Calculus AB and Calculus BC

Answers Explained

Multiple-Choice

Part A

1. (C) Use the Rational Function Theorem.

2. (D) Image

3. (A) Since f (1) = 0 and f changes from negative to positive there, f reaches a minimum at x = 1. Although f (2) = 0 as well, f does not change sign there, and thus f has neither a maximum nor a minimum at x = 2.

4. (D) Image

5. (E) Image

6. (D) The graph must look like one of these two:

Image

7. (E) F (x) = 3 cos x cos 3x − sin x sin 3x.
Image

8. (B) Image

9. (C) Let Image Then f increases for 1 < x < 2, then begins to decrease. In the figure above, the area below the x-axis, from 2 to 3, is equal in magnitude to that above the x-axis, hence, Image

10. (D) P (x) = 2g(x) · g (x)

11. (E) Note that H(3) = f −1 (3) = 2. Therefore

Image

12. (D) Note that the domain of y is all x such that |x| Image 1 and that the graph is symmetric to the origin. The area is given by

Image

13. (E) Since

y = 2(x − 3)−2 and = Image

y ″ is positive when x < 3.

14. (D) Image represents the rate of change of mass with respect to time; y is directly proportional to x if y = kx.

15. (B) Image

16. (E) [cos (x2)] = −sin (x2) · 2x. The missing factor 2x cannot be created through introduction of constants alone.

17. (E) As the water gets deeper, the depth increases more slowly. Hence, the rate of change of depth decreases: Image

Image

18. (D) The graph of f is shown in the figure above; f is defined and continuous at all x, including x = 1. Since

Image

f (1) exists and is equal to 2.

19. (B) Since |x − 2| = 2 − x if x < 2, the limit as Image

20. (A) Image

Note that the distance covered in 6 seconds is Image the area between the velocity curve and the t-axis.

21. (C) Acceleration is the slope of the velocity curve, Image

22. (D) Particular solutions appear to be branches of hyperbolas.

23. (A) Differentiating implicitly yields 2xyy + y2 − 2y + 12y2 y = 0. When y = 1, x = 4. Substitute to find y .

24. (C)

Image

25. (A) Image

26. (B) Separate to get Image Since −(−1) = 1 + C implies that C = 0, the solution is Image

This function is discontinuous at x = 0. Since the particular solution must be differentiable in an interval containing the initial value x = 1, the domain is x > 0.

27. (E)
Image

28. (C) Note that Image for all x.

Part B

29. (B) At x = 3, the equation of the tangent line is y − 8 = −4(x − 3), so f (x) Image −4(x − 3) + 8. f (3.02) Image −4(0.02) + 8.

30. (C)
Image

The velocity is graphed in [−1,11] × [−15, 5]. The object reverses direction when the velocity changes sign, that is, when the graph crosses the x-axis. There are two such reversals − at x = a and at x = b.

31. (D) The sign diagram shows that f changes from increasing to decreasing at x = 4

Image

and thus f has a maximum there. Because f increases to the right of x = 0 and decreases to the left of x = 5, there are minima at the endpoints.

32. (D) Since f decreases, increases, then decreases, f ″ changes from negative to positive, then back to negative. Hence, the graph of f changes concavity at x = 2 and x = 3.

33. (B)
Image

On the curve of f (x) − exx2, the two points labeled are (0,1) and (1, e − 1).

The slope of the secant line is Image Find c in [0,1] such that, f (c) = e − 2, or f (c) − (e − 2) = 0. Since f (x) = ex − 2x, c can be calculated by solving 0 = ex − 2x − (e − 2). The answer is 0.351.

34. (B)

Image

Use disks; then ΔV = πR2 H = π(ln y)2 Δy. Note that the limits of the definite integral are 1 and 2. Evaluate the integral

Image

Alternatively, use shells*; then ΔV = 2πRHT = 2πx(2 − ex) Δx. Here, the upper limit of integration is the value of x for which ex = 2, namely, ln 2. Now evaluate

Image

* No question requiring the use of the shells method will appear on the AP exam.

35. (C) Note that the rate is people per minute, so the first interval width from midnight to 6 A.M. is 360 minutes. The total number of people is estimated as the sum of the areas of six trapezoids:

Image

36. (C) Image so v = 3t3 + c.

Since v = 1 when t = 0, c = 1.

Now Image so s = t3 + t + c.

Since, s = 3 when t = 0, c = 3; then s = t3 + t + 3.

37. (A) Let u = x2. Then

Image

38. (C)
Image

To find a, the point of intersection of y = x2 and y = cos (x), use your calculator to solve the equation x2 − cos (x) = 0. (Store the value for later use; a ≈ 0.8241.)

As shown in the diagram above, ΔA = (cos (x) − x2) Δx.

Evaluate the area: Image

39. (D) If x = 2t + 1, then Image When t = 0, x = 1; when t = 3, x = 7.

40. (E) Use A(t) = A0 ekt, where the initial amount A0 is 50. Then A(t) = 50ekt. Since 45 g remain after 9 days, 45 = 50ek·9, which yields Image

To find t when 20 g remain, solve Image for t.

Thus, Image

Image

41. (C) See the figure above. Since x2 + y2 = 262, it follows that

Image

at any time t. When x = 10, then y = 24 and it is given that Image

Hence, Image

42. (E) Let u = 2x and note that Image

Then Image

43. (D) Image

v(0) = −π + 3; so v(0) = π − 3.

44. (C) Image

45. (D) Let y = (x3 − 4x2 + 8)ecos(x2). The equation of the tangent at point (2, y (2)) is yy(2) = y (2)(x − 2). Note that y(2) = 0. To find the y-intercept, let x = 0 and solve for y: y = −2y (2). A calculator yields y = 4.161.

Free-Response

Part A

AB 1/BC 3. (a) Image (Review Chapter 3)

(b) After 15 minutes the rate at which grain is leaking is slowing down at the rate of one half a cubic foot per minute per minute. (Review Chapter 3)

(c) Let h = the height of the cone and r = its radius. The cone’s diameter is given to be 5h, so Image and the cone’s volume,

Image

Then

Image

At t = 10 the table shows Image and it is given that h = 3; thus:

Image

(Review Chapter 4)

(d) Use one of these Riemann sums:

Image

(Review Chapter 6)

AB/BC 2. (a) Graph y = (x + ex) sin (x2) in [1, 3] × [−15, 20], Note that y represents velocity v and x represents time t.

Image

(b) The object moves to the left when the velocity is negative, namely, on the interval p < t < r. Use the calculator to solve (x + ex)(sin (x2)) = 0; then p = 1.772 and r = 2.507. The answer is 1.772 < t < 2.507.

(c) As the object moves to the left (with v(t) negative), the speed of the object increases when its acceleration v (t) is also negative, that is, when v(t) is decreasing. This is true when, for example, t = 2.

(d) The displacement of an object from time t1, to time t2, is equal to

Image

Evaluate this integral on the calculator; to three decimal places the answer is 4.491. This means that at t = 3 the object is 4.491 units to the right of its position at t = 1, given to be x = 10. Hence, at t = 3 the object is at x = 10 + 4.491 = 14.491.

(Review Chapter 8)

Part B

AB 3. One possible graph of h is shown; it has the following properties:

• continuity on [−4,4],

• symmetry about the y-axis,

• roots at x = −1, 1,

• horizontal tangents at x = −2, 0, 2,

• points of inflection at x = −3, −2, −1, 1, 2, 3,

• corners at x = −3, 3.

Image

(Review Chapter 4)

AB 4. (a) Draw elements as shown. Then

Image

Image

(b) Image

(c) Revolving the element around the x-axis generates disks. Then

Image

(Review Chapter 7)

AB 5. (a) The differential equation Image is separable:

Image

If y = 0 when x = 2, then Image thus c = 3, and Image

Solving for y gives the solution: Image

Note that Image is defined only if Image

Image only if the numerator and denominator have the same sign.

Image

Since the particular solution must be continuous on an interval containing the initial point x = 2, the domain is Image (Review Chapter 9)

(b) Since Image the function Image has a horizontal asymptote at y = ln 3. (Review Chapter 2)

AB/BC 6. (a) Image

Image

(c) The line tangent to the graph of A at x = 6 passes through point (6, A(6)) or (6, 9π). Since A (x) = f (x), the graph of f shows that A (6) = f (6) = 6. Hence, an equation of the line is y − 9π = 6(x − 6).

(d) Use the tangent line; then A(x) = y ≈ 6(x − 6) + 9π, so A(7) ≈ 6(7 − 6) + 9π = 6 + 9π.

(e) Since f is increasing on [0,6], f is positive there. Because f (x) = A (x), f (x) = A ″(x); thus A is concave upward for [0,6]. Similarly, the graph of A is concave downward for [6,12], and upward for [12,18]. There are points of inflection on the graph of A at (6,9π) and (12,18π).

Answers Explained

The explanations for questions not given below will be found in the answer section for Calculus AB Diagnostic Test. Some questions in Section I of Diagnostic Tests AB 1 and BC 1 are identical. Explanations of the answers for Questions 1–6, not given below, will be found in Section I of Calculus AB Diagnostic Test Answers 1–6.

Multiple-Choice

Part A

7. (D) The volume is given by Image an improper integral.

Image

9. (B) Let y = xx and take logarithms. ln Image this function has the indeterminate form ∞/∞. Apply L’Hôpital’s rule:

Image

So ye0 or 1.

12. (A) Use the Parts Formula with u = x and dv = ex dx. Then du = dx and v = ex, and

Image

15. (B) The arc length is given by the integral Image dx which is

Image

16. (E) Separating variables yields Image Integrating gives ln x = kt + C. Since x = 2 when t = 0, ln 2 = C. Then ln x = kt + ln 2. Using x = 6 when t = 1, it follows that ln 6 = k + ln 2, so k = ln 6 − ln 2 = Image

17. (B) y = x + 2x ln x and Image

18. (D) Image

19. (C) At (0, 2), Image With step size Image the first step gives Image

where Image so the next step produces Image

24. (B) Since function f is increasing on the interval [2,6], rectangles based on left endpoints of the subintervals will all lie completely below the curve, and thus have smaller areas than any of the other sums or the definite integral.

