.4
1.
2.
3.
4.
5.
6.
7.
t<0
t>0
t=0
8.
1.
2.
3.
4.
5.
Both
Neither
Right end behavior model
Left end behavior model
(a) Continuous at x П­ c
(b) Discontinuous at x П­ c;
removable discontinuity
(c) Discontinuous at x П­ c;
removable discontinuity
(d) Discontinuous at x П­ c;
jump discontinuity
(e) Discontinuous at x П­ c;
infinite discontinuity
(f) Discontinuous at x П­ c;
infinite discontinuity
6. Possible answers:
1
1
(a) f1 x2 =
(b) f1x 2 = 2
x
x -4
(c) f(x) П­ tan x
7. Possible answers:
(a) f(x) П­ 3x2 ПЄ5
0, if x is rational
(b) f1 x 2 = e
1, if x is irrational
[–4, 4] by [–3, 3]
140
Answers
В© 2007 Pearson Education, Inc.
5223_ans_138-156 04/10/06 11:18 AM Page 141
Sections 3.1–3.3
1. (a)
3p
2
p
+1
2
13
(d)
36
p
(f)
8
113
(h)
2
(b)
(c) - 4
3
4
(e) -p
(g)
2
9
2. (a) At t П­ 1 sec and at t L П­ 4.25 sec
(b) 5 ПЅ t Х… 6
(c) t П­ 3 sec
(d)
2. (a)
(b) x = -2, x = 1
3. (a) None
(b) x П­ 0
(c) x П­ 0
(d) x П­ 0
(e) x Х… 0, x П­ 1
(f) x П­ 0
4. Corner points, cusp points, very sharply
changing points, as well as points at which the
function is discontinuous or the tangent line
is vertical.
Section 3.4
1. (a) 5 ПЅ t ПЅ 7
(b) 0 ПЅ t ПЅ 1, 3 ПЅ t ПЅ 4
(c) 1 ПЅ t ПЅ 3, 4 ПЅ t ПЅ 5
(d) Speed(in./sec)
3. (a) 108 cm
(b) 18 cm/sec
1 2
(c) v(t) П­ t + 4t; a 1 t 2 = t + 4
2
(d) none
4. (a)
(b)
(c)
(d)
5. (a)
(b)
ПЄ10 ft/sec; ПЄ10 ft/sec; ПЄ29 ft/sec
t L 1 sec, t L 2 sec
t L 3.5 sec
t П­ ПЄ3, x П­ 228; t П­ 7, x П­ ПЄ272
6
4
[–400, 400] by [–1, 2]
2
1
2
3
4
5
6
7
t(sec)
Section 3.7
–2
dy
dx
2. (a)
(b)
(c)
3.
–4
1.
–6
Speed(in./sec)
6
4
2
1
2
3
4
5
6
7
y2 - 5x4 - 4x3y
x4 - 2xy - 3y2
ПЄ1
ПЄ1
4
=
t(sec)
–2
–4
–6
(e) 3 ПЅ t ПЅ 4
В© 2007 Pearson Education, Inc.
4. No; The expression is undefined at (0, 0).
5. 0
6. ПЄ1
Answers
141
5223_ans_138-156 04/10/06 11:18 AM Page 142
7. No; f Вї (0) does not exist because the right- and
left-hand derivatives at x П­ 0 are not equal.
8. 2; 3
9. f(x) П­ 0
Section 3.8
1.
2.
4. 1eln 2 2 x = ex ln 2; ex ln 2 1 ln 2 2 = 2 x 1 ln 2 2 ; ln 2
5. 1 eln 3 2 x = ex ln 3; ex ln 3 1 ln 3 2 = 3x 1 ln 3 2 ; ln 3
6. 1 eln x 2 x = ex ln x;
1
ex ln x a x # + ln x b = xx 1 1 + ln x 2
x
Section 4.1
1. (a) x П­ a, c, g, k
(b) x П­ h
3.
4.
(c) x П­ b, d, f, h Х… x Х… i, x П­ j, k, m
(d) [b, f ], [ j, k], [m, q )
(e) (ПЄ q , b], [f, h), [i, j], [k, m]
(f) x П­ f, h Х… x Х… i, x П­ k
5. a, c, e, f
6. e, f
7. Each is the inverse function of the other.
1
8. gВї 1x2 =
fВї 1g1x2 2
9.
(g) One possible answer: x П­ f
(Answer may depend on assumptions
made regarding the end behavior of the
function.)
(h) x П­ b, h Х… x Х… i, x П­ j, x П­ m
(i) None
(j) (d, e), (l, n)
(k) (ПЄ q , b), (b, d), (e, h), ( j, l), (n, q )
(l) x П­ d, e, l, n
2. (a) y П­ 0 (and possibly y П­ p for some positive constant p)
(b) x = h
3. x = b, h, i, j
1
7
x 2
2
11
1
11. y = x +
4
4
10. y =
12. y = -2x - 2
Section 3.9
3x +h - 3x
3x # 3h - 3x
3x 1 3h - 1 2
, lim
, lim
hS0
hS0
hS0
h
h
h
1. lim
2.
