AP Calculus BC Practice Exam 2

5 Steps to a 5 AP Calculus AB & BC, 2012-2013 Edition (2011)

STEP 5. Build Your Test-Taking Confidence

ANSWER SHEET FOR MULTIPLE-CHOICE QUESTIONS

Section I—Part A

Directions:

Use the answer sheet provided in the previous page. All questions are given equal weight. There is no penalty for unanswered questions. Unless otherwise indicated, the domain of a function f is the set of all real numbers. The use of a calculator is not permitted in this part of the exam.

1. If f (3) = − 2 and f′(3) = 5, then the equation of the tangent to the curve y = f(x) at x = 3 is

(A) y = 5 x − 2

(B) y = 5 x − 17

(D) y = −2 x − 17

(E) y = 5 x + 3

2. On the interval − 3 < x < − 1, the curve is

(A) increasing and concave up.

(B) increasing and concave down.

(D) decreasing and concave down.

(E) horizontal.

3. The relative extremum of the function , a > 0 is

(A) relative maximum at (− 2a, 2a^⅔).

(B) relative maximum at (− 2a, 0).

(D) relative minimum at (0, − 2a).

(E) no relative extrema.

4. A function is defined for all real numbers and has the property f(x + h) − f(x) = 4x²h + 2xh − 6x³h². Find f′(3).

(A) − 18

(B) − 3

(D) 3

(E) 42

5. For what positive value of k is the line y = −9x + k tangent to the curve y = x³ − 6x²?

(A) − 27

(B) 0

(D) 5

(E) 4

6. The tangent to the parabola y = ax² at x = p intersects the y-axis at

(A) (− ap², 0)

(B) (− 2a, 0)

(D) (0, ap²)

(E) (0, − ap)

7. If h(x) = f(x)g(x), then h′(1) =

(A) 1

(B) 2

(D) 4

(E) 5

8. If h(x) = f(g(x)), then h′(2) =

(A) − 2

(B) − 1

(D) 1

(E) 2

9. If f(x) is a continuous function and f(5) = 2 and f′(5) = 3, then f (5.01) is approximately

(A) 0.02

(B) 0.03

(D) 2.03

(E) 3.02

The table show some of the values of differentiable functions f and g and their derivatives. Use this information for questions 10 and 11.

10. Evaluate

(A) ln 2

(B) − ln 2

(D) 1

(E) Does not exist

11. A cube is expanding so that its surface area is increasing at a constant rate of 4 square inches per second. How fast is the volume increasing at the instant when the surface area is 24 square inches?

(A) 2 cubic inches per second

(B) 4 cubic inches per second

(D) 16 cubic inches per second

(E) 24 cubic inches per second

12. If , and , then

(A) 2

(B) 8

(D) 10

(E) 11

13. An antiderivative for is

14. is

(A) Indeterminate

(B) 1

15.

16. If x³ + 3xy + 2y³ = 17, then in terms of x and y,

17. If f(x) = 3tan⁵(2x), then f′(x) is:

(A) 3 sec¹⁰(2x)

(B) 30 sec¹⁰(2x)

(D) 30 tan⁴(2x)

(E) 30 tan⁴(2x) sec²(2x)

18. If f′(x) = cos x sin² x and f (0) = 2, then is

(A) 0

(B) −1

(E) 2

19. The particular solution for the differential equation satisfying the initial condition y (0) = 1 is

(D) y = 1 − x²

(E) y = 1 + x

20. Let f(x) be a differentiable function on the closed interval [1, 3]. The average value of f′(x) on [1, 3] is

(A) f(3) − f(1)

(B) 2f(3) − f(1))

(D) f(3) − f(1)

21. Which of the following improper integrals converge?

(A) I only

(B) III only

(D) II and III

(E) I and III

22. A particle moves in the xy-plane so that its coordinates at time t are given by x = t² and y = 4 − t³. At t = 1, its acceleration vector is

(A) ⟨2, − 3⟩

(B) ⟨2, − 6⟩

(D) ⟨2, 6⟩

(E) ⟨1, − 2⟩

23. If , then

(A) tan x

(D) x² tan x

(E) tan² x

24.