Part B

29. (B) Set

Image

30. (E) S is the region bounded by y = sec x, the y-axis, and y = 4.

Image

We send region S about the x-axis. Using washers, Δ V = π(R2r2) Δx. Symmetry allows us to double the volume generated by the first-quadrant portion of S. So for V we have

Image

A calculator yields 108.177.

31. (A) Use the Ratio Test:

Image

which equals zero if x ≠ 1. The series also converges if x = 1 (each term equals 0).

32. (C) The absolute value function f (x) = |x| is continuous at x = 0, but f (0) does not exist.

33. (B) The Maclaurin series is

Image

When an alternating series satisfies the Altering Series Test, the sum is approximated by using a finite number of terms, and the error is less than the first term omitted. On the interval −π Image x Image π, the maximum error (numerically) occurs when x = π. Since

Image

four terms will suffice to assure no error greater than 0.1.

34. (B)

Image

Graph x = 4 − t2 and y = 2t for −3 ≤ t ≤ 3 in the window [−1, 5] × [−1, 5]. Now ΔA = x Δy; the limits of integration are the two points where the curve cuts the y-axis, that is, where x = 0. In terms of t, these are t1 = 2 and t2 = + 2. So

Image

35. (E) Image

36. (B) Image

37. (C) Image = cos θ dx = cos θ dθ, sin−1 0 = 0, Image

38. (C) The power series for ln (1 − x), if Image

39. (B) Solve by separation of variables; then

Image

Use P(0) = 200; then c = 1000, so P(x) = 1200 − 1000e−0.16t. Now P(2) = 473.85.

40. (E) As the water gets deeper, the depth increases more slowly. Hence, the rate of change of depth decreases: d2 h/dt2 < 0.

45. (E) The first quadrant area is Image (3 sin 2θ)2 dImage ≈ 3.53.

Free-Response

Part A

1. (a) Use the Ratio Test:

Image

The radius of convergence is 1. At the endpoint x = 1, Image

Since

Image

this series converges by the Alternating Series Test. Thus Image converges for positive values 0 < x ≤ 1.

(b) Because Image satisfies the Alternating Series Test, the error in approximation after n terms is less than the magnitude of the next term. The calculator shows that Image at n = 5 terms.

(c) Image is a negative series. Therefore the error will be larger than the magnitude of the first omitted term, and thus less accurate than the estimate for f (0.5).

2. See solution for AB-2.

Part B

3. See solution for AB 1.

4. (a) Using the differential equation, evaluate the derivative at each point, then sketch a short segment having that slope. For example, at (−1, −1), Image draw a segment at (−1, −1) that decreases steeply. Repeat this process at each of the other points. The result is shown below.

Image

(b) At (0, −1), Image For Δx = 0.5 and Image Δy = 0, so move to (0 + 0.5, −1 + 0) = (0.5, −1).

At (0.5, −1), Image Thus, for Δx = 0.5 and Image Δy = 1.

Move to (0.5 + 0.5, −1 + 1) = (1,0), then f (1) ≈ 0.

(c) The differential equation Image is separable:

Image

It is given that f passes through (0, −1), so −1 = tan(02 + c) and Image

The solution is Image

5. (a) To find the y-intercepts of the graph of P(t) = (9 − t2,2t), let x = 9 − t2 = 0, and solve: t = −3, 3. Then Image and P(3) = (0, 8).

Draw a horizontal element of area as shown in the graph. Then:

Image

(b) Image

(c) Use disks. Then ΔV = πx2 Δy,

Image

6. See solution for AB-6.

Answers Explained

1. (C) f (−2) = (−2)3 − 2(−2) − 1 = −5.

2. (E) The denominator, x2 + 1, is never 0.

3. (D) Since x − 2 may not be negative, x Image 2. The denominator equals 0 at x = 0 and x = 1, but these values are not in the interval x ≥ 2.

4. (E) Since g(x) = 2, g is a constant function. Thus, for all f (x), g(f (x)) = 2.

5. (D) f (g(x)) = f (2) = −3.

6. (B) Solve the pair of equations

Image

Add to get A; substitute in either equation to get B. A = 2 and B = 4.

7. (C) The graph of f (x) is symmetrical to the origin if f (−x) = −f (x). ln (C), f (−x) = (−x)3 + 2(−x) = −x3 − 2x = −(x3 + 2x) = −f (x).

8. (C) For g to have an inverse function it must be one-to-one. Note, that although the graph of y = xe−x2 is symmetric to the origin, it is not one-to-one.

9. (B) Note that Image the sine function varies from −1 to 1 as the argument varies from Image

10. (E) The maximum value of g is 2, attained when cos x = −1. On [0,2π], cos x = −1 for x = π.

11. (C) f is odd if f (−x) = −f (x). ln (C), f (−x) = (−x)3 + 1 = −x3 + 1 ≠ −f (x)

12. (B) Since f (q) = 0 if q = 1 or q = −2, f (2x) = 0 if 2x, a replacement for q, equals 1 or −2.

13. (B) f (x) = x(x2 + 4x + 4) = x(x + 2)2; f (x) = 0 for x = 0 and x = −2.

14. (E) Solving simultaneously yields (x + 2)2 = 4x; x2 + 4x + 4 = 4x; x2 + 4 = 0.

There are no real solutions.

15. (A) The reflection of y = f (x) in the y-axis is y = f (−x).

16. (B) If g is the inverse of f, then f is the inverse of g. This implies that the function f assigns to each value g(x) the number x.

17. (D) Since f is continuous, then, if f is negative at a and positive at b, f must equal 0 at some intermediate point. Since f (1) = −2 and f (2) = 13, this point is between 1 and 2.

18. (D) The function sin bx has period Image Then Image

19. (A) Since ln q is defined only if q > 0, the domain of ln cos x is the set of x for which cos x > 0, that is, when 0 < cos x Image 1. Thus − ∞ < ln cos x Image 0.

20. (E) Image implies ImageThen Image and 3 = b1/2. So 32 = b.

21. (E) Interchange x and y: x = y3 + 2.

Solve for y: Image

22. (D) Since f (1) = 0, x − 1 is a factor of f. Since f (x) divided by x − 1 yields x2x − 2, f (x) = (x − 1) (x + 1) (x − 2); the roots are x = 1,−1, and 2.

23. (B) If Image then − ∞< tan x < ∞ and 0 < etanx < ∞.

24. (A) The reflection of f (x) in the x-axis is −f (x).

25. (C) f (x) attains its maximum when Image does. The maximum value of the sine function is 1; the smallest positive occurrence is at Image Set Image equal to Image

26. (A) arccos Image

27. (A) Interchange x and y: x = 2ey

Solve for y: Image

Thus Image

28. (C) The function in (C) is not one-to-one since, for each y between Image (except 0), there are two x’s in the domain.

29. (D) The domain of the In function is the set of positive reals. The function g(x) > 0 if x2 < 9.

30. (C) Since the domain of f (g) is (−3, 3), ln (9 − x2) takes on every real value less than or equal to ln 9.

31. (A) Substituting t2 = x − 3 in y(t) = t2 + 4 yields y = x + 1.

32. (D) Using the identity Image

33. (D) 2 cos 5Image = 0 when Image

34. (C) If 2 + 2 cos Image = 3, then Image

35. (B) For polar functions x = r cos Image. Solving (Image − 2 cos Image) cos Image = 2 yields Image ≈ 5.201, and thus y = r sin Image = (5.201 − 2 cos 5.201)sin 5.201.

Answers Explained

1. (B) The limit as x → 2 is 0 ÷ 8.

2. (D) Use the Rational Function Theorem. The degrees of P(x) and Q(x) are the same.

3. (C) Remove the common factor x − 3 from numerator and denominator.

4. (A) The fraction equals 1 for all nonzero x.

5. (D) Note that Image

6. (B) Use the Rational Function Theorem.

7. (A) Use the Rational Function Theorem.

8. (E) Use the Rational Function Theorem.

9. (C) The fraction is equivalent to Image the denominator approaches ∞

10. (D) Since Image therefore, as x → −∞ the fraction → +∞

11. (D) Image

12. (B) Image

13. (B) Because the graph of y = tan x has vertical asymptotes at Image the graph of the inverse function y = arctan x has horizontal asymptotes at Image

14. (C) Since Image (provided x ≠ 3), y can be defined to be equal to 2 at x = 3, removing the discontinuity at that point.

15. (B) Note that Image

16. (C) As x → 0, Image takes on varying finite values as it increases. Since the sine function repeats, Image oscillates, taking on, infinitely many times, each value between −1 and 1. The calculator graph of Y1 = sin(1/X) exhibits this oscillating discontinuity at x = 0.

17. (A) Note that, since Image both x = 2 and Image are vertical asymptotes. Also, Image is a horizontal asymptote.

18. (B) Image Use the Rational Function Theorem.

19. (B) Since |x| = x if x > 0 but equals −x if x < 0, Image while Image

20. (E) Note that x Image can be rewritten as Image and that, as x → ∞, Image

21. (A) As x → π, (π − x) → 0.

22. (C) Since f (x) = x + 1 if x ≠ 1, Image exists (and is equal to 2).

23. (B) Image for all x ≠ 0. For f to be continuous at x = 0, Image must equal f (0). Image

24. (B) Only x = 1 and x = 2 need be checked. Since Image for x ≠ 1, 2, and Image = −3 = f (1), f is continuous at x = 1. Since Image does not exist, f is not continuous at x = 2.

25. (C) As x → ±∞, y = f (x) → 0, so the x-axis is a horizontal asymptote. Also, as x → ±1, y → ∞, so x = ±1 are vertical asymptotes.

26. (C) As x → ∞, Image the denominator (but not the numerator) of y equals 0 at x = 0 and at x = 1.

27. (D) The function is defined at 0 to be 1, which is also Image

28. (D) See Figure N2–1.

29. (E) Note, from Figure N2–1, that Image

30. (E) As x → ∞, the function sin x oscillates between −1 and 1; hence the limit does not exist.