3. 0.693; 1.099
142
Answers
В© 2007 Pearson Education, Inc.
5223_ans_138-156 04/10/06 11:18 AM Page 143
4. x = b, d, e, h Х… x Х… j, x = l, n
5.
Section 4.2
1. (a)
(b)
The Mean Value Theorem does not apply,
because the function is not differentiable
at x П­ d.
(c)
(b)
(d)
(e)
(c)
3. (a)
(b)
(d) The variation of Rolle’s Theorem does
not apply because f1a2 Z f1 b2 .
2. (a)
It appears that the graph is not differentiable at x = d. The Mean Value Theorem
does not apply.
Every point between a and b would be an
appropriate c in the Mean Value
Theorem, since the tangent line coincides
with the secant line for every c.
The Mean Value Theorem does not apply,
because the function is not differentiable
at x = 0.
Applying the Mean Value Theorem with
a = 0 and b = 4, there is at least one
value of c in (0, 4) such that
f 14 2 - f 1 02
5 -3
1
fВї 1 c2 =
=
= .
4 -0
4 -0
2
Since fВї 1 x2 is continuous and
differentiable, we may apply the Mean
Value Theorem to fВї 1 x2 with a = 1 and
b = 3. There is at least one value of c in
(1, 3) such that
f– 1 c2 =
В© 2007 Pearson Education, Inc.
fВї 1 4 2 - fВї 10 2
-1 - 3
=
= -1.
4 -0
4 -0
Answers
143
5223_ans_138-156 04/10/06 11:18 AM Page 144
2.
Section 4.3
1.
3.
2. One possible answer:
4.
5.
3.
4. (a) Origin; -f (x)
(c) 0
(e) П®2a, 0
(b) 0
(d) П®5a
(f) (2a, 6a)
6. c
8. a
10. (a) 0
(c) 0 ПЅ x, ПЅ 1
(e)
7. d
9. b
(b) Increasing
(d) None
11. (a) 0
(b) 0
1
(d)
2
Section 5.2
1
1.
1
2
x dx
30
2.
1
30
2
1x + x2 2 dx
1
1
1
dx
dx or
2 30 1 + x
30
31 2x
2
2
1
1
5.
dx 6.
dx
30 1 + 2x
30 1 + 2x
1 4 1
1 51
dx 8.
dx
7.
2 30 1 + x
2 31 x
9. Yes, because they can all be represented by
the same limit of Riemann sums.
3.
11 + 2x dx 4.
Sections 5.3–5.4
1.
144
Answers
(c) 1
(e)
3
2
1 3
, 6 x 6 2
2 2
1
3
(g) 6 x 6
2
2
(h)
(f) 0 6 x 6
В© 2007 Pearson Education, Inc.
5223_ans_138-156 04/10/06 11:18 AM Page 145
12. (a) 0
(c) 2b
13.
(b) 2b
b -a
. The inscribed rectangular
n
area approximation is
4. Let Вўx =
I = 3f1x0 2 + f1x1 2 + f1x2 2 + p + f1 xn-12 4 Вўx,
and the circumscribed rectangular area
approximation is
C = 3f1x1 2 + f1x2 2 + f1x3 2 + p + f1xn 2 4 Вўx.
Section 5.4
1
1.
1 + x2
Therefore,
1
1 +C
= B f1 x0 2 + 2f 1 x1 2 + 2f 1x2 2 + p
2
2
2. 2 14x2 + 1
+ 2f 1 xn-1 2 + f 1 xn 2 R Вўx
sin x
4.
2 1x
3. 0
2
x
7. 16
6. 115
1
8.
2
5. -
Section 5.5
=
b -a
B f1 x0 2 + 2f 1 x1 2 + 2f 1x2 2 + p
2n
+ 2f1 xn - 1 2 + f 1xn 2 R .
Sections 6.1–6.2
1.
2
1.
x3 dx П­ 4. The trapezoidal approximation
30
is greater since y П­ x3 is always concave up
for 0 Х… x Х… 2 and the top side of each
trapezoid lies above the curve.
20
2.
10
x
99
x dx П­ . The trapezoidal
dx П­ 1
2
32 4
31
approximation would equal the integral
2.
answer since the function being integrated
represents a straight line which the trapezoids
would exactly fit.
4
3. (a)
30
1x dx =
16
b + 211 2 + 2 a
b + 21 12 2
2
2
110
114
+ 2a
b + 21 132 + 2 a
b + 2R
2
2
(b) B0 + 2 a
1
4
16
3
12
(c) Note that f1x2 = 1x is concave down.
Since the Trapezoidal Rule approximation
is based on straight segments which lie
below the graph of y П­ f(x) and the
Simpson’s Rule approximation is based on
curves which nearly match the graph
of y ϭ f(x), expect the Simpson’s Rule
approximation to be larger. To confirm,
observe that S L 5.3046 and T L 5.2650.