(A) −π

(B) π

(E) π + 1

25. Which of the following equations has a graph that is a circle with radius 2 and center at the origin?

II. x = cos t, y = sin t; 0 ≤ t ≤ 2π

III. r = 4 cos θ; 0 ≤ θ ≤ π

(A) I

(B) II

(D) All of these

(E) None of these

26. If r = f(θ) is continuous and non-negative for α ≤ x ≤ β, then the area enclosed by the polar curve r = f(θ) and the lines θ = α and θ = β is given by

27. The first five nonzero terms in the series expansion of e^−4x about x = 0 are

28. The general solution of is a family of

(A) Lines

(B) Circles

(D) Hyperbolas

(E) Parabolas

Section I—Part B

Directions:

Use the same answer sheet for Part A. Please note that the questions begin with number 76. This is not an error. It is done to be consistent with the numbering system of the actual AP Calculus BC Exam. All questions are given equal weight. There is no penalty for unanswered questions. Unless otherwise indicated, the domain of a function f is the set of all real numbers. If the exact numerical value does not appear among the given choices, select the best approximate value. The use of a calculator is permitted in this part of the exam.

76. If y = 5^(7−2x²⁾, then at x = 1 is approximately

(A) 5

(B) 25

(D) 5029.494

(E) −20117.974

77. Given f(x) = 2^x and 2^3.03 ≈ 8.168, using differentials, determine which of the following is closest to f′(3)?

(A) 3.2

(B) 4.6

(D) 8.2

(E) 9.1

78. The rate of growth of a population is proportional to the population. If the population in 2000 was 3 million and in 2010, the population was 3.21 million, what is the predicted population for the year 2020?

(A) 3.315 million

(B) 3.435 million

(D) 7.913 million

(E) 11.609 million

79. The function f(x) = 5x³ − 7x² − 5 has which of these relative extrema?

(A) Relative maximum at (0.933, − 7.033)

(B) Relative maximum at (− 5, 0)

(D) Relative minimum at (0.933, − 7.033)

(E) No relative maxima or minima

80. The slope of the tangent line to the curve r = 4 − 3 cos θ at is approximately

(A) 3.284

(B) 0.524

(D) 0.027

(E) − 4.837

81. A particle moves in the xy-plane in such a way that its path is defined by x = e^t cos t and y = e^t sin 2t. Find the speed of the particle when .

(A) 10.757

(B) − 4.810

(D) 14.431

(E) 4.810

82. At what value of x does the graph of have a point of inflection?

(A) x = 1

(B) x = 2

(D) x = 4

(E) x = 5

83.

(A) e

(B) 1 + e

84. The base of a solid is the region enclosed by the ellipse 4x² + y² = 1. If all plane cross sections perpendicular to the x-axis are semicircles, then its volume is

85. The area under the curve bounded by the line x = 0 and x = 3 is approximately

(A) 0.633

(B) 1.937

(D) 4.667

(E) 5.333

86. Which of the following is the interval of convergence for the power series given by

(A) (− 2, 2)

(B) (− 1, 5)

(D) (− 4, 0)

(E) (− ∞, ∞)

87. Find the coordinates of the point where the line tangent to y = 4 − 3x − x² at x = 1 intersects the axis of symmetry of the parabola.

(A) (5, 0)

88. A rectangular field is fenced off along the bank of a river, and no fence is required along the river. If the material for the fence costs $5.00 per foot for the two ends and $7.50 per foot for the side parallel to the river, find the largest possible area that can be enclosed with $9,000 worth of fence.