31. (A) Note that Image if x ≠ 0 and that Image

32. (A) Image

33. (E) Verify that f is defined at x = 0, 1, 2, and 3 (as well as at all other points in [−1,3]).

34. (C) Note that Image However, f (2) = 1. Redefining f (2) as 0 removes the discontinuity.

35. (B) The function is not continuous at x = 0, 1, or 2.

36. (B) Image

37. (E) As x → 0, arctan Image

The graph has a jump discontinuity at x = 0. (Verify with a calculator.)

38. (D) No information is given about the domain of f except in the neighborhood of x = −3.

39. (E) As x → 0+, Image and therefore y → 0. As x → 0, Image → −∞, so Image and therefore Image Because the two one-sided limits are not equal, the limit does not exist. (Verify with a calculator.)

40. (A) Image but f (−1) = 2. The limit does not exist at a = 1 and f (2) does not exist.

41. (B) Image

42. (D) Image but since these two limits are not the same, Image does not exist.

Answers Explained

Many of the explanations provided include intermediate steps that would normally be reached on the way to a final algebraically simplified result. You may not need to reach the final answer.

NOTE: the formulas or rules cited in parentheses in the explanations are given.

1. (E) By the Product Rule, (5),

y = x5 (tan x) + (x5) (tan x).

2. (A) By the Quotient Rule, (6),

Image

3. (B) Since y = (3 − 2x)1/2, by the Power Rule, (3),

Image

4. (B) Since y = 2(5x + 1)−3, y = −6(5x + 1)−4 (5).

5. (E) Image

6. (D) Rewrite: Image

7. (A) Rewrite: y = (x2 + 2x − 1)1/2; then Image (x2 + 2x − 1)−1/2 (2x + 2).

8. (D) Use the Quotient Rule:

Image

9. (C) Since

Image

10. (E) Use formula (18): Image

11. (A) Use formulas (13), (11), and (9):

Image

12. (D) By the Quotient Rule,

Image

13. (D) Since Image ln (x2 + 1)

Image

14. (C) Image

15. (A) Since Image(−csc 2x cot 2x · 2).

16. (A) y = ex (−2 sin 2x) + cos 2x(−ex).

17. (C) y = (2 sec x)(sec x tan x).

18. (E) Image The correct answer is 3 ln2 x + ln3 x.

19. (B) Image

20. (C) Image

21. (D) Let y be Image then 3x2 − 3y2 y = 0; Image

22. (A) 1 − sin (x + y)(1 + y ) = 0; Image

23. (D) cos x + sin y · y′ = 0; Image

24. (B) 6x − 2(xy + y) + 10yy = 0; y (10y − 2x) = 2y − 6x.

25. (A) Image

26. (E) f (x) = 4x3 − 12x2 + 8x = 4x(x − 1)(x − 2).

27. (E) f (x) = 8x−1/2; Image

28. (A) f (x) = 3 ln x; Image Replace x by 3.

29. (D) 2x + 2yy = 0; Image

30. (E) Image Replace t by 1.

31. (D) Image

32. (D) y = ex · 1 + ex (x − 1) = xex;

y ″ = xex + ex and y ″(0) = 0 · 1 + 1 = 1.

33. (E) When simplified, Image

34. (B) Since (if sin t ≠ 0)

Image = −2 sin t = −4 sin t cos t and Image

then Image Thus:

Image

BC ONLY

NOTE: Since each of the limits in Questions 35–39 yields an indeterminate form of the type Image we can apply L’Hôpital’s Rule in each case, getting identical answers.

35. (C) The given limit is the derivative of f (x) = x6 at x = 1.

36. (B) The given limit is the definition for f (8), where f (x) = Image

Image

37. (B) The given limit is f (e), where f (x) = ln x.

38. (B) The given limit is the derivative of f (x) = cos x at x = 0; f (x) = − sin x.

39. (B) Image but f (1) = 4.

Thus f is discontinuous at x = 1, so it cannot be differentiable.

40. (E) Image so the limit exists. Because g(3) = 9, g is continuous at x = 3. Since Image

41. (E) Since Image f (0) is not defined; f (x) must be defined on (−8,8).

42. (A) Note that f (0) = Image = 0 and that f (x) exists on the given interval. By the MVT, there is a number, c, in the interval such that f (c) = 0. If c = 1, then 6c2 − 6 = 0. (−1 is not in the interval.)

43. (B) Since the inverse, h, of f (x) = Image is h(x) = Image then h (x) = Image Replace x by 3.

44. (D) After 50(!) applications of L’Hôpital’s Rule we get Image which “equals” ∞. A perfunctory examination of the limit, however, shows immediately that the answer is ∞. In fact, Image for any positive integer n, no matter how large, is ∞.

45. (C) cos(xy)(xy + y) = 1; x cos(xy)y = 1 − y cos(xy);

Image

NOTE: In Questions 46–50 the limits are all indeterminate forms of the type Image We have therefore applied L’Hôpital’s Rule in each one. The indeterminacy can also be resolved by introducing Image which approaches 1 as a approaches 0. The latter technique is presented in square brackets.

46. (B) Image

[Using sin 2x = 2 sin x cos x yields Image cos x = 2 · 1 · 1 = 2.]

47. (C) Image

[We rewrite Image As x → 0, so do 3x and 4x; the fraction approaches 1 · 1 · Image]

48. (E) Image

[We can replace 1 − cos x by Image getting

Image

49. (D) Image

[Image as x (or πx) approaches 0, the original fraction approaches π · 1 · Image = π]

50. (C) The limit is easiest to obtain here if we rewrite:

Image

51. (B) Since x − 3 = 2 sin t and y + 1 = 2 cos t,

(x − 3)2 + (y + 1)2 = 4.

This is the equation of a circle with center at (3,−1) and radius 2. In the domain given, −π ≤ t ≤ π, the entire circle is traced by a particle moving counterclockwise, starting from and returning to (3, −3).

52. (C) Use L’Hôpital’s Rule; then

Image

53. (A) Image

54. (D) Image

55. (E) Image

56. (C) Since Image

57. (B) (f + 2g) (3) = f (3) + 2g (3) = 4 + 2(−1)

58. (B) (f · g) (2) = f (2) · g (2) + g(2) · f (2) = 5(−2) + 1(3)

59. (E) Image

60. (D) Image

61. (C) Image

62. (A) M (1) = f (g(1)) · g (1) = f (3)g (1) = 4(−3).

63. (B) [f (x3)] = f (x3)·3x2, so P (1) = f (13)·3·12 = 2·3.

64. (D) f (S(x)) = x implies that f (S(x)) · S (x) = 1, so

Image

65. (E) Since g (a) exists, g is differentiable and thus continuous; g (a) > 0.

66. (C) Near a vertical asymptote the slopes must approach ±∞.

67. (A) There is only one horizontal tangent.

68. (D) Use the symmetric difference quotient; then

Image

69. (E) Since the water level rises more slowly as the cone fills, the rate of depth change is decreasing, as in (C) and (E). However, at every instant the portion of the cone containing water is similar to the entire cone; the volume is proportional to the cube of the depth of the water. The rate of change of depth (the derivative) is therefore not linear, as in (C).

70. (C) The only horizontal tangent is at x = 4. Note that f (1) does not exist.

71. (E) The graph has corners at x = 1 and x = 2; the tangent line is vertical at x = 6.

72. (B) Consider triangle ABC: AB = 1; radius AC = 2; thus, BC = Image and AC has m = −Image The tangent line is perpendicular to the radius.

Image

73. (D) The graph of y = x + cos x is shown in window [−5,5] × [−6,6]. The average rate of change is represented by the slope of secant segment Image There appear to be 3 points at which tangent lines are parallel to Image

Image

74. (C) Image

75. (A) Since an estimate of the answer for Question 74 is f (2) ≈ −5, then

Image

76. (B) When x = 3 on g−1, y = 3 on the original half-parabola. 3 = x2 − 8x + 10 at x = 1 (and at x = 7, but that value is not in the given domain).

Image

77. (E) f satisfies Rolle’s Theorem on [2,10].

78. (C) The diagrams show secant lines (whose slope is the difference quotient) with greater slopes than the tangent line. In both cases, f is concave upward.

Image

79. (C) (f ο g) at x = 3 equals f (g(3)) · g (3) equals cos u (at u = 0) times 2x (at x = 3) = 1 · 6 = 6.

80. (E) Here f (x) equals Image

81. (A) Image

82. (A) Image

83. (B) Note that f (g(x)) = Image

84. (B) Sketch the graph of f (x) = 1 − |x|; note that f (−1) = f (1) = 0 and that f is continuous on [−1,1]. Only (B) holds.

85. (C) Since f (x) = 6x2 − 3, therefore Image also, f (x), or 2x3 − 3x, equals −1, by observation, for x = 1. So h (−1) or Image (when x = 1) equals Image

86. (D) Image

87. (B) Since f (0) = 5, Image

88. (D) The given limit is the derivative of g(x) at x = 0.

89. (B) The tangent line appears to contain (3,−2.6) and (4,−1.8).

90. (D) f (x) is least at the point of inflection of the curve, at about 0.7.

91. (C) Image

92. (B) By calculator, f (0) = 1.386294805 and Image

93. (E) Now Image

94. (B) Note that any line determined by two points equidistant from the origin will necessarily be horizontal.

Image

95. (D) Note that Image f (h(x)) = f (h(x)) · h (x) = g(h(x)) · h (x) = g(sin x) · cos x.

96. (E) Since f (x) = 3xx3, then f (x) = 3x ln 3 − 3x2. Furthermore, f is continuous on [0,3] and f is differentiable on (0,3), so the MVT applies. We therefore seek c such that Image Solving 3x ln 3 − 3x2 = Image with a calculator, we find that c may be either 1.244 or 2.727. These values are the x-coordinates of points on the graph of f (x) at which the tangents are parallel to the secant through points (0,1) and (3,0) on the curve.

97. (A) The line segment passes through (1,−3) and (2,−4).

Image

Use the graph of f (x), shown above, for Questions 98–101.

