Take a Diagnostic Exam - Determine Your Test Readiness - 5 Steps to a 5 AP Calculus AB & BC

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

STEP 2. Determine Your Test Readiness

Chapter 3. Take a Diagnostic Exam

IN THIS CHAPTER

Summary: Get started in your review by working out the problems in the diagnostic exam. Use the answer sheet to record your answers. After you have finished working the problems, check your answers with the answer key. The problems in the diagnostic exam are presented in small groups matching the order of the review chapters. Your results should give you a good idea of how well you are prepared for the AP Calculus Exams at this time. Note those chapters that you need to study the most, and spend more time on them. Good luck. You can do it.

Key Ideas

Image

Image Work out the problems in the diagnostic exam carefully.

Image Check your work against the given answers.

Image Determine your areas of strength and weakness.

Image Identify and mark the pages that you must give special attention.

DIAGNOSTIC TEST ANSWER SHEET

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3.1 Getting Started!

Taking the Diagnostic Test helps you assess your strengths and weaknesses as you begin preparing for the AP Calculus exam. The questions in the Diagnostic Test contain both multiple-choice and open-ended questions. They are arranged by topic, and designed to review concepts tested on the AP Calculus exam. All questions in the diagnostic test should be done without the use of a graphing calculator, except in a few cases where you need to find the numerical value of a logarithmic or exponential function. Towards the end of the test, you will find a group of questions that cover topics unique to the BC exam. These include all the material from Chapter 14 as well as selected topics from other chapters.

3.2 Diagnostic Test

Chapter 5

1. A function f is continuous on [−2, 0] and some of the values of f are shown below.

Image

If f(x) = 2 has no solution on [−2, 0], then b could be

(A) 3

(B) 2

(C) 1

(D) 0

(E) −2

2. Evaluate Image.

3. If

Image

4. If f(x) = |2xex|, what is the value of Image

Chapter 6

5. If f (x)= −2 csc (5x), find Image.

6. Given the equation y = (x + 1)(x − 3)2, what is the instantaneous rate of change of y at x = −1?

7. What is Image?

Chapter 7

8. The graph of f is shown in Figure D-1. Draw a possible graph of f′ on (a, b).

Image

Figure D-1

9. The graph of the function g is shown in Figure D-2. Which of the following is true for g on (a, b)?

I. g is monotonic on (a, b).

II. g′ is continuous on (a, b).

III. g″ >0 on (a, b).

Image

Figure D-2

10. The graph of f is shown in Figure D-3 and f is twice differentiable, which of the following statements is true?

Image

Figure D-3

(A) f(10) < f′(10) < f″(10)

(B) f″(10) < f′(10) < f(10)

(C) f′(10) < f(10) < f″(10)

(D) f′(10) < f″(10) < f(10)

(E) f″(10) < f(10) < f′(10)

11. The graph of f′, the derivative of f, is shown in Figure D-4. At what value(s) of x is the graph of f concave up?

Image

Figure D-4

12. How many points of inflection does the graph of y = sin(x2) have on the interval [−π, π]?

13. If Image and the graph of f is shown in Figure D-5, which of the graphs in Figure D-6 on the next page is a possible graph of g?

Image

Figure D-5

14. The graphs of f′, g′, p′, and q′ are shown in Figure D-7 on the next page. Which of the functions f, g, p, or q have a point of inflection on (a, b)?

Chapter 8

15. When the area of a square is increasing four times as fast as the diagonals, what is the length of a side of the square?

16. If g(x) = |x2 − 4x − 12|, which of the following statements about g is/are true?

I. g has a relative maximum at x = 2.

II. g is differentiable at x = 6.

III. g has a point of inflection at x = −2.

Chapter 9

17. Given the equation Image what is an equation of the normal line to the graph at x = 5?

18. What is the slope of the tangent to the curve y = cos(x y) at x = 0?

Image

Figure D-6

Image

Figure D-7

19. The velocity function of a moving particle on the x-axis is given as v(t) = t2t. For what values of t is the particle’s speed decreasing?

20. The velocity function of a moving particle is Image for 0 ≤ t ≤ 6. What is the maximum acceleration of the particle on the interval 0 ≤ t ≤ 6?

21. Write an equation of the normal line to the graph of y = x3 for x ≥ 0 at the point where f′(x) = 12 and y = f (x).

22. At what value(s) of x do the graphs of Image and y = −x2 have perpendicular tangent lines?

23. Given a differentiable function f with Image and Image. Using a tangent line to the graph at Image, find an approximate value of, Image.

Chapter 10

24. Evaluate Image.

25. If f (x) is an antiderivative of Image and f(0) = ln (2), find f(ln 2).

26. Find the volume of the solid generated by revolving about the x-axis on the region bounded by the graph of y = sin 2x for 0 ≤ x ≤ π and the line Image.