(A) 270,000 square feet

(B) 27,000 square feet

(D) 450 square feet

(E) 270 square feet

89. Let R be the region in the first quadrant enclosed by the graph of , the line x = 7, the x-axis and the y-axis. The volume of the solid generated when R is revolved about the x-axis is

90. If and y = −1 when x = 1, then when x = 2, y =

91. For all x, if , then f′(x) =

92. Find the equation of the tangent line to y = x³ + 2x at its point of inflection.

(A) y = 2x

(B) y = 3x + 2

(D) y = 2x − 3

(E) y = −2x + 5

STOP. AP Calculus BC Practice Exam 2 Section I—Part B

Section II—Part A

Directions:

Show all work. You may not receive any credit for correct answers without supporting work. You may use an approved calculator to help solve a problem. However, you must clearly indicate the setup of your solution using mathematical notations and not calculator syntax. Calculators may be used to find the derivative of a function at a point, compute the numerical value of a definite integral, or solve an equation. Unless otherwise indicated, you may assume the following: (a) the numeric or algebraic answers need not be simplified; (b) your answer, if expressed in approximation, should be correct to 3 places after the decimal point; and (c) the domain of a function f is the set of all real numbers.

1. Let R represent the region in the first quadrant bounded by the graphs of y = cos x, y = 1 − cos x, and the y-axis.

(A) Find the area of region R.

(B) Find the volume of the solid created when region R is revolved about the x-axis.

(C) Find the volume of the solid that has R as its base if each of the cross sections is an isosceles triangle of height x.

2. A drum containing 100 gallons of oil is punctured by a nail and begins to leak at the rate of 10 sin gallons/minute where t is measured in minutes and 0 ≤ t ≤ 10.

(A) How much oil to the nearest gallon leaked out after t = 6 minutes?

(B) What is the average amount of oil leaked out per minute from t = 0 to t = 6tothe nearest gallon?

(D) At what value of t to the nearest minute will there be 40 gallons of oil remaining in the drum?

STOP. AP Calculus BC Practice Exam 2 Section II—Part A

Section II—Part B

Directions:

The use of a calculator is not permitted in this part of the test. When you have finished this part of the test, you may return to the problems in Part A of Section II and continue to work on them. However, you may not use a calculator. You should show all work. You may not receive any credit for correct answers without supporting work. Unless otherwise indicated, the numeric or algebraic answers need not be simplified, and the domain of a function fis the set of all real numbers.

Figure D-11

3. A population is modeled by a function P that satisfies the logistic differential equation

(A) If P(0) = 4, what is , what is

(B) If P (0) = 4, for what value of P is the population growing fastest?

(C) A different population is modeled by a function Q that satisfies the separable differential equation . Find Q(t) if Q(0) = 3.

(D) For the function found in part (C), what is

4. The graph of f′, the derivative of the function f, for − 4 ≤ x ≤ 6 is shown above. (See Figure D-11.)

(A) At what value(s) of x does f have a relative maximum value? Justify your answer.

(B) At what value(s) of x does f have a relative minimum value? Justify your answer.

(D) At what value(s) of x, if any, does the graph of f″ have a point of inflection? Justify your answer.

(E) Draw a possible sketch of f (x), if f (− 2) = 3.

5. The MacLaurin series for the function f is given by

on its interval of convergence.

(A) Find the interval of convergence.

(B) Find the first four terms and the general term for the MacLaurin series for

6. Given the equation x²y² = 4:

(A) Find .

(B) Write an equation of the line tangent to the graph of the equation at the point (1, − 2).

(D) The line is tangent to the curve P. Find the coordinates of point P.

STOP. AP Calculus BC Practice Exam 2 Section II—Part B

Answers to BC Practice Exam 2—Section I

Part A

1. B

2. D

3. C

4. E

5. E

6. C

7. B

8. B

9. D

10. D

11. A

12. A

13. E

14. D

15. B

16. A

17. E

18. D

19. A

20. E

21. E

22. B

23. C

24. B

25. E

26. E

27. C

28. D

Part B

76. E

77. C

78. B

79. D

80. A

81. A

82. B

83. B

84. A

85. D

86. B

87. D

88. A

89. E

90. B

91. D

92. A

Answers to BC Practice Exam 2—Section II

Part A

1. (A) 0.685

(B) 2.152

2. (A) 38 gallons

(B) 6 gallons

(D) 8 minutes

Part B

3. (A) 20; 20

(B) when P = 10

(D) 0

4. (A) x = 2

(B) x = 5 and x = −2

(D) x = 0 and x = 4

(E) See solution.