98. (E) f (x) = 0 when the slope of f (x) is 0; that is, when the graph of f is a horizontal segment.

99. (E) The graph of f (x) jumps at each corner of the graph of f (x), namely, at x equal to −3, −1, 1,2, and 5.

100. (D) On the interval (−6,−3), f (x) = Image

101. (B) Verify that all choices but (B) are true. The graph of f (x) has five (not four) jump discontinuities.

102. (C) The best approximation to f (0.10) is Image

103. (D)

Image

The average rate of change is represented by the slope of secant segment Image There appear to be 3 points at which the tangent lines are parallel to Image

Answers Explained

1. (D) Substituting y = 2 yields x = 1. We find y implicitly.

3 y2 y − (2 xyy + y2) = 0; (3 y2 − 2xy)y y2 = 0.

Replace x by 1 and y by 2; solve for y .

2. (A) 2yy − (xy + y) − 3 = 0. Replace x by 0 and y by −1; solve for y .

3. (E) Find the slope of the curve at Image Image The equation is Image

4. (B) Since y = ex (1 − x) and ex > 0 for all x, y = 0 when x = 1.

5. (D) The slope y = 5x4 + 3x2 − 2. Let g = y . Since g (x) = 20x3 + 6x = 2x(10x2 + 3), g (x) = 0 only if x = 0. Since g ″ (x) = 60x2 + 6, g ″ is always positive, assuring that x = 0 yields the minimum slope. Find y when x = 0.

6. (C) Since 2x − 2yy = 0, Image At (4, 2), y = 2. The equation of the tangent at (4, 2) is y − 2 = 2(x − 4).

7. (D) Since Image the tangent is vertical for x = 2y. Substitute in the given equation and solve for y.

8. (D) Since Image therefore, dV = 4πr2 dr. The approximate increase in volume is dV ≈ 4π(32)(0.1) in3.

9. (C) Differentiating implicitly yields 4x − 3y2 y = 0. So Image The linear approximation for the true value of y when x changes from 3 to 3.04 is

Image

Since it is given that, when x = 3, y = 2, the approximate value of y is

Image

or

Image

10. (B) We want to approximate the change in area of the square when a side of length e increases by 0.01e. The answer is

A (e)(0.01e) or 2e (0.01e).

11. (D) Since V = e3, V = 3e2. Therefore at e = 10, the slope of the tangent line is 300. The change in volume is approximately 300(±0.1) = 30 in.3

12. (E) f (x) = 4x3 − 8x = 4x(x2 − 2). f = 0 if x = 0 or Image

f ″(x) = 12x2 − 8; f ″ is positive if x = Image negative if x = 0.

13. (C) Since f ″(x) = 4(3x2 − 2), it equals 0 if Image Since f ″ changes sign from positive to negative at Image and from negative to positive at Image both locate inflection points.

14. (A) The domain of y is {x | x Image 2}. Note that y is negative for each x in the domain except 2, where y = 0.

15. (B) f (x) changes sign (from negative to positive) as x passes through zero only.

16. (E) The graph must be decreasing and concave downward.

17. (B) The graph must be concave upward but decreasing.

18. (B) The distance is increasing when v is positive. Since Image = 3(t − 2)2, v > 0 for all t ≠ 2.

19. (D) The speed = |v|. From Question 18, |v| = v. The least value of v is 0.

20. (A) The acceleration Image From Question 18, a = 6(t − 2).

21. (E) The speed is decreasing when v and a have opposite signs. The answer is t < 2, since for all such t the velocity is positive while the acceleration is negative. For t > 2, both v and a are positive.

22. (B) The particle is at rest when v = 0; v = 2t(2t2 − 9t + 12) = 0 only if t = 0. Note that the discriminant of the quadratic factor (b2 − 4ac) is negative.

23. (D) Since a = 12(t − 1)(t − 2), we check the signs of a in the intervals t < 1, 1 < t < 2, and t > 2. We choose those where a > 0.

24. (A) From Questions 22 and 23 we see that v > 0 if t > 0 and that a > 0 if t < 1 or t > 2. So both v and a are positive if 0 < t < 1 or t > 2. There are no values of t for which both v and a are negative.

25. (D) See the figure, which shows the motion of the particle during the time interval −2 ≤ t ≤ 4. The particle is at rest when t = 0 or 3, but reverses direction only at 3. The endpoints need to be checked here, of course. Indeed, the maximum displacement occurs at one of those, namely, when t = −2.

Image

26. (C) Since v = 5t3 (t + 4), v = 0 when t = −4 or 0. Note that v does change sign at each of these times.

27. (E) Since Image

28. (A) Note that Image

29. (B) Image

30. (D) The slope of the curve is the slope of v, namely, Image At Image the slope is equal to

Image

31. (C) Since Image

Image

32. (B) See Figure N4–22. Replace the printed measurements of the radius and height by 10 and 20, respectively. We are given here that Image and that Image

Image

Replace h by 8.

33. (D) Image Since y = 0 if x = 1 and changes from negative to positive as x increases through 1, x = 1 yields a minimum. Evaluate y at x = 1.

34. (A) The domain of y is −∞ < x < ∞. The graph of y, which is nonnegative, is symmetric to the y-axis. The inscribed rectangle has area A = 2xex2. Thus Image which is 0 when the positive value of x is Image This value of x yields maximum area. Evaluate A.

35. (B) See the figure. If we let m be the slope of the line, then its equation is y − 2 = m(x − 1) with intercepts as indicated in the figure.

Image

The area A of the triangle is given by

Image

Then Image and equals 0 when m = ±2; m must be negative.

36. (C) Let q = (x − 6)2 + y2 be the quantity to be minimized. Then

q = (x − 6)2 + (x2 − 4);

q = 0 when x = 3. Note that it suffices to minimize the square of the distance.

37. (E) Minimize, if possible, xy, where x2 + y2 = 200 (x, y > 0). The derivative of the product is Image which equals 0 for x = 10. The derivative is positive to the left of that point and negative to the right, showing that x = 10 yields a maximum product. No minimum exists.

38. (C) Minimize Image Since

Image

q = 0 if x = 3. Since q is negative to the left of x = 3 and positive to the right, the minimum value of q occurs at x = 3.

39. (A) The best approximation for Image when h is small is the local linear (or tangent line) approximation. If we let Image then Image and Image The approximation for f (h) is f (0) + f (0) · h, which equals Image

40. (A) Since f (x) = ex (1 − x), f (0) > 0.

41. (E) The graph shown serves as a counterexample for A−D.

Image

42. (D) Since V = 10Imagew, Image = 10(8 · −4 + 6 · 2).

43. (E) We differentiate implicitly: 3x2 + x2 y + 2xy + 4y = 0. Then Image At (3, −2), Image

44. (D) Since ab > 0, a and b have the same sign; therefore f ″(x) = 12ax2 + 2b never equals 0. The curve has one horizontal tangent at x = 0.

45. (C) Since the first derivative is positive, the function must be increasing. However, the negative second derivative indicates that the rate of increase is slowing down, as seen in table C.

46. (B) Since Image therefore, at t = 1, Image Also, x = 3 and y = 2.

47. (A) Let f (x) = x1/3, and find the slope of the tangent line at (64, 4). Since Image If we move one unit to the left of 64, the tangent line will drop approximately Image unit.

48. (D) Image

49. (E) ekh Image ek·0 + kek·0 (h − 0) = 1 + kh

50. (E) Since the curve has a positive y-intercept, e > 0. Note that f (x) = 2cx + d and f ″(x) = 2c. Since the curve is concave down, f ″(x) < 0, implying that c < 0. Since the curve is decreasing at x = 0, f (0) must be negative, implying, since f (0) = d, that d < 0. Therefore c < 0, d < 0, and e> 0.

51. (A) Since the slope of the tangent to the curve is Image the slope of the normal is Image

52. (E) The slope Image at the given point Image and y = 1. The equation is therefore

y − 1 = −1(x + 2) or x + y + 1 = 0.

53. (C) Image

54. (E) Since f < 0 on 5 ≤ x < 7, the function decreases as it approaches the right endpoint.

55. (B) For x < 5, f > 0, so f is increasing; for x > 5, f is decreasing.

56. (D) The graph of f being concave downward implies that f ″ < 0, which implies that f is decreasing.

57. (D) Speed is the magnitude of velocity; at t = 3, speed = 10 ft/sec.

58. (D) Speed increases from 0 at t = 2 to 10 at t = 3; it is constant or decreasing elsewhere.

59. (E) Acceleration is positive when the velocity increases.

60. (D) Acceleration is undefined when velocity is not differentiable. Here that occurs at t = 1, 2, 3.

61. (A) Acceleration is the derivative of velocity. Since the velocity is linear, its derivative is its slope.

62. (C) Positive velocity implies motion to the right (t < 2); negative velocity (t > 2) implies motion to the left.

63. (B) The average rate of change of velocity is Image

64. (E) The slope of y = x3 is 3x2. It is equal to 3 when x = ±1. At x = 1, the equation of the tangent is

y − 1 = 3(x − 1) or y = 3x + 2.

At x = −1, the equation is

y + 1 = 3(x + 1) or y = 3x + 2.

65. (C) Let the tangent to the parabola from (3, 5) meet the curve at (x1, y1). Its equation is y − 5 = 2x1 (x − 3). Since the point (x1, y1) is on both the tangent and the parabola, we solve simultaneously:

y1 − 5 = 2x1 (x1 − 3) and Image

The points of tangency are (5, 25) and (1, 1). The slopes, which equal 2x1, are 10 and 2.

66. (E) Image

67. (D) The graph of f (x) = x sin x − cos x is drawn here in the window [0,4] × [−3,3]:

Image

A local maximum exists at x = 0, where f changes from positive to negative; use your calculator to approximate a.

68. (C) f ″ changes sign when f changes from increasing to decreasing (or vice versa). Again, use your calculator to approximate the x-coordinate at b.

69. (E) Eliminating t yields the equation Image

70. (B) Image

71. (A) Since Image We note that, as t increases through 2, the sign of |v| changes from negative to positive, assuring a minimum of |v| at t = 2. Evaluate |v| at t = 2.