В© 2007 Pearson Education, Inc.
x
1
dt = ln x
31 t
1
1
(b)
dt = 0
31 t
1
1
(c)
dt = -ln x
3x t
4. (a)
3. (a)
Answers
145
5223_ans_138-156 04/10/06 11:18 AM Page 146
(b)
Sections 7.1–7.5
60
R1 t 2 dt
(b) 842.5 cm
30
2. (a) L 4.6; 4.6 inches of rain fell during the
24 hours beginning at midnight.
(b) L 1.2; 1.2 inches of rain fell between
4 A.M. and noon.
(c) L 2.5; 2.5 inches of rain fell between
8 A.M. and 8 P.M.
3. $2,500,000
4. 3200 ft-lb
1. (a)
(c)
5. (a)
Section 7.2
b
1.
3a
f1 x 2 dx
c
2. (b)
3a
b
3.
3a
c
4.
f1 x 2 dx +
3a
+
6. (a) ln 0 x 0
(b) ln 11 + x2
(c) -ln 1 1 – x2
7. If t = au, then dt = adu. Therefore, when
t = a, u = 1 and when t = ax, u = x.
ax
x
x
1
1 #
du
adu =
dt =
3a t
31 u
31 au
ax
ax
1
dt = ln t ` = ln 1ax 2 - ln 1 a 2
a
3a t
x
x
du
= ln u ` = ln x - ln 1 = ln x
1
31 u
Therefore, ln (ax) ПЄ ln (a) П­ ln x, so
ln (ax) П­ ln x П© ln (a).
1
dy
=
8. (a)
(b) y = ln x + C
dx
x
1
dy
9. (a)
=
dx
x
(b) The equations are the same. Since ln 2x =
ln 2 + ln x, the general solution y =
1
dy
ln x + C of the equation
= includes
dx
x
the particular solution y = ln 2x.
146
Answers
3c
6. 2
30
32
7.
3
f1 x 2 dx
3g 1 x 2 - f1 x2 4 dx
3c
d
5.
3c
3 f1 x2 - g 1 x2 4 dx
b
(c)
b
3f 1 x2 - g 1 x 2 4 dx
3g 1 y 2 - f 1 y2 4 dy
a
f1 y 2 dy or
8. (a) D1 t 2 =
a
3-a
f1 y 2 dy
t
1 R1 x2 - E1 x2 2 dx
30
(b) - 105.5 or a deficit of 105.5 muds
(c) The deficit disappears in 2039.
Section 7.3
1. (a)
(b)
В© 2007 Pearson Education, Inc.
5223_ans_138-156 04/10/06 11:18 AM Page 147
(c) px4
(d) px 4dx
2
(e) p
x4 dx
30
(f)
2. (a)
(b)
(d) p 1 4 –y2 dy
(c) 4p - py
4
(e) p
30
14 - y2 dy
3. (a)
(c)
32p
5
(f) 8p
(b)
1 4 - 2y 2
a
b = 12 - y2 2
2
12
2
(d) (2 ПЄ y)2 dy
(f)
4. (a)
(b)
(c)
(d)
(e)
30
1 2 - y 2 2 dy
b
3a
xf1x2 dx
Sections 9.1–9.5
1
1
6 x Х…
2
2
1
1
1
1
(b) f a b = +
- p
4
2
8
24
1. (a) -
+
+
5
x2n-1 d + p
1 2n - 1 2 !
Note that, if n is replaced by n + 2 in the general term for f Р‰(x), we obtain the opposite of
the general term for f(x). This means that
f Љ(x) = -f(x), so y = f(x) solves yЉ + y = 0.
(b) All real numbers
(c) f1 x2 = 3 cosx + sinx
8
3
f(x)
2px
2px f 1x2
2px f1x 2 dx
(e) 2p
3
(c) ln a b
2
(d) fВї 1 x2 = 2 - 4x + 8x2 - p
+ 1 -1 2 n + 1 1 2 2 1 2x 2 n - 1 + p ;
1
4
fВї a b =
4
3
2. (a) 7 - 3 (x - 3) + 6 (x - 3)2
(b) 8.44; 0.162
3
(c) -2 + 7 (x - 3) - (x - 3)2 + 2 (x - 3)3
2
2
3
4
(d) 7x - 3x + 6x
5
3
5
3. (a) fВї 1 x2 = 5 - 3x - x2 + x3 + x4
2!
3!
4!
3 5
5 6
- x - x + p
5!
6!
3
+ 1 -12 n c
x2n - 1
1 2n - 1 2!
5
+
x2n d + p
1 2n2 !
3
5
3
f– 1x 2 = -3 - 5x + x2 + x3 - x4
2!
3!
4!
5 5
- x + p
5!
3
+ 1 -12n c
x2n-2
1 2n - 22 !
Section 10.2
1. (a) 1 - 12, 2 2 ,1 12, 2 2
(b) (0, 0)
16 3
16 3
(c) (0, 0), a
, b, a , b
2 2
2 2
(d)
1 - 1 2 n+1 1 n
a b + p
n
2
В© 2007 Pearson Education, Inc.