5. (A) (− ∞, ∞)

(D) (− 2, 1)

Solutions to BC Practice Exam 2—Section I

Section I Part A

1. The correct answer is (B).

At the point (3, − 2), m = f′(3) = 5, so the equation of the tangent line is y + 2 = 5(x − 3) or y = 5x − 17.

2. The correct answer is (D).

If , then y′ =

Since the numerator is always negative and the denominator is always positive, y′ is always negative and so the function is monotonically decreasing.

On the interval − 3 < x < − 1, − 13 ≤ 5x + 2 ≤ − 3 ⇒ (5x + 2)³ < 0 ⇒ y″ < 0. Therefore, the function is concave down on the given interval.

3. The correct answer is (C).

If , . Set the derivative equal to zero.

The second derivative,

evaluated at ⇒ x = −2a, is

Therefore, the point (−2a, 2a^⅔) is a relative minimum.

4. The correct answer is (E).

Find f′(3) = 4(3)² + 2(3) = 42.

5. The correct answer is (E).

y = x³ − 6x² ⇒ y′ = 3x² − 12x. Slope of the tangent line is − 9, so 3x² − 12x = −9, and solving gives x = 3 or x = 1. For each value, determine the equation of the tangent line. For x = 3, y = 27 − 54 = −27 and y + 27 = −9(x − 3) which gives k = 0. Reject this possibility since k must be positive. For x = 1, y = 1 − 6 = −5 and the tangent is y + 5 = −9(x − 1), which implies k = 4.

6. The correct answer is (C).

y = ax² ⇒ y′ = 2ax. Evaluate when x = p to find the slope of the tangent line is m = 2ap. The point of tangency is (p, ap²), so the equation of the tangent line is y − ap² = 2ap(x − p) or y = 2apx − ap². This tangent intersects the y-axis at (0, − ap²).

7. The correct answer is (B).

If h(x) = f(x)g(x), then

h′(x) = f(x)g′(x) + g(x) f′(x) and

h′(1) = f (1)g′(1) + g (1) f′(1)

= (0)(5) + (− 1)(− 2) = 2.

8. The correct answer is (B).

If h(x) = f(g(x)), then

h′(x) = f′(g(x))g′(x) and

h′(2) = f′(g (2))g′(2)

⇒ h′(2) = f′(3)g′(2)

⇒ h′(2) = (− 1)(1) = − 1.

9. The correct answer is (D).

If f(x) is a continuous function and

f(5) = 2 and f′(5) = 3, then

f (5.01) = f(5) + 0.01 f′(5)

= 2 + 0.01(3) = 2.03.

10. The correct answer is (D).

Simplify the expression as ,

which is equivalent to .

Thus,

11. The correct answer is (A).

At the moment the surface area is 24 square inches, A = 6x² = 24 ⇒ x² = 4 ⇒ x = 2, so the edge of the cube is 2 inches. The surface area is changing so

and when the edge is 2 inches, .

The volume V = s³ is changing at the rate

. When the surface area is

24 square inches and the edge is 2 inches,

. The volume increasing

at 2 cubic inches per second.

12. The correct answer is (A).

13. The correct answer is (E).

Integrate by partial fractions.

A(x − 1) + B(x − 2) = 1 ⇒ Ax + Bx = 0

⇒ A = − B. Substituting, − A − 2B = 1

⇒ − A + 2A = 1 and so A = 1, B = − 1. Then the integral becomes

14. The correct answer is (D).

Since θ → 0 ⇒ 1 − cos θ → 0

and θ → 0 ⇒ 2 sin² θ → 0,

and is indeterminate.

Differentiate

and .