72. (C) The direction of a is Image the acceleration is always directed downward. Its magnitude, Image is 2 for all t.

73. (D) Using the notations vx, vy, ax, and ay, we are given that Image where k is a constant. Then

Image

74. (E) Image

75. (B) The rate of change of the distance from the origin with respect to time is given by Image

76. (B) In parametric form, x = r cos Image = 6 cos 2Image cos Image; hence:

Image

77. (B) A local minimum exists where f changes from decreasing (f < 0) to increasing (f > 0). Note that f has local maxima at both endpoints, x = 0 and x = 5.

78. (D) See Answer 68.

79. (D) At x = a, f changes from increasing (f ″ > 0) to decreasing (f ″ < 0). Thus f changes from concave upward to concave downward, and therefore has a point of inflection at x = a. Note that f is differentiable at a (because f (a) exists) and therefore continuous at a.

80. (C) We know that Image

Image

81. (E) The equation of the tangent is y = −2x + 5. Its intercepts are Image and 5.

82. (D) See the figure. At noon, car A is at O, car B at N; the cars are shown t hours after noon. We know that Image Using s2 = x2 + y2, we get

Image

At 1 P.M., x = 30, y = 40, and s = 50.

Image

83. (B) Image (from Question 82) is zero when Image Note that x = 90 − 60t and y = 40t.

84. (B) Maximum acceleration occurs when the derivative (slope) of velocity is greatest.

85. (B) The object changes direction only when velocity changes sign. Velocity changes sign from negative to positive at t = 5.

86. (D) From the graph, f (2) = 3, and we are told the line passes through (2,10). We therefore have f (x) Image 10 + 3(x − 2) = 3x + 4.

87. (C) At x = 1 and 3, f (x) = 0; therefore f has horizontal tangents.

For x < 1, f > 0; therefore f is increasing.

For x > 1, f < 0, so f is decreasing.

For x < 2, f is decreasing, so f ″ < 0 and the graph of f is concave downward.

For x > 2, f is increasing, so f ″ > 0 and the graph of f is concave upward.

88. (C) Note that Image at Q, R, and T. At Q, Image at T, Image

89. (D) Only at S does the graph both rise and change concavity.

90. (E) Only at T is the tangent horizontal and the curve concave down.

91. (C) Since f (6) = 4, the equation of the tangent at (6, 30) is y − 30 = 4(x − 6). Therefore f (x) Image 4x + 6 and f (6.02) Image 30.08.

92. (C) Image

Answers Explained

All the references in parentheses below are to the basic integration formulas. In general, if u is a function of x, then du = u (x) dx.

1. (C) Use, first, formula (2), then (3), replacing u by x.

2. (E) Hint: Expand. Image

3. (A) By formula (3), with u = 4 − 2t and Image

Image

4. (D) Rewrite: Image

5. (E) Rewrite:

Image

Use (3).

6. (B) Rewrite:

Image

Using (3) yields Image

7. (A) This is equivalent to Image Use (4).

8. (E) Rewrite as Image Use (3).

9. (D) Use (5) with u = 3x; du = 3 dx: Image

10. (A) Use (4). If u = 1 + 4x2, du = 8x dx: Image

11. (D) Use (18). Let u = 2x; then du = 2 dx: Image

12. (C) Rewrite as Image Use (3) with n = −2.

13. (B) Rewrite as Image Use (3) with Image

Note carefully the differences in the integrands in Questions 10–13.

14. (C) Use (17); rewrite as Image

15. (B) Rewrite as Image Use (3).

Compare the integrands in Questions 14 and 15, noting the difference.

16. (A) Divide to obtain Image Use (2), (3), and (4). Remember that Image whenever k ≠ 0.

17. (E) Image

(Note the Binomial Theorem with n = 3 to expand (x − 2)3.)

18. (D) The integral is equivalent to Image Integrate term by term.

19. (A) Integrate term by term.

20. (D) Division yields

Image

21. (E) Use formula (4) with u = 1 − Image = 1 − y1/2. Then Image Note that the integral can be written as Image

22. (E) Rewrite as Image and use formula (3).

23. (B) The integral is equal to Image Use formula (6) with u = 2θ; du = 2 dθ.

24. (C) Use formula (6) with Image

25. (A) Use formula (5) with u = 4t2; du = 8t dt; Image

26. (A) Using the Half-Angle Formula (23) with α = 2x yields Image

27. (E) Use formula (6): Image

28. (B) Integrate by parts. Let u = x and dv = cos x dx. Then du = dx and v = sin x. The given integral equals Image

29. (D) Replace Image by sec2 3u; then use formula (9): Image

30. (C) Rewrite using u = 1 + sin x and du = cos x dx as Image

Use formula (3).

31. (B) The integral is equivalent to Image Use formula (12).

32. (E) Use formula (13) with Image

33. (A) Replace sin 2x by 2 sin x cos x; then the integral is equivalent to

Image

where u = 1 + cos2 x and du = −2 sin x cos x dx. Use formula (3).

34. (D) Rewriting in terms of sines and cosines yields

Image

35. (E) Use formula (7).

36. (C) Replace Image by csc2 2x and use formula (10): Image

37. (E) Let u = tan−1 y; then integrate Image The correct answer is Image

38. (A) Replacing sin 2θ by 2 sin θ cos θ yields

Image

39. (C) Image

40. (B) Rewrite as Image and use formula (8).

41. (E) Use formula (4) with u = ex − 1; du = ex dx.

42. (D) Use partial fractions; find A and B such that

Image

Then x − 1 = A(x − 2) + Bx.

Set x = 0: −1 = −2A and Image

Set x = 2: 1 = 2B and Image

So the given integral equals

Image

43. (A) Use formula (15) with u = x2; du = 2x dx; Image

44. (B) Use formula (15) with u = sin θ; du = cos θ dθ.

45. (C) Use formula (6) with u = e; du = 2e dθ: Image

46. (B) Use formula (15) with Image

47. (C) Use the Parts Formula. Let u = x, dv = e−x dx; du = dx, v = −ex. Then,

Image

48. (C) See Example 44.

49. (D) The integral is of the form Image

50. (A) The integral has the form Image Use formula (18), with u = ex, du = ex dx.

51. (C) Let u = ln v; then Image Use formula (3) for Image

52. (E) Hint: ln Image ln x; the integral is Image

53. (B) Use parts, letting u = ln x and dv = x3 dx. Then Image and Image The integral equals Image

54. (B) Use parts, letting u = ln Image and dv = dx. Then Image and v = Image. The integral equals Image ln Image

55. (B) Rewrite ln x3 as 3 ln x, and use the method of Answer 54.

56. (D) Use parts, letting u = ln y and dv = y−2 dy. Then Image and Image The Parts Formula yields Image

57. (E) The integral has the form Image where Image

58. (A) By long division, the integrand is equivalent to Image

59. (C) Image use formula (18) with u = x + 1.

60. (D) Multiply to get Image

61. (C) See Example 45. Replace x by θ.

62. (E) The integral equals Image it is equivalent to Image where u = 1 − ln t.

63. (A) Replace u by x in the given integral to avoid confusion in applying the Parts Formula. To integrate Image let the variable u in the Parts Formula be x, and let dv be sec2 x dx. Then du = dx and v = tan x, so

Image

64. (D) The integral is equivalent to Image Use formula (4) on the first integral and (18) on the second.

65. (D) The integral is equivalent to Image Use formula (17) on the first integral. Rewrite the second integral as Image and use (3).

66. (E) Rewrite: Image

67. (B) Hint: Divide, getting Image

68. (D) Letting u = sin θ yields the integral Image Use formula (18).

69. (E) Use integration by parts, letting u = arctan x and dv = dx. Then

Image

The Parts Formula yields

Image

70. (B) Hint: Note that

Image

Or multiply the integrand by Image recognizing that the correct answer is equivalent to −ln|ex − 1|.

71. (D) Hint: Expand the numerator and divide. Then integrate term by term.

72. (C) Hint: Observe that e2 ln u = u2.

73. (A) Let u = 1 + ln y2 = 1 + 2 ln |y|; integrate Image

74. (B) Hint: Expand and note that

Image

Use formulas (9) and (7).

75. (E) Multiply by Image The correct answer is tan θ − sec θ + C.

76. (D) Note the initial conditions: when t = 0, v = 0 and s = 0. Integrate twice: v = 6t2 and s = 2t3. Let t = 3.

77. (D) Since y = x2 − 2, Image Replacing x by 1 and y by −3 yields Image

78. (D) When t = 0, v = 3 and s = 2, so

v = 2t + 3t2 + 3 and s = t2 + t3 + 3t + 2.

Let t = 1.

79. (C) Let Image then

v = at + C. (*)

Since v = 75 when t = 0, therefore C = 75. Then (*) becomes

v = at + 75

so

0 = at + 75 and a = −15.

80. (A) Divide to obtain Image Use partial fractions to get

Image

Answers Explained

1. (C) The integral is equal to

Image

2. (B) Rewrite as Image This equals

Image

3. (E) Rewrite as

Image

4. (B) This integral equals

Image

5. (D) Image

6. (A) Rewrite as

Image

7. (D) Image

8. (A) Image

9. (C) Image

10. (D) You get Image = −(e−1 − 1).

11. (B) Image

12. (B) Evaluate Image which equals Image (0 − 1).

13. (E) Image

14. (C) If x = 2 sin θ, Image = 2 cos θ, dx = 2 cos θ dθ. When x = 1, Image

when x = 2, Image The integral is equivalent to Image

15. (D) Evaluate Image This equals Image

16. (A) Image

17. (C) Use the Parts Formula with u = x and dv = ex dx. Then du = dx and v = ex.

The result is Image

18. (E) Image

19. (A) Image

20. (E) Image

21. (C) Image

22. (A) Image

23. (E) Evaluate ln Image getting ln (e + 1) − ln 2.