Answers
147
5223_ans_138-156 04/10/06 11:18 AM Page 148
p
p
13 3
b = rВї a b = h ,- i
6
6
2
2
p
p
1 3 13
a a b = r– a b = h , i
6
6
2
2
p
3 3 13
i
(b) r a b = h ,
6
2 2
2. (a) v a
2. (a)
(b)
(c)
(c) minimum t = 0, p, 2p
p 3
maximum t = , p
2 2
2 3
3. (a) 8t , t – 19
(b)
(c) rВї 11 2 = 82, 39
(d) r– 1t 2 = 82, 6t 9, r– 102 = 82, 0 9
(e) There exist none since the horizontal
component of the vector is always nonzero.
4. (a) r 1t 2 = 8C1et, C2et 9 where C1, C2 are
arbitrary constants.
(b) r 1t2 = 8C1e-t, C2e-t 9 where C1, C2 are
arbitrary constants.
Section 10.3
1. (a)
148
3. (a)
(b)
4.
5. (a)
(b)
(b)
(c)
(c)
Answers
В© 2007 Pearson Education, Inc.
5223_ans_138-156 04/10/06 11:18 AM Page 149
6. (a)
(b)
7. (a) x = (1 + cos u) cos u, y = (1 + cos u) sin u
dx
= - sin u - 2sin u cos u
(b)
du
= - sin u - sin 2u;
dy
= cos u + cos2u - sin2u
du
= cos u + cos 2u
cos u + cos 2u
(c)
- sin u - sin 2u
(d) y = x + 1
(e) y = 1 - x
8. (a) x = -cos2 u, y = -cos u sin u
dx
= 2 sin u cos u = sin 2u,
(b)
du
dy
= -cos2 u + sin2 u = -cos 2u
du
- cos 2u
= -cot 2u
(c)
sin 2u
p 3p
(d) ,
4 4
p
(e) 0, , p
2
9. To find point of intersection: 2 cos u = 2 sin u
p
or at the pole.
u =
4
For r = 2 cos u,
dy
2 cos2 u - 2 sin2 u
2 cos 2u
=
=
= -cot 2u
dx
-4 sin u cos u
-2 sin 2u
dy
p
= 0 at u = .
dx
4
For r = 2 sin u,
4 sin u cos u
dy
=
= tan 2u
dx
2 cos 2u
p
dy
is undefined (vertical slope) at u = .
dx
4
Therefore, the slopes of the tangents at
p
u = indicate the lines are perpendicular.
4
В© 2007 Pearson Education, Inc.
At the pole:
r = 2 cos u and r = 2 sin u intersect for different
values at u, namely 0 = p>2 and u = 0,
respectively. Slope of the tangent to r = 2 cos u
at 0 = p>2 is undefined. Slope of the tangent to
r = 2 sin u at u = 0 is 0. Therefore, the slopes
indicate the tangent are perpendicular at the pole.
p 3p
3
3
, sin-1 , p - sin-1
10. (a) u = ,
2 2
5
5
(b)
11.
The point closest to the pole is at (1, p), and
the line tangent to r = 2 + cos u at this point
is vertical. For the “dimpling effect” we are
interested in any other vertical tangents
symmetrically situated on the curve about
u = p. The only other vertical tangent occurs
when u = 0. Therefore, there is no dimple
around u = p.
12. 1 Х… a 6 2
Group Activity Explorations
Chapter 1 Review
Here is one possible solution:
A: 4
B: 6
C: 10
D: 18
E: 1
F: y = 4x + 1
1
1
G: H: y = - x + 18
4
4
1
I: 1
J:
4
K: -1
L: Denmark
M: 8
N: 1
O: tan
P: p
Q: 0
R: 0
S: 1
T: 162
U: 9
V: SINE
W: U
X: 1
Special message: You won!
Answers
149
5223_ans_138-156 04/10/06 11:18 AM Page 150
Section 2.3
A
1. When x takes on only irrational values,
f(x) П­ 0 and lim 0 = 0.
Section 4.4
1.
xS0
2. When x takes on only rational values, f(x) П­ x
and lim x = 0.
xS0
3. lim 0 = 0 and lim a = a.
xSa
xSa
4. Since lim 0 = lim a only when a П­ 0, f(x) is
xSa
xSa
continuous only at x П­ 0.
B
1. When x is any integer multiple of p, sin x П­ 0.
sin px, if x is rational
2. f 1x 2 = e
0,
if x is irrational
3. f (x) П­ int x (the step function).
2. This makes sense because a car consumes
more fuel when traveling very slow or very fast
compared to traveling at a moderate speed.
3. The critical point is approximately (36.84, 0.95).
The consumption rate C (in gallons/hour) is
minimized.
y
C1 v 2 gallons>hour
C 1v 2
=
=
gallons>mile
4.
x
v miles>hour
v
5. One possible answer:
C
x2 - 4,
0,
x3 - x,
2. f 1x 2 = e
0,
px
,
sin
2
•
f
1
x2
=
3.