Then, by L’Hôpital’s Rule,

15. The correct answer is (B).

∫ xf(x) dx requires integration by parts. Let

u = f(x), du = f′(x) dx, dv = x dx, and

. Then .

16. The correct answer is (A).

Differentiate x³ + 3xy + 2y³ = 17 implicitly.

17. The correct answer is (E).

Rewrite 3 tan⁵(2x) as 3 [tan(2x)]⁵.

Thus, f′(x) = 15 [tan(2x)]⁴ [sec²(2x)](2)

= 30 [tan(2x)]⁴ [sec²(2x)].

18. The correct answer is (D).

Letting u = sin x gives us du = cos x dx

19. The correct answer is (A).

Separate as

Integrate to produce sin⁻¹(y)

= sin⁻¹(x) + C. With the initial condition

y(0)= 1, sin⁻¹(1) = sin⁻¹(0) + C, .

. Thus .

20. The correct answer is (E).

The average value of f′(x) on [1, 3] is

21. The correct answer is (E).

22. The correct answer is (B).

and .

At t = 1, the acceleration vector is ⟨2, −6t⟩|_t₌₁ = ⟨2, −6⟩.

23. The correct answer is (C).

Integrate by parts. Let u = x,

du = dx, dv = sec² x dx, and v = tan x.

Then = x tan x − ∫ tan x dx = x tan x + ln |cos x| + C.

Comparing to the given information, tells us that f(x) = x tan x.

24. The correct answer is (B).

The integrand of

can be factored as

dx, which can be recognized as .

Integrate by parts with u = x, du = dx,

dv = sin x dx, and v = − cos x. Then

25. The correct answer is (E).

Consider each curve.

(I) ; − 2 ≤ x ≤ 2 defines a semicircle, not a circle. (II) x = cos t, y = sin t; 0 ≤ t ≤ 2π is a circle, but its radius is 1. (III) r = 4 cos θ;0 ≤ θ ≤ π is a circle of radius 2, but its center is at (2, 0). Therefore, none of these equations defines a circle of radius 2 with center at the origin.

26. The correct answer is (E).

The area enclosed by a polar curve is

27. The correct answer is (C).

Begin with the known MacLaurin series

and substitute −4x for x.

28. The correct answer is (D).

separates to y dy = 4x dx, which

integrates to

⇒ 4x² − y² = − 2C. This form is the family of hyperbolas.

Section I—Part B

76. The correct answer is (E).

If , then ln y = (7 − 2x²) ln 5.

Differentiating implicitly, .

Evaluating at x = 1 produces .

77. The correct answer is (C).

If f(x) = 2^x, f(3) = 8 and f′(x) = 2^x ln 2, so

f′(3) = 8 ln 2. We could estimate f (3.03)

≈ f (3) + 0.03 f′(3), and since we know

2^3.03 ≈ 8.168, 8.168 ≈ 8 + 0.03 f′(3). Solve

78. The answer is (B).

Separate as . Integrate

to produce

ln |P| = kt + c1. The initial condition tells us

P = c²e^kt ⇒ 3 = c²e⁰ ⇒ c2 = 3 so P = 3e^kt. In

2010, the population was 3.21 million. Thus

3.21 = 3e^10k ⇒ 1.07 = e^10k

⇒ ln 1.07 = 10k ⇒ 0.00677 = k. Therefore,

P = 3e^0.00677t and P = 3e^0.00677(20) = 3.435.

79. The correct answer is (D).

The function f(x) = 5x³ − 7x² − 5

⇒ f′(x) = 15x² − 14x. Set the first derivative equal to zero. 15x² − 14x = 0

⇒ x (15x − 14) = 0 ⇒ x = 0 or .

Evaluate the second derivative

f″(x) = 30x − 14 at each of the critical points.

f′(0) = − 14

⇒ max ⇒

80. The correct answer is (A).

. Since

and

81. The correct answer is (A).

82. The correct answer is (B).

Set the second derivative equal to zero and

solve.

⇒ 6x − 12 = 0 ⇒ x = 2. Also y″ < 0

for x < 2 and y″ > 0 for x > 2.