24. (C) Note that dx = sec2 θ dθ and that Image Be sure to express the limits as values of θ: 1 = tan θ yields Image

25. (B) If Image then u2 = x + 1, and 2u du = dx. When you substitute for the limits, you get Image Since u ≠ 0 on its interval of integration, you may divide numerator and denominator by it.

26. (D) Image f (x) dx = f (0) − f (8) = 11 − 7 = 4

27. (D) On [0,6] with n = 3, Δx = 2. Heights of rectangles at x = 1, 3, and 5 are 5, 9, and 5, respectively; M(3) = (5 + 9 + 5)(2).

28. (D) Image

29. (E) For L(2) use the circumscribed rectangles: Image

for R(2) use the inscribed rectangles: Image

30. (D) On [0, 1] f (x) = cos x is decreasing, so R < L. Furthermore, f is concave downward, so T < A.

31. (D) On the interval [−1,3] the area under the graph of y = |x| is the sum of the areas of two triangles: Image

32. (E) Note that the graph y = |x + 1| is the graph of y = |x| translated one unit to the left. The area under the graph y = |x + 1| on the interval [−3,2] is the sum of the areas of two triangles: Image

33. (C) Because Image is a semicircle of radius 8, its area is 32π. The domain is [−8,8], or 16 units wide. Hence the average height of the function is Image

34. (C) The average value is equal to Image

35. (C) The average value is equal to Image

36. (E) The average value is Image The integral represents the area of a trapezoid: Image (5 + 3) · 10 = 40. The average value is Image(40).

37. (B) Since x2 + y2 = 16 is a circle, the given integral equals the area of a semicircle of radius 4.

38. (B) Use a graphing calculator.

39. (D) A vertical line at x = 2 divides the area under f into a trapezoid and a triangle; hence, Image Thus, the average value of f on [0,6] is Image There are points on f with y-values of Image in the intervals [0,2] and [2,4],

40. (B) Image

41. (D) g (x) = f (2x) · 2; thus g (1) = f (2) · 2

42. (C) Image (Why 14? See the solution for question 39.)

43. (C) This is the Mean Value Theorem for Integrals.

44. (D) This is theorem (2). Prove by counterexamples that (A), (B), (C), and (D) are false.

45. (A) This is a restatement of the Fundamental Theorem. In theorem (1), interchange t and x.

46. (D) Apply theorem (1), noting that

Image

47. (E) Let Image and u = x2; then

Image

By the Chain Rule, Image where theorem (1) is used to find Image Replace u by x2.

48. (E) Since dx = −4 sin θ dθ, you get the new integral Image Use theorem (4) to get the correct answer.

49. (C) Since dx = 2a sec2 θ dθ, you get 8πa3 Image Use the fact that cos2 θ sec2 Image = 1.

50. (D) Use the facts that dx = sin t dt, that t = 0 when x = 0, and that Image when Image

51. (E) The expression for L(5) does not multiply the heights of the rectangles by Δx = 0.2

52. (D) The average value is Image

53. (B) Image

Answers Explained

AREA

We give below, for each of Questions 1–17, a sketch of the region, and indicate a typical element of area. The area of the region is given by the definite integral. We exploit symmetry wherever possible.

1. (C)

Image

2. (C)

Image

3. (A)

Image

4. (D)

Image

5. (D)

Image

6. (C)

Image

7. (E)

Image

8. (A)

Image

9. (A)

Image

10. (D)

Image

11. (D)

Image

12. (C)

Image

13. (D)

Image

14. (A)

Image

15. (B)

Image

16. (B)

Image

17. (C)

Image

VOLUME

A sketch is given below, for each of Questions 18–27, in addition to the definite integral for each volume.

18. (E)

Image

19. (D)

Image

20. (A)

Image

21. (C)

Image

22. (D)

Image

23. (B)

Image

24. (A)

Image

25. (D)

Image

26. (B)

Image

27. (E)

Image

ARC LENGTH

28. (C) Note that the curve is symmetric to the x-axis. The arc length equals Image

29. (A) Integrate Image Replace the integrand by sec x, and use formula (13) to get Image

IMPROPER INTEGRALS

30. (A) The integral equals Image

31. (E) Image So the integral diverges to infinity.

32. (B) Redefine as Image

33. (A) Rewrite as Image Each integral converges to 3.

34. (E) Image Neither of the latter integrals converges; therefore the original integral diverges.

35. (C) Evaluate Image

36. (A)

Image

37. (C)

Image

38. (E)

Image

39. (B)

Image

40. (D)

Image

41. (B)

Image

42. (A)

Image

AREA

43. (C)

Image

44. (D)

Image

45. (E) T(4) = Image (1.62 + 2(4.15) + 2(7.5) + 2(9.0) + 12.13).

46. (B) Imagedθ = 1.571, using a graphing calculator.

47. (C) The small loop is generated as θ varies from Image (C) uses the loop’s symmetry.

VOLUME

48. (D)

Image

49. (B)

Image

50. (C)

Image

51. (B)

Image

52. (C)

Image

53. (A)

Image

54. (D)

Image

ARC LENGTH

55. (D) From (3), we obtain the length:

Image

56. (D) Note that the curve is symmetric to the x-axis. Use (2).

57. (B) Use (3) to get the integral:

Image

IMPROPER INTEGRALS

58. (C) The integrand is discontinuous at x = 1, which is on the interval of integration.

59. (D) The integral in (D) is the sum of two integrals from −1 to 0 and from 0 to 1. (see Example 29). Both diverge. Note that (A), (B), and (C) all converge.

60. (E) Choices (A), (C), and (D) can be shown convergent by the Comparison Test; the convergence of (B) is shown in Example 24.

Answers Explained

1. (D) Velocity Image and changes sign both when t = 1 and when t = 3.

2. (A) Since v > 0 for 0 Image t Image 2, the distance is equal to Image

3. (E) The answer is 8. Since the particle reverses direction when t = 2, and v > 0 for t > 2 but v < 0 for t < 2, therefore, the total distance is

Image

4. (E) Image so there is no change in position.

5. (B) Since v = sin t is positive on 0 < t Image 2, the distance covered is

Imagesin t dt = 1 − cos 2.

6. (D) Image.

7. (D) The velocity v of the car is linear since its acceleration is constant:

Image

8. (D) Image Since R(0) = 0,1,

c1 = 0 and c2 = 1.

9. (A) a = v(t) = 1,1 for all t.

10. (B) Image

11. (B) Since R = x, y, its slope is Image since Image its slope is Image

If R is perpendicular to v, then Image so

Image and x2 + y2 = k (k > 0).

Since (4, 3) is on the curve, the equation must be

x2 + y2 = 25.

12. (D) Image since Image and e° + c2 = 0; hence c1 = 2 and c2 = −1.

13. (B) The object’s position is given by Image Since the object was at the origin at Image and 4 · 1 + c2 = 0, making the position Image When t = 0, x(0) = −2, y(0) = −4.

14. (D) Image

15. (B) We want the accumulated number of people to be 100:

Image

This occurs at h = 2 hours after 8 A.M.

16. (C) Image

17. (A) Image

18. (A) The number of new people who hear the rumor during the second week is

Image

Be careful with the units! The answer is the total change, of course, in F(t) from t = 7 to t = 14 days, where F (t) = t2 + 10t.

19. (B) Total gallons = Image

20. (A) Be careful! The number of cars is to be measured over a distance of x (not 20) mi. The answer to the question is a function, not a number. Note that choice (C) gives the total number of cars on the entire 20-mi stretch.

21. (C) Since the strip of the city shown in the figure is at a distance r mi from the highway, it is Image mi long and its area is Image The strip’s population is approximately 2(12 − 2r) Image Δr. The total population of the entire city is twice the integral Image as it includes both halves of the city.

Image

22. (C)

Image

The population equals ∑ (area · density). We partition the interval [0,10] along a radius from the center of town into 5 equal subintervals each of width Δr = 2 mi. We will divide Winnipeg into 5 rings. Each has area equal to (circumference × width), so the area is 2πrk Δr or 4πrk. The population in the ring is

(4πrk)· (density at rk) = 4πrk · f (rk).

A Riemann sum, using left-hand values, is 4π · 0 · 50 + 4π · 2 · 45 + 4π · 4 · 40 + 4π · 6 · 30 + 4π · 8 · 15 = 4π(90 + 160 + 180 + 120) Image 6912 hundred people—or about 691,200 people.

23. (E) The total amount dumped after 7 weeks is

Image

24. (B) The total change in temperature of the roast 20 min after it is put in the refrigerator is

Image

Since its temperature was 160°F when placed in the refrigerator, then 20 min later it is (160 − 89.7)°F or about 70°F. Note that the temperature of the refrigerator (45°F) is not used in answering the question because it is already “built into” the cooling rate.

25. (A) Let T be the number of weeks required to release 9 tons. We can use parts to integrate Image then substitute the limits. We must then set the resulting expression equal to 9 and solve for T. A faster, less painful alternative is to use a graphing calculator to solve the equation

Image

The answer is about 10.2 weeks.

26. (D) Note that the curve is above the x-axis on [0, 1], but below on [1, 3], and that the areas for x < 0 and x > 3 are unbounded.

Image

Using the calculator, we get

Image

27. (E) The FTC yields total change:

Image

28. (C) The total change (increase) in population during the second hour is given by Image The answer is 1046.

29. (C) Call the time in hours t and the function for visitors/hour R(t). Then the area under the curve represents the number of visitors V. We will estimate the time k when Image using a Riemann sum.

The table shows one
approach, based on
the Midpoint Rule.

Hour

Visitors/hour (midpoint est.)

Total visitors since noon

Noon–1 P.M.

5

5

1−2 P.M.

25

30

2−3 P.M.

70

100

3−4 P.M.

120

220

We estimate that there had been about 100 visitors by 3 P.M. and 220 by 4 P.M., so the 200th visitor arrived between 3 and 4 P.M.

30. (B) Image

31. (C) We partition [0, 2] into n equal subintervals each of time Δt hr. Since the 18-wheeler gets (4 + 0.01v) mi/gal of diesel, it uses Image

Since it covers v · Δt mi during Δt hr, it uses Image

Since Image we see that the diesel fuel used in the first 2 hr is

Image

Answers Explained

1. (C) v(t) = 2t2t + C; v(1) = 3; so C = 2.