0,
1. f 1x 2 = e
if x is rational
if x is irrational
if x is rational
if x is irrational
if x is rational
if x is irrational
1
D f (x) is continuous when x = ; , where n is a
n
positive integer.
Section 3.1
f1x + h 2 - f1x 2
h
f1x2 + f1h2 + 2xh - f 1 x2
= lim
hS0
h
f1h2 + 2xh
f1h2
2xh
= lim
= lim
+
hS0
hS0 h
h
h
= 10 + 2x
One possible answer: f(x) = x2 + 10x
One possible answer: f(x) = x2 + 10x + 1
f(x) = x2 + 10x + C; C is any real number.
f 10 + h 2 = f102 + f1h 2 + 2 # 0 # h
f 1h 2 = f102 + f 1h2 + 0
0 = f102 + 0
f10 2 = 0
f(x) = x2 + 10x
f(x П© h) = (x + h)2 + 10(x + h)
= x2 + 2xh + h2 + 10x + 10h
= x2 + 10x + h2 + 10h + 2xh
= f(x) + f(h) + 2xh
f (x) = 2x2 + 10x 9. f(x) = x2 + 20x
A
f 1x2 = x 2 + Bx
2
1. fВї 1 x2 = lim
hS0
2.
3.
4.
5.
6.
7.
8.
10.
150 Answers
The slope of this line represents the number
of gallons per mile at the chosen speed.
6. Gallons per mile is not minimized at the
critical point because the slope of the line
through a point on the graph and the origin is
not minimized at the critical point.
7. When the slope of the line through a point on
the graph and the origin is minimized, fuel consumption will be minimized at that point. This
occurs when the line is tangent to the graph.
C1 v 2
CВї 1 v 2 =
8.
v
1.5
3
-7
3
4v * 10 - 0.02 = v * 10-7 - 0.02 +
v
1.5
4v3 * 10-7 = v3 * 10-7 +
v
1.5
3
-7
3v * 10 =
v
1.5
4
v =
3 * 10-7
1>4
1.5
b
L 47.287
v = a
3 * 10-7
9. One possible answer: A calculator cannot
determine the velocity that optimizes fuel
consumption given this equation because the
minimum value of the graph does not
correspond to the optimum velocity.
10. One possible answer: When a shark starts at
zero velocity, it must expend extra energy to
В© 2007 Pearson Education, Inc.
5223_ans_138-156 04/10/06 11:18 AM Page 151
overcome inertia. Once it is moving it requires
less energy to stay moving, but it must expend
increasingly more energy against the water
resistance as it swims faster.
11. One possible answer: The graph shows us that
a shark is expending less energy when it is
moving slowly than when it is staying still in
the water.
12. One possible answer: It is true that the shark
has to swim faster to minimize calories per
mile than to minimize calories per hour. It
would always make sense to minimize calories per hour (requiring less feeding) if that
were an option, but the shark does not have
that luxury. It must cover great distances to
“graze” for food. That is why it depends on
the circumstances. When it is hunting, it
should minimize calories per mile; when it is
resting, it should minimize calories per hour.
Section 5.4
1. P = (h,r)
2.
2
r
3. V = p (radius)2(height) = p a xk b Вўx
h
h
n
pr2 2
pr2 2
4. lim a 2 xk Вўx =
2 x dx
nSq
30 h
k =1 h
5. Since we are evaluating the integral with
pr2
respect to x, 2 is a constant. It is always safe
h
to move a constant factor of the integrand
outside of the integral.
h
pr2
pr2 h3
1
x2 dx = 2 .
= pr2h
6. 2
h 30
h
3
3
7. radius = r; height = h
8. Since we are adding the same discs, the
Riemann sum is unchanged:
h
n
pr2 2
pr2 2
lim a 2 xk Вўx =
2 x dx
nSq
30 h
k =1 h
1
9. V = pr2h
3
В© 2007 Pearson Education, Inc.
Section 6.1
I.
dy
= x + y, the slope should be zero
1. For
dx
dy
= x - y, the slope
where y = -x. For
dx
should be zero where y = x.
dy
= x + y,
2. Graph A is the slope field of
dx
dy
= x - y.
and graph B is the slope field of
dx
3. As you move to the right, x increases, so the
values of x + y and x - y are both increasing.
II.
2
1. The slope is always positive because 1.1x and
2
1.1y are always positive.
2. Graph A shows slopes that depend on x but
not on y, and graph B shows slopes that depend on y but not on x.
dy
2
= 1.1x , and
3. Graph A is the slope field of
dx
dy
2
= 1.1y .
graph B is the slope field of
dx
4. For either differential equation, the solution
that passes through (0, 0) would be an odd
function, while all other solutions would be
neither odd nor even.
III.