Chapter 11

27. Evaluate Image.

28. If Image, find k.

29. If Image.

30. If f′(x) = g(x) and g is a continuous function for all real values of x, then Image is

Image

(B) f (2) − f (0).

(C) f (6) − f (0).

Image

(E) 3 f (6) − 3 f (0).

31. Evaluate Image.

32. If a function f is continuous for all values of x, which of the following statements is/are always true?

Image

33. If Image on Image, find the value(s) of x where g has a local minimum.

Chapter 12

34. The graph of the velocity function of a moving particle is shown in Figure D-8. What is the total distance traveled by the particle during 0 ≤ t ≤ 6?

Image

Figure D-8

35. The graph of f consists of four line segments, for −1 ≤ x ≤ 5 as shown in Figure D-9.

What is the value of Image

Image

Figure D-9

36. Find the area of the region enclosed by the graph of y = x2x and the x-axis.

37. If Image for all real values of k, then which of the graphs in Figure D-10 could be the graph of f?

Image

Figure D-10

38. The area under the curve Image from x = 1 to x = k is 8. Find the value of k.

39. For 0 ≤ x ≤ 3π, find the area of the region bounded by the graphs of y = sin x and y = cos x.

40. Let f be a continuous function on [0, 6] that has selected values as shown below:

Image

Using three midpoint rectangles of equal widths, find an approximate value of Image

Chapter 13

41. What is the average value of the function y = e−4x on [-ln 2, ln 2]?

42. If Image and at x = π, y = 2, find a solution to the differential equation.

43. Water is leaking from a tank at the rate of f(t) = 10 ln (t + 1) gallons per hour for 0 ≤ t ≤ 10, where t is measured in hours. How many gallons of water have leaked from the tank at exactly after 5 hours?

44. Carbon-14 has a half-life of 5730 years. If y is the amount of Carbon-14 present and y decays according to the equation Image, where k is a constant and t is measured in years, find the value of k.

45. What is the volume of the solid whose base is the region enclosed by the graphs of y = x2 and y = x + 2 and whose cross sections perpendicular to the x-axis are squares?

Image

Note: The questions below focus on topics covered only on the BC exam.

Chapter 6

46. Evaluate Image.

Chapter 7

47. Find the rectangular equation of the curve defined by x = 1 + e-t and y = 1 + et.

Chapter 9

48. An object moves in the plane on a path given by x = 4t2 and Image Find the acceleration vector when t = 4.

49. Find the equation of the tangent line to the curve defined by x = 2t + 3, y = t2 + 2t at t = 1.

Chapter 10

50. Evaluate Image.

51. Evaluate Image.

Chapter 11

52. Evaluate Image.

Chapter 12

53. Find the total area bounded by the curves r = 2 cos θ and r = 2 sin θ.

54. Determine the length of the curve defined by x = 3tt3 and y = 3t2 from t = 0 to t = 2.

Chapter 13

55. The growth of a colony of bacteria in a controlled environment is modeled by Image. If the initial population is 100, find the population when t = 5.

56. If Image and y = 3 when x = 2, approximate y when x = 3.

Chapter 14

57. Determine whether the series Image converges or diverges.

58. For what values of x does the series Image converge absolutely?

59. Find the Taylor series expansion of Image about the point x = 2.

60. Find the MacLaurin series for Image.

3.3 Answers to Diagnostic Test

1. A

2. −1/2

3. Does not exist.

4. 2

5. Image

6. 16

7. 2

8. See Figure DS-3.

9. II & III

10. C

11. x < x2

12. 8

13. A

14. q

15. Image

16. I

17. y = −4x + 22

18. 0

19. Image

20. 12

21. Image

22. 1.370

23. 2.983

24. Image

25. ln 3

26. 1.503

27. 2

28. {−2, 5}

29. 0

30. A

31. Image

32. I&III

33.

34. 50 feet

35. 2

36. Image

37. D

38. 132/3

39. 5.657

40. 76

41. Image

42. y = −2 cos x

43. 57.506

44. Image

45. Image

Image

Topics Unique to the BC Exam

46. Image

47. Image

48. Image

49. y = 2x − 7

50. Image

51. x 2 sin x + 2x cos x − 2 sin x + C

52. 1

53. Image

54. 14

55. 514.325

56. 2.415

57. Converges

58. −1 < x < 1

59. Image

60. Image

3.4 Solutions to Diagnostic Test

Chapter 5

1. See Figure DS-1.

If b = 2, then x = −1 would be a solution for f(x) = 2.

If b = 1, 0 or −2, f(x) = 2 would have two solutions.

Thus, b = 3, choice (A).

Image

Figure DS-1

2.

Image

3

Image

Since Image does not exist.

4.