2. (B) If a(t) = 20t3 − 6t, then

v(t) = 5t4 − 3t2 + C1,

s(t) = t5t3 + C1 t + C2,

Since

s(−1) = −1 + 1 − C1 + C2 = 2

and

s(1) = 1 − 1 + C1 + C2 = 4,

therefore

2C2 = 6, C2, = 3,

C1 = 1.

So

v(t) = 5t4 − 3t2 + 1.

3. (D) From Answer 2, s(t) = t5t3 + t + 3, so s(0) = C2 = 3.

4. (B) Since a(t) = −32, v(t) = −32t + 40, and the height of the stone s(t) = −16t2 + 40t + C. When the stone hits the ground, 4 sec later, s(t) = 0, so

Image

5. (C) From Answer 4

s(t) = −16t2 + 40t + 96.

Then

s ′(t) = −32t + 40,

which is zero if t = 5/4, and that yields maximum height, since s ″(t) = −32.

6. (E) The velocity v(t) of the car is linear, since its acceleration is constant and

Image

7. (B) Since v = 100 − 20t, s = 100t − 10t2 + C with s(0) = 0. So s(1) = 100 − 10 = 90 ft.

8. (A) Image

9. (A) The odometer measures the total trip distance from time t = a to t = b (whether the car moves forward or backward or reverses its direction one or more times from t = a to t = b). This total distance is given exactly by Image

10. (E) (A), (B), (C), and (D) are all true. (E) is false: see Answer 9.

11. (A) Integrating yields Image + C or y2 = x2 + 2C or y2 = x2 + C , where we have replaced the arbitrary constant 2C by C ′.

12. (C) For initial point (−2,1), x2y2 = 3. Rewriting the d.e. y dy = x dx as Image reveals that the derivative does not exist when y = 0, which occurs at Image Since the particular solution must be differentiable in an interval containing x = −2, the domain is Image

13. (E) We separate variables. Image The initial point yields ln Image hence c = −2. With y > 0, the particular solution is ln Image

14. (C) We separate variables. Image The particular solution is −ey = x − 2.

15. (B) The general solution is Image when x = 4 yields C = 0.

16. (E) Since Image it follows that

ln y = ln x + C or ln y = ln x + ln k;

so y = kx.

17. (E) Image hence the general solution is y = kex, k ≠ 0.

18. (A) We rewrite and separate variables, getting Image The general solution is

Image

19. (C) We are given that Image The general solution is ln |y| = 3 ln |x| + C.

Thus, |y| = c |x3 |; y = ±c x3. Since y = 1 when x = 1, we get c = 1.

20. (E) The d.e. Image reveals that the derivative does not exist when x = 0. Since the particular solution must be differentiable in an interval containing initial value x = 1, the domain is x > 0.

21. (E) The general solution is y = k ln |x| + C, and the particular solution is y = 2 ln |x| + 2.

22. (D) We carefully(!) draw a curve for a solution to the d.e. represented by the slope field. It will be the graph of a member of the family y = sin x + C. At the right we have superimposed the graph of the particular solution y = sin x − 0.5.

Image

23. (B)

Image

It’s easy to see that the answer must be choice (A), (B), or (C), because the slope field depends only on x: all the slope segments for a given x are parallel. Also, the solution curves in the slope field are all concave up, as they are only for choices (A) and (B). Finally, the solution curves all have a minimum at x = 2, which is true only for differential equation (B).

24. (E) The solution curve is y = tan x, which we can obtain from the differential equation y ′ = 1 + y2 with the condition y(0) = 0 as follows:

Image

Since y(0) = 0, C = 0. Verify that (A) through (D) are incorrect.

NOTE: In matching slope fields and differential equations in Questions 25–29, keep in mind that if the slope segments along a vertical line are all parallel, signifying equal slopes for a fixed x, then the differential equation can be written as y ′ = f (x). Replace “vertical” by “horizontal” and “x” by “ y” in the preceding sentence to obtain a differential equation of the form y ′ = g(y).

25. (B) The slope field for y ′ = y must by II; it is the only one whose slopes are equal along a horizontal line.

26. (D) Of the four remaining slope fields, IV is the only one whose slopes are not equal along either a vertical or a horizontal line (the segments are not parallel). Its d.e. therefore cannot be either of type y ′ = f (x) or y ′ = g(y). The d.e. must be implicitly defined—that is, of the form y ′ = F(x,y). So the answer here is IV.

27. (C) The remaining slope fields, I, III, and V, all have d.e.’s of the type y ′ = f (x). The curves “lurking” in III are trigonometric curves—not so in I and V.

28. (A) Given y ′ = 2x, we immediately obtain the general solution, a family of parabolas, y = x2 + C. (Trace the parabola in I through (0, 0), for example.)

29. (E) V is the only slope field still unassigned! Furthermore, the slopes “match” ex2: the slopes are equal “about” the y-axis; slopes are very small when x is close to −2 and 2; and e−x2 is a maximum at x = 0.

30. (A) From Answer 25, we know that the d.e. for slope field II is y ′ = y. The general solution is y = cex. For a solution curve to pass through point (0, −1), we have −1 = ce0; and c = −1.

31. (C) Euler’s method for y ′ = x, starting at (1, 5), with Δx = 0.1, yields

x

y

(SLOPE) *· Δx

= Δ y

1

5

1 · (0.1)

= 0.1

*The slope is x.

1.1

5.1

(1.1) · (0.1)

= 0.11

1.2

5.21

32. (B) We want to compare the true value of y(1.2) to the estimated value of 5.21 obtained using Euler’s method in Solution 31. Solving the d.e. Image yields Image and initial condition y(1) = 5 means that Image or C = 4.5. Hence Image + 4.5 = 5.22. The error is 5.22−5.21 = 0.01.

33. (A) Slopes depend only on the value of y, and the slope field suggests that y ′ = 0 whenever y = 0 or y = −2.

34. (D) The slope field suggests that the solution function increases (or decreases) without bound as x increases, but approaches y = 1 as a horizontal asymptote as x decreases.

35. (D) We separate variables to get Image We integrate:

Image With t = 0 and s = 1, C = 0. When Image we get Image

36. (B) Since Image and ln R = ct + C. When t = 0, R = R0; so ln R0 = C or ln R = ct + ln R0. Thus

ln R − ln R0 = ct; ln Image

37. (D) The question gives rise to the differential equation Image where P = 2P0 when t = 50. We seek Image for t = 75. We get ln Image with ln 2 = 50k; then

Image

38. (A) We let S equal the amount present at time t; using S = 40 when t = 0 yields ln Image Since, when t = 2, S = 10, we get

Image

39. (A) We replace g(x) by y and then solve the equation Image We use the constraints given to find the particular solution Image

40. (C) The general solution of Image is ln y = x + C or y = cex. For a solution to pass through (2, 3), we have 3 = ce2 and c = 3/e2 Image 0.406.

41. (C) At a point of intersection, y ′ = x + y and x + y = 0. So y ′ = 0, which implies that y has a critical point at the intersection. Since y ″ = 1 + y ′ = 1 + (x + y) = 1 + 0 = 1, y ″ > 0 and the function has a local minimum at the point of intersection. [See Figure N9–5, showing the slope field for y ′= x + y and the curve y = exx − 1 that has a local minimum at (0, 0).]

42. (A) Although there is no elementary function (one made up of polynomial, trigonometric, or exponential functions or their inverses) that is an anti-derivative of F ′(x) = ex2, we know from the FTC, since F(0) = 0, that

Image

To approximate Image use your graphing calculator

For upper limits of integration x = 50 and x = 60, answers are identical to 10 decimal places. Rounding to three decimal places yields 0.886.

43. (C) Logistic growth is modeled by equations of the form Image where L is the upper limit. The graph shows L = 1000, so the differential equation must be Image Only equation C is of this form (k = 0.003).

44. (D) We start with x = 3 and y = 100. At x = 3, Image moving us to x = 3 + 2 = 5 and y = 100 + 5 = 105. From there Image so when x = 5 + 2 = 7 we estimate y = 105 + (−1) = 104.

45. (C) We separate the variables in the given d.e., then solve:

Image

Since y(0) = 180, ln 112 = c. Then

Image

When t = 10, y = 68 + 112e−1.1 Image 105°F.

46. (E) The solution of the d.e. in Question 45, where y is the temperature of the coffee at time t, is

Image

47. (D) If Q is the concentration at time t, then Image We separate variables and integrate:

Image

We let Q(0) = Q0. Then

Image

We now find t when Q = 0.1Q0:

Image

48. (E) Please read the sections on Applications for Restricted Growth and Applications for Logistic Growth in this chapter for the characteristics of the logistic model.

49. (D) (A), (B), (C), and (E) are all of the form y ′ = ky(ay).

50. (B) The rate of growth, Image is greatest when its derivative is 0 and the curve of Image is concave down. Since

Image

which is equal to 0 if Image or 500, animals. The curve of y ′ is concave down for all P, since

Image

so P = 500 is the maximum population.

51. (A) The description of temperature change here is an example of Case II: the rate of change is proportional to the amount or magnitude of the quantity present (i.e., the temperature of the corpse) minus a fixed constant (the temperature of the mortuary).

52. (C) Since (A) is the correct answer to Question 51, we solve the d.e. in (A) given the initial condition T(0) = 32:

Image

BC ONLY

Answers Explained

1. (B) Image converges to −1.

2. (C) Note that Image

3. (D) The sine function varies continuously between −1 and 1 inclusive.

4. (E) Note that Image is a sequence of the type sn = rn with |r| < 1; also that Image by repeated application of L’Hôpital’s rule.

5. (C) Image

6. (E) The harmonic series Image is a counterexample for (A), (B), and (C). Image shows that (D) does not follow.

7. (B) Image

8. (A) Image

BC ONLY

9. (B) Find counterexamples for statements (A), (C), and (D).

10. (D) Image the general term of a divergent series.