1. Because both equations depend on y but not
on x.
dy
= sin (y3), and
2. Graph A is the slope field of
dx
dy
= sin (y2).
graph B is the slope field of
dx
You can tell because sin (y2) is an even function of y, meaning that the slope shown at any
point (x, y) is the same as the slope shown at
(x, ПЄy). Since sin (y3) is an odd function of
y, the slope shown at any point (x, y) is the
opposite of the slope shown at (x, ПЄy).
3. No. Both cos (y2) and cos (y3) are even functions of y.
IV.
dy
= cos 1xy 2 ,
1. Graph A is the slope field of
dx
dy
= sin 1xy 2 .
and graph B is the slope field of
dx
2. Since sin (0) = 0, the slopes representing
dy
= sin 1 xy 2 should be zero along the axes,
dx
and since cos (0) = 1, the slopes representing
dy
= cos 1 xy 2 should be 1 along the axes.
dx
Answers
151
5223_ans_138-156 04/10/06 11:18 AM Page 152
3. Since the sine function is odd, changing the sign
of x or y should change the sign of the slope, as
shown in graph B. Since the cosine function is
even, the slope shown at any point should be the
same as the slope at any corresponding point on
the other side of either axis, as shown in graph A.
Section 8.4
1. Let f1 x2 = e-1x >22 and g 1 x2 = e-x # f and g are
2
continuous functions for all x, 0 Х… f (x) Х…
q
g(x) for all x Х† 2, and
32
So, by Theorem 6 (Direct Comparison Test),
Section 7.3
r1
xk
xk
= ; r1 = r2
2.
r2
h
h
q
e-1x >22 dx also converges.
2
32
q
e-1x >22 dx. Since f 1 x2 = e-1x >22 is
2
Let a =
32
2
q
an even function,
2
b =
3. Rotate the cone 90Вє about the x-axis, then use
the same similar triangles as in step 2.
4. Since the cross section is scaled both
xk
horizontally and vertically by a factor of ,
h
xk 2
the area is scaled by a factor of a b . For
h
example, a triangle of height a and base b
which is scaled both horizontally and
vertically by a factor of r will have area
1
1
1rb 2 1 ra 2 = a ba b r2, which is the original
2
2
area scaled by a factor of r2.
5. The volume of the slab equals the area of its
base times its height. Since the area of its base
x 2
is A a k b , and its height is вЊ¬x, the volume
h
x 2
of the slab is A a k b Вўx.
h
n
n
x 2
6. lim a Vk = lim a A a k b Вўx
nSq
nSq
h
k=1
k=1
h
A 2
=
2 x dx
30 h
7. Since we are evaluating the integral with
A
respect to x, 2 is a constant, so we can move
h
it outside of the integral by the Constant
Multiple Rule for Definite Integrals.
h
A
A x3 h
A
h3
1
x2 dx = 2 B R = 2 *
= Ah
2
h 30
h 3 0
h
3
3
1 2
13 2
9. a. V = s h
b. V =
sh
3
12
13 2
c. V =
sh
2
8.
152 Answers
e-x dx converges.
e
3-2
q
- 1x2>22
e-1x >22 dx = a. Now let
2
32
dx . We have
e-1x >22dx = a + b + a = 2a + b.
2
3-q
q
Therefore
4
2.
e-1x >22 dx converges.
2
3-q
e-1x >22 dx L 2.50646950
2
3-4
6
e-1x >22 dx L 2.50662827
2
3-6
8
e1x >22 dx L 2.50662828
2
3-8
40
3.
e-1x >22 dx L
60
2
3-40
e-1x >22dx
2
3-60
80
e-1x >22 dx L 2.506628275.
2
L
3-80
4. It appears that
N
e-1x >22 dx L 2.50662875.
2
lim
NSq
3-N
We cannot conclude that this is an exact value
of the integral.
5. No
6. (2.506628275)2 П­ 6.283185307
6.283185307
= 3.141592654;
7.
2
q
e-1x >22dx = 12p
2
8.
9.
10.
11.
3-q
0
120 units wide
2
e-60 >2 L 0
All 100 of the rectangles will have
area L 120 # 0 = 0; The sum is 0.
В© 2007 Pearson Education, Inc.
5223_ans_138-156 04/10/06 11:18 AM Page 153
12. 150 rectangles would each be 80 units wide. 200 rectangles would each be 60 units wide. The
MRAM sum would still be 0.
13. If all of the rectangles have a calculated height of zero, then the sum of all of the rectangle areas
will also be zero.
Section 9.2
1. TO 1x 2 = x +
2.
x3
x5
x7
x 2n+1
x2
x4
x6
x2n
+
+
+p +
+ p ; TE 1x 2 = 1 +
+
+
+p +
+ p
3!
5!
7!
12n + 1 2!
2!
4!
6!
12n 2!
T 1x2 + T1 -x 2
2
a1 + x +
=
x2
x3
xn
1 -x2 2
1 -x2 3
1 -x2 n
+
+ p +
+ p b + a 1 + 1 -x2 +
+
+ p + 1 -12 n
+ pb
2!
3!
n!
2!
3!
n!