83. The correct answer is (B).

The series is a geometric series, so

Compare the series to the known

MacLaurin series , and

Therefore, .

84. The correct answer is (A).

The ellipse 4x² + y² = 1 has x −intercepts

. The semicircles will have radius

of and area of

Then the volume is

85. The correct answer is (D).

86. The correct answer is (B).

The series will converge when

. This

means

⇒ − 3 < x − 2 < 3 ⇒ − 1 < x < 5. When

x = − 1, the series becomes

, which diverges. When

x = 5, the series becomes

, which diverges. Therefore,

the interval of convergence is (− 1, 5).

87. The correct answer is (D).

If y = 4 − 3x − x², ⇒ y′ = − 3 − 2x. The slope

of the tangent line is y′(1) = − 3 − 2(1) = − 5, and the point is (1, 0), since

y (1) = 4 − 3(1) − (1)² = 0. The equation of the

tangent line is y − 0 = − 5(x − 1) or

y = − 5x + 5. The axis of symmetry of the

parabola is . Evaluating the

tangent equation when gives

. The line

y= − 5x + 5 intersects the axis of symmetry at

88. The correct answer is (A).

Let l be the length of the section of fence

parallel to the river and w be the length of the

two remaining sides. The cost of the fence is

C = 5(2w) + 7.50(l) = 9000. Set cost equal to

9000 and solve for l. 5(2w) + 7.50(l) = 9000

Differentiate and set the

derivative equal to zero.

⇒ 9000 − 20w = 0 ⇒ w = 450. If the width is

450, the length ,

dA″ < 0; so the maximum area is

450 · 600 = 270,000.

89. The correct answer is (E).

Using your calculator you obtain 93π/5.

90. The correct answer is (B).

Separate into , and

integrate

If y = − 1 when x = 1,

Thus . If x = 2,

91. The correct answer is (D).

If , then

92. The correct answer is (A).

y = x³ + 2x ⇒ y′ = 3x² + 2 ⇒ y″ = 6x. Set the second derivative equal to zero. y″ = 6x = 0 ⇒ x = 0. When x < 0, y″ < 0, and when x > 0, y″ > 0. The point of inflection is (0, 0) and the slope at that point is . The equation of the tangent line to the curve at its point of inflection is y − 0 = 2(x − 0) ⇒ y = 2x.

Solutions to BC Practice Exam 2—Section II

Section II—Part A

1. (A) Determine the point of intersection of the two curves, either by calculator, or algebraically. cos x = 1 − cos x , so is the x −coordinate of the point of intersection. The area of region R is

(B) The volume of the solid created when the region R is revolved about the x −axis can be found by the method of washers.

Integrate and evaluate.

(C) To find the volume of the solid that has R as its base, and cross sections that are isosceles triangles of height x, recognize that the area of each cross section is , so the volume is . Integrate by parts, with u = x, du = dx, dv = (2 cos x − 1)dx, and v = 2 sin x − x.

Or simply use your calculator and obtain the same results.

2. (A) The amount of oil leaked out after 6 minutes

(B) Average amount of oil leaked out per minute from t = 0to t = 6:

(D) Let a be the value of t:

Section II—Part B

3. If

Integrate

. Exponentiate for

and isolate P. Thus,

(A) If P(0) = 4,

, so and

(B) The rate of growth is given by

. To find the maximum value of the growth function, take the derivative of the function.

Set

When P = 10, .

The population is growing fastest, at a rate of 125%, when the population is 10.

and integrate.

(D) For the function found in part (C), as t

increases, so

; therefore,

4. (A) (See Figure DS-15.) Since f increases on (− 2, 2) and decreases on (2, 5), f has a relative maximum at x = 2.

Figure DS-15

(B) Since f decreases on (− 4, − 2) and increases on (− 2, 2), f has a relative minimum at x = − 2. Since f decreases on (2, 5) and increases on (5, 6), f has a relative minimum at x = 5.