Image

If x ≤ 0, f′(x) = 2ex + ex (2x) = 2ex + 2xex

Image

Chapter 6

5.

Image

6. y = (x + 1)(x − 3)2;

Image

Image

7.

Image

Chapter 7

8. See Figure DS-2 on the next page.

9. I. Since the graph of g is decreasing and then increasing, it is not monotonic.

II. Since the graph of g is a smooth curve, g′ is continuous.

III. Since the graph of g is concave upward, g″ > 0.

Thus, only statements II and III are true.

10. The graph indicates that (1) f(10) = 0, (2) f′(10) < 0, since f is decreasing; and (3) f″(10) > 0, since f is concave upward. Thus, f′(10) < f(10) < f″(10), choice (C).

Image

Figure DS-2

11. See Figure DS-3.

The graph of f is concave upward for x < x2.

Image

Figure DS-3

12. See Figure DS-4.

Enter y1 = sin(x2). Using the [Inflection] function of your calculator, you obtain four points of inflection on [0, π]. The points of inflection occur at x = 0.81, 1.81, 2.52, and 3.07. Since y1 = sin (x2) is an even function, there is a total of eight points of inflection on [−π, π]. An alternate solution is to enter Image. The graph of y2 crosses the x-axis eight times, thus eight zeros on [−π, π].

Image

Figure DS-4

13. Since Image, g′(x) = f(x).

See Figure DS-5.

The only graph that satisfies the behavior of g is choice (A).

Image

Figure DS-5

14. See Figure DS-6.

A change of concavity occurs at x = 0 for q. Thus, q has a point of inflection at x = 0. None of the other functions has a point of inflection.

Image

Figure DS-6

Chapter 8

15. Let z be the diagonal of a square. Area of a

Image

Let s be a side of the square. Since the diagonal z = 4, s2 + s2 = z2 or 2s2 = 16. Thus, s2 = 8 or Image.

16. See Figure DS-7.

The graph of g indicates that a relative maximum occurs at x = 2; g is not differentiable at x = 6, since there is a cusp at x = 6, and g does not have a point of inflection at x = −2, since there is no tangent line at x = −2. Thus, only statement I is true.

Image

Figure DS-7

Chapter 9

17.

Image

Slope of normal line = negative reciprocal of Image.

Equation of normal line:

y − 2 = −4(x − 5) ⇒ y = −4(x − 5) + 2 or y = −4x + 22.

18.

Image

Thus, the slope of the tangent at x = 0 is 0.

19. See Figure DS-8.

v(t) = t2t

Set v(t) = 0 ⇒ t(t − 1) = 0

t = 0 or t = 1

a(t) = v′(t) = 2t − 1.

Image

Since v(t) < 0 and a(t) > 0 on Image, the speed of the particle is decreasing on Image.

Image

Figure DS-8

20.

Image

See Figure DS-9.

The graph indicates that the maximum acceleration occurs at the endpoint t = 6. a(t) = t2 − 4t and a(6) = 62 − 4(6) = 12.

Image

Figure DS-9

21.

Image

Slope of normal = negative reciprocal of slope of tangent Image.

At x = 2, y = x3 = 23 = 8; (2, 8) Image.

Image or Image.

22.

Image

Perpendicular tangents

Image

Using the [Solve] function on your calculator, you obtain Image.

23. Image is on the graph.

Image ⇒ slope of the tangent at Image is −1.

Equation of tangent line:

Image

Thus,

Image

Chapter 10

24.

Image

Image

You can check the answer by differentiating your result.

25. Let u = ex + 1; du = ex dx.

Image

Thus, f(x) = ln (ex + 1) and f(ln 2)

= ln (eln2 + 1) = ln (2 + 1)

= ln 3.

26. See Figure DS-10.

To find the points of intersection, set

Image

Volume of solid

Image

Using your calculator, you obtain: Volume of solid ≈ (0.478306)π ≈ 1.50264 ≈ 1.503.

Image

Figure DS-10

Chapter 11

27.

Image

28.

Image

Set k2 − 3k − 4 = 6 ⇒ k2 − 3k − 10 = 0

⇒ (k − 5)(k + 2) = 0 ⇒ k = 5 or k = −2.

Image

You can check your answer by evaluating Image and Image.

29.

Image

30. Let u = 3x; du = 3dx or Image.

Image

Image

Thus, the correct choice is (A).

31.

Image

32.

Image

The statement is true, since the upper and lower limits of the integrals are in sequence, i.e. ac = abc.

II.

Image

The statement is not always true.

III.

Image

The statement is true.

Thus, only statements I and III are true.

33. Since Image, then g′(x) = 2 sin x.

Set g′(x) = 0 ⇒ 2 sin x = 0 ⇒ x = π or 2π g″(x) = 2 cos x and g″(π) = 2 cos π = −2 and g″(2π) = 1.