2
x3
x2
x3
x2
+
+ pb + a1 - x +
+ pb
2!
3!
2!
3!
2
a1 + x +
=
2x2
2x4
+
+ p
x2
2!
4!
x4
x2n
= 1 +
=
+
+ p +
+ p = TE 1 x2
2
2!
4!
1 2n2 !
T 1x2 - T1 -x 2
3.
2
2 +
a1 + x +
=
x2
x3
xn
1 -x2 2
1 -x2 3
1 -x2 n
+
+ p +
+ p b - a 1 + 1 -x2 +
+
+ p + 1 -12 n
+ pb
2!
3!
n!
2!
3!
n!
2
x2
x3
x2
x3
+
+ pb - 1 + x +
+ p
2!
3!
2!
3!
=
2
3
5
2x
2x
2x +
+
+ p
3!
5!
x3
x5
x2n+1
=
= 1 +
+
+ p +
+ p = TO 1 x2
2
3!
5!
1 2n + 1 2 !
TE 1x2 is similar to cos x except that every other term in the expansion of cos x is negative.
TO 1x2 is similar to sin x except that every other term in the expansion of sin x is negative.
ex - e-x
sinh x =
2
x
-x
x
e +e
e - e-x
ex + e-x + ex - e-x
2ex
+
=
=
= ex
2
2
2
2
f1x 2 + f1 -x 2
f 1x 2 - f 1 -x 2
f1 x2 + f1 - x2 + f1 x2 - f1 -x2
2f 1 x 2
+
=
=
= f 1x 2
2
2
2
2
f1x 2 - f1 -x 2
f 1 -x2 - f1 x2
Furthermore, fO 1x2 =
is odd because fO 1 -x2 =
2
2
f1x 2 - f1 - x 2
f 1 x2 + f 1 -x2
= = -fO 1 x2 , and fE 1 x2 =
is even because
2
2
a1 + x +
4.
5.
6.
7.
8.
fE 1 -x2 =
f 1 -x2 + f1 x2
f 1x2 + f1 -x2
= -fE 1 x2 .
= 2
2
В© 2007 Pearson Education, Inc.
Answers
153
5223_ans_138-156 04/10/06 11:18 AM Page 154
1
1 +x
1
1 -x
+
+
f 1 x 2 + f 1 -x2
2
1 -x
1 +x
1 1 + x2 1 1 - x2
1 1 + x 2 1 1 - x2
=
=
=
9. fE 1x 2 =
2
2
2
1 - x2
1
1
=
2
1 - x2
#
1
1
1 +x
1 -x
f1x2 - f1 -x2
1 -x
1 +x
11 + x2 11 - x2
11 + x2 11 - x2
2x
fO 1x2 =
=
=
=
2
2
2
1 - x2
#
1
x
=
2
1 - x2
x
= x + x3 + x5 + x7 + p + x2n+1 + p
1 - x2
1
= 1 + x2 + x4 + x6 + p + x2n + p
1 - x2
1
= 1 + x + x2 + x3 + x4 + p + xn + p
1 - x
x2
x3
x4
xn
11. ln 1x + 12 = x +
+ p + 1 -12 n-1
+p
2
3
4
n
x2
x4
x6
x2n
ln 11 - x2 = - p - p
2
4
6
2n
1 + x
x3
x5
x2n-1
ln a
b = x +
+
+ p +
+ p
A1 - x
3
5
2n - 1
10.
Section 10.3
2. If n is odd, then there are n petals. If n is even,
then there are 2n petals.
3. If n is odd, then each petal is traced twice over
the interval 0 Х… u Х… 2p. So, for all positive
integers, n, there are 2n petals traced over the
interval 0 Х… u Х… 2p.
4. NINT((2 sin (2x))2, x, 0, 2p)/2 L 6.283185307.
We integrate over the interval 0 Х… x Х… 2p to
include the entire graph.
5. 2p
6. There are three times as many petals, but the
area is the same.
7. NINT ((2 sin (3x))2, x, 0, p)/2 L 3.141592654.
We integrate over the interval 0 Х… x Х… p so
that we include each petal only once. The
exact area is p.
2p, n even
8. A = e
p,
n odd
9. One petal is traced over the interval
p
0 Х… u Х… , so
n
p>n
1
A =
12 sin 1nu 2 2 2 du
2 30
p>n
1
=
4 sin2 1nu 2 du
2 30
p>n
=2
154
30
Answers
sin2 1nu 2 du
10. Let u = nu and du = n du.
du
= du.
So, 0 Х… u Х… p and
n
p>n
2
30
sin 1 nu2 du = 2
2
p
sin2 1 u2
du
n
30
p
2
=
sin2 1 u2 du.
n 30
p
2
A
sin2 1 u2 du = .
11. The area of 1 petal is
n 30
n
If n is odd, there are n petals and the area is
A
n a b = A. If n is even, there are 2n petals
n
A
and the area is 2n a b = 2A.
n
p
p
u
sin 2u
2
R = p.
12. A = 2 sin 1 u2 du = 2 B 2
4
0
30
So, if n is odd the area is p, and if n is even
the area is 2p.
Sample Advanced Placement Tests
AB: Section I Answers
1.
5.
9.
13.
17.
21.
B
B
A
B
D
B
2. E
6. C
10. E
14. E
18. A
22. E
3. C
7. D
11. D
15. D
19. D
23. C
4. E
8. C
12. D
16. D
20. B
24. C
В© 2007 Pearson Education, Inc.
5223_ans_138-156 04/10/06 11:18 AM Page 155
25.
29.
33.
37.
41.
45.
A
E
C
B
A
D
26. E
30. E
34. E
38. D
42. D
27. C
31. A
35. E
39. B
43. A
28.
32.
36.
40.
44.
(d)
B
E
C
B
D
6. (a)
y
AB: Section II Answers
3
1. (a) 2 - 5p cos pt
5
5
(b) t2 + cos pt p
p
(c) 0 6 t 6 0.885 or 2.418 6 t 6 2.500
(d) L 4.658
5
y +
2 1x
2. (a) x + 3y2
15
4
(b) y = - x +
7
7
(c) 1.8
(d) 5 10.6 + 0.6y + y3 = 11; f1 0.62 L 1.821
(e) No; ПЄ0.1 is not in the domain of f (x) because 1 -0.1 is not a real number.
4
3. (a) 64ph 1t2 + pC r1t2 D 3
3
(b) Volume = 64p11 + ln 24 2 L 840.048 in3
at t = 23 sec
(c) -80p L -251.327 in3>sec
3
(d) Radius = 1
48118 - ln4 2 L 9.273 in.;
5
rate =
L 0.233 in.>sec
3 6118 - ln 4 2 4 2>3
4. (a) 12, 82
(b)
(c)
2
3
14 1
4x - x3 2 dx = 8
8
3
a1
y -
30
30
8
(d) p
30
y3
b dy
256
3
B11
y + 12 2 - a
2
or 2p
2
y3
+ 1 b R dy
256
3
1x + 12 14 1
4x - x3 2 dx
30
5. (a) 5
(b) -1 6 x 6 2, since g(x) is increasing for
these vaues of x.
x
(c) -3 +
30
g1t2 dt
В© 2007 Pearson Education, Inc.
2
1
–3
(b)
–2
x
–1
dy
= 1y + 52 1x + 22
dx
dy
= 1 x + 2 2 dx
y +5
dy
=
1 x + 2 2 dx
3
3y + 5
x2
ln 0 y + 5 0 =
+ 2x + C
2
2
ex >2 + 2x + C = y + 5
y = C ex >2 + 2x - 5
f1 0 2 = 2 . . . C = 7
2
y = 7e x >2 + 2x - 5
2
(c) Domain: all real numbers
Range: g Пѕ ПЄ5
BC: Section I Answers
1.
5.
9.
13.
17.
21.
25.
29.
33.
37.
41.
45.
B
D
C
A
B
C
B
B
E
E
D
E
2. C
6. B
10. C
14. B
18. B
22. B
26. E
30. D
34. D
38. D
42. D
3. E
7. A
11. C
15. C
19. C
23. A
27. B
31. D
35. E
39. D
43. A
4. A
8. B
12. D
16. B
20. C
24. A
28. A
32. D
36. C
40. E
44. B
Answers
155
5223_ans_138-156 04/10/06 11:19 AM Page 156
BC: Section II Answers
4
1. (a) 64ph1t 2 + p3 r1t2 4 3
3
(b) - 80p L - 251.327in3>sec
5
L 0.233 in.>sec
(c)
36 118 - ln4 2 4 2>3
(d) Volume = 64p 123 - ln 24 2 L 3985.439 in3
at t = 23 sec
2. (a)
(b)
(c)
(d)
6 + 8x + 15x2 + 8x3
8.264; maximum error is 0.005.
6x + 8x3 + 15x5 + 8x7 + . . .
5 + 6x + 4x2 + 5x3 + 2x4 + . . .
2p
1
r2 du = 4.5p L 14.137
2 30
k sin ku sin u - 12 + cos ku 2 cos u
(b)
k sin ku cos u + 1 2 + cos ku 2 sin u
p
(c)
4
3. (a) Area =
5. (a) (2, 8)
(b)
(c)
2
3
141
4x - x3 2 dx = 8
8
3
a1
y -
30
30
8
(d) p
30
3
c11
y + 122 - a
2
2p
y3
b dy
256
2
y3
+ 1 b d dy or
256
3
1x + 12 141
4x - x3 2 dx
30
6. (a) 8
(b) 1
x
1
(c) gВї a + 2 b
9
3
(d) x Пѕ 6, since this is where g a
x
+ 2 b is
3
decreasing.
(e)
4. (a) 875, 409; magnitude П­ 85
(b) 163 units
8
(c)
3t
8
(d) - 4
9t
156 Answers
В© 2007 Pearson Education, Inc.

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