Thus g has a local minimum at x = 2π. You can also approach the problem geometrically by looking at the area under the curve. See Figure DS-11.

Image

Figure DS-11

Chapter 12

34.

Image

35.

Image

36. To find points of intersection, set y = x2x = 0

x (x − 1) = 0 ⇒ x = 0 or x = 1.

See Figure DS-12.

Image

Figure DS-12

Image

37. Image is an odd function, i.e., f(x) = − f(−x). Thus the graph in choice (D) is the only odd function.

38.

Image

Since A = 8, Image

= 12 ⇒ k3/2 = 13 or k = 132/3.

39. See Figure DS-13.

Image

Figure DS-13

Using the [Intersection] function of the calculator, you obtain the intersection points at x = 0.785398, 3.92699, and 7.06858.

Image

You can also find the area by:

Image

40. Width of a rectangle Image. Midpoints are x = 1, 3, and 5 and f(1) = 2, f(3) = 10 and f(5) = 26.

Image

Chapter 13

41.

Image.

Let u = −4x; du = −4dx, or Image.

Image

Image

42.

Image

Thus, y = −2 cos x.

43. Amount of water leaked

Image

Using your calculator, you obtain 10(6 ln 6 − 5) which is approximately 57.506 gallons.

44.

Image

Half-life = 5730 ⇒ Image

when t = 5730.

Thus, Image.

Image

45. See Figure DS-14.

Image

Figure DS-14

To find points of intersection, set x2 = x + 2 ⇒ x2x−2 = 0 ⇒ x = 2 or x = −1.

Area of cross section = ((x + 2) − x2)2.

Volume of solid, Image.

Using your calculator, you obtain: Image.

Image

Topics Unique to the BC Exam

Chapter 6

46. By L’Hôpital’s Rule, Image

Chapter 7

47. Solve x = 1 + et for t. x − 1 = et ⇒ − ln (x − 1) = t. Substitute in y = 1 + et.

Image

Chapter 9

48. Position is given by x = 4t2 and Image, so velocity is Image and Image. The acceleration will be Image, Image. Image get the acceleration vector Image.

49. The slope of the tangent line is Image and Image, Image, so Image

x = 5, y = 3, ⇒ (5, 3),

So, the equation of the tangent line is y − 3 = 2(x − 5) ⇒ y = 2(x − 5) + 3 ⇒ y = 2x − 7.

Chapter 10

50. Image Use a partial fraction decomposition with Image, which gives Image and Image. Then the integral becomes

Image

Image

51. Integrate Image x2 cos x dx by parts with u = x2,

du = 2x dx, dv = cos x dx, and v = sin x.

The integral becomes

= x2 sin xImage sin x (2x) dx

= x2 sin x − 2 Image x sin x dx.

Use parts again for the remaining integral, letting u = x, du = dx, dv = sin x dx, and v = − cos x. The integral

Image simplifies to

Image

and the final integration gives you

= x2 sin x + 2 x cos x − 2 sin x + C.

Chapter 11

52.

Image

Chapter 12

53. The intersection of the circles r = 2 cos θ and r = 2 sin θ can be found by adding the area swept out by r = 2 sin θ for Image and the area swept by r = 2 cos θ for Image.

Image

54. Differentiate Image and Image The length of the curve from t = 0 to t = 2 is

Image

Chapter 13

55. Separate and simplify

Image

Integrate with a partial fraction decomposition.

Image

ln |P | − ln |4000 − P | = .35t + C1

Image

Population at t = 0 is 100, so

Image

The population model is

Image

When t = 5, Image

56. If Image and y = 3 and x = 2, approximate y when x = 3. Use Euler’s

Method with an increment of 0.5.

y(2) = 3 and Image so

Image

Image

Image

Chapter 14

57. The series Image is a series with positive terms, which can be compared to the series Image. Image and Image is a p-series with p = 4, and therefore convergent. Image is term by term smaller than Image and so Image converges.

58. The series Image is an alternating series with general term Image. Using the ratio test for absolute convergence, we have Image Image. The series will converge absolutely when |x| < 1 ⇒ − 1 < x < 1. We do not consider the end points since the question asks for absolute convergence.

59. Investigate the first few derivatives of

Image. Image, Image, Image, and, in general,

Image Image.

Evaluate the derivatives at x = 2. Image,

Image

and, in general, Image.

The Taylor series is Image

Image

60. Begin with the MacLaurin series for ex.

If f(x) = ex, then f′(x) = ex, f″(x) = ex, and fn(x) = ex.

Thus Image

Replacing x by −x 2, we have Image

Thus, Image.

3.5 Calculate Your Score

Short-Answer Questions

Questions 1–45 for AP Calculus AB

Image

Questions 1–60 for AP Calculus BC

Image

AP Calculus AB/BC Diagnostic Exam

Image