11. (D) (A), (B), (C), and (E) all converge; (D) is the divergent geometric series with r = −1.1.

12. (D) Image

13. (A) If Image then f (0) is not defined.

14. (C) Image unless x = 3.

15. (B) The integrated series is Image See Question 27.

16. (E) Image

17. (A) Image

18. (E) The series satisfies the Alternating Series Test, so the error is less than the first term dropped, namely, Image (see (5)), in the chart for Common Maclaurin Series in Chapter 10 of this eBook. So n 500.

19. (D) Note that the Taylor series for tan−1 x satisfies the Alternating Series Test and that Image then the first omitted term, Image is negative. Hence P7 (x) exceeds tan−1 x.

20. (E) Now the first omitted term, Image is positive for x < 0. Hence P9 (x) is less than tan−1 x.

21. (A) If Image converges, so does Image where m is any positive integer; but their sums are probably different.

BC ONLY

22. (E) Each series given is essentially a p-series. Only in (E) is p > 1.

23. (C) Use the Integral Test.

24. (C) The limit of the ratio for the series Image is 1, so this test fails; note for (E) that

Image

25. (B) Image does not equal 0.

26. (E) Note the following counterexamples:

(A) Image

(B) Image

(C) Image

(D) Image

27. (C) Since Image the series converges if |x| < 1. We must test the end-points: when x = 1, we get the divergent harmonic series; x = −1 yields the convergent alternating harmonic series.

28. (A) Image for all x ≠ −1; since the given series converges to 0 if x = −1, it therefore converges for all x.

29. (B) The differentiated series is Image

Image

30. (B) See Example 52.

31. (D) Note that every derivative of ex is e at x = 1. The Taylor series is in powers of (x − 1) with coefficients of the form Image

32. (D) For f (x) = cos x around Image

33. (C) Note that ln q is defined only if q > 0, and that the derivatives must exist at x = a in the formula for the Taylor series.

34. (A) Use

Image

Image Or use the series for ex and let x = −0.1.

BC ONLY

35. (C) Image

Or generate the Maclaurin series for esin x.

36. (E) (A), (B), (C), and (D) are all true statements.

37. (A) Image

38. (C) Image

Since the series converges when Image that is, when Image the radius of convergence is Image

39. (E) The Maclaurin series sin x = Image converges by the Alternating Series Test, so the error |R4 | is less than the first omitted term. For x = 1, we have Image

40. (D) Image

Answers Explained

Part A

1. (D) If f (x) = Image for x ≠ 0 and f (0) = 0 then,

Image

thus this function is continuous at 0. In (A), Image does not exist; in (B), f has a jump discontinuity; in (C), f has a removable discontinuity; and in (E), f has an infinite discontinuity.

2. (C) To find the y-intercept, let x = 0; y = − 1.

3. (A) Image

4. (D) The line x + 3y + 3 = 0 has slope Image a line perpendicular to it has slope 3.

The slope of the tangent to y = x2 − 2x + 3 at any point is the derivative 2x − 2. Set 2x − 2 equal to 3.

5. (A) Image is f′ (1), where Image Or simplify the given fraction to Image

6. (E) Because p ″(2) < 0 and p ″(5) > 0, p changes concavity somewhere in the interval [2,5], but we cannot be sure p ″ changes sign at x = 4.

7. (C) Image

Save time by finding the area under y = |x − 4| from a sketch!

Image

8. (A) Since the degrees of numerator and denominator are the same, the limit as x→∞ is the ratio of the coefficients of the terms of highest degree: Image

9. (D) On the interval [1, 4], f′ (x) = 0 only for x = 3. Since f (3) is a relative minimum, check the endpoints to find that f (4) = 6 is the absolute maximum of the function.

10. (C) To find lim f as x → 5 (if it exists), multiply f by Image

Image

and if x ≠ 5 this equals Image So lim f (x) as x → 5 is Image For f to be continuous at x = 5, f (5) or c must also equal Image

11. (D) Evaluate Image

12. (A) Image From the equation given, y = esin x.

13. (D) If f (x) = x cos x, then f ′(x) = −x sin x + cos x, and

Image

14. (C) If y = ex ln x, then Image which equals e when x = 1. Since also y = 0 when x = 1, the equation of the tangent is y = e(x − 1).

15. (B) v = 4(t − 2)3 and changes sign exactly once, when t = 2.

16. (C) Evaluate Image

17. (C) Image

18. (C) Since v = 3t2 + 3, it is always positive, while a = 6t and is positive for t > 0 but negative for t < 0. The speed therefore increases for t > 0 but decreases for t < 0.

19. (A) Note from the figure that the area, A, of a typical rectangle is

Image

Image

For y = 2, Image Note that Image is always negative.

20. (B) If S represents the square of the distance from (3, 0) to a point (x, y) on the curve, then S = (3 − x)2 + y2 = (3 − x)2 + (x2 − 1). Setting Image yields the minimum distance at Image

21. (D) Image

22. (D) See the figure. Since the area, A, of the ring equals π (y22y12),

A = π [(6xx2)2x4] = π [36x2 − 12x3 + x4x4 ]

and Image = π (72x − 36x2) = 36πx (2 − x).

It can be verified that x = 2 produces the maximum area.

Image

23. (A) This is of type Image

Image

24. (A) About the y-axis; see the figure. Washer.

Image

25. (E) Separating variables, we get y dy = (1 − 2x) dx. Integrating gives

Image

or

y2 = 2x − 2x2 + k

or

2x2 + y2 − 2x = k.

26. (E) 2(5) + Image

Image

27. (E) Image

28. (D) Use L’Hôpital’s Rule or rewrite the expression as Image

29. (D) For f (x) = tan x, this is Image

30. (E) The parameter k determines the amplitude of the sine curve. For f = k sin x and g = ex to have a common point of tangency, say at x = q, the curves must both go through (q, y) and their slopes must be equal at q. Thus, we must have

k sin q = eq and k cos q = eq,

and therefore

sin q = cos q.

Image

The figure shows Image

Image

31. (D) We differentiate implicitly to find the slope Image

Image

At (3, 1), Image The linearization is Image

32. (C) Image

33. (A) About the x-axis. Disk.

Image

34. (C) Let f (x) = ax; then Image ln a = ln a.

35. (E) Image is a function of x alone; curves appear to be asymptotic to the y-axis and to increase more slowly as |x| increases.

36. (D) The given limit is equivalent to

Image

where Image The answer is Image

37. (B) Image

38. (C) In the figure, the curve for y = ex has been superimposed on the slope field.

Image

39. (C) The general solution is y = 3 ln|x2 − 4| + C. The differential equation Image reveals that the derivative does not exist for x = ±2. The particular solution must be differentiable in an interval containing the initial value x = −1, so the domain is −2 < x < 2.

40. (A) The solution curve shown is y = ln x, so the differential equation is Image

Image

41. (D) Image = sec θ; dx = sec2 θ; 0 x 1; so 0 ≤ θ ≤ Image

42. (C) The equations may be rewritten as Image = sin u and y = 1 − 2 sin2 u,

giving Image

43. (D) Use the formula for area in polar coordinates,

Image

then the required area is given by

Image

(See polar graph 63 in the Appendix.)

44. (C) Image

45. (A) The first three derivatives of Image The first four terms of the Maclaurin series (about x = 0) are 1, + 2x, Image

Note also that Image represents the sum of an infinite geometric series with first term 1 and common ratio 2x. Hence,

Image

46. (D) We use parts, first letting u = x2, dv = e−x dx; then du = 2x dx, v = −e−x and

Image

Now we use parts again, letting u = x, dv = ex dx; then du = dx, v = −ex and

Image

Alternatively, we could use the Tic-Tac-Toe Method:

Image

Then Image

47. (E) Use formula (20) in the Appendix to rewrite the integral as

Image

48. (E) The area, A, is represented by Image

49. (D) Image

50. (C) Check to verify that each of the other improper integrals converges.

51. (D) Note that the integral is improper.

Image

See Example 26.

52. (C) Let Image Then ln y = −x ln x and

Image

Now apply L’Hôpital’s Rule:

Image

So, if Image ln y = 0, then Image

53. (D) The speed, |v|, equals Image and since x = 3yy2,

Image

Then |v| is evaluated, using y = 1, and equals Image

54. (A) This is an indeterminate form of type Image use L’Hôpital’s Rule:

Image

55. (E) We find A and B such that Image

After multiplying by the common denominator, we have

3x + 2 = A(x − 4) + B(x + 3).

Substituting x = −3 yields A = 1, and x = 4 yields B = 2; hence,

Image

56. (B) Since Image

Then Image

Note that Image so the integral is proper.

57. (D) We represent the spiral as P(θ) = (θ cos θ, θ sin θ). So

Image

Part B

58. (D) Since h is increasing, h ′ ≥ 0. The graph of h is concave downward for x < 2 and upward for x > 2, so h ″ changes sign at x = 2, where it appears that h′ = 0 also.

59. (C) I is false since, for example, f ′(−2) = f ′(1) = 0 but neither g(−2) nor g(1) equals zero.

II is true. Note that f = 0 where g has relative extrema, and f is positive, negative, then positive on intervals where g increases, decreases, then increases.

III is also true. Check the concavity of g: when the curve is concave down, h < 0; when up, h > 0.

60. (A) If Image

61. (D) Image represents the area of the same region as Image translated one unit to the left.

62. (D) According to the Mean Value Theorem, there exists a number c in the interval [1,1.5] such that Image Use your calculator to solve the equation Image for c (in radians).

63. (E) Here are the relevant sign lines:

Image

We see that f ′ and f ″ are both positive only if x > 1.

64. (E) Note from the sign lines in Question 63 that f changes from decreasing to increasing at x = 1, so f has a local minimum there.

Also, the graph of f changes from concave up to concave down at x = 0, then back to concave up at Image hence f has two points of inflection.

65. (C) The derivatives of ln (x + 1) are Image

The nth derivative at Image

66. (C) The absolute-value function f (x) = |x| is continuous at x = 0, but f ′(0) does not exist.

67. (C) Let F ′(x) = f (x); then F ′(x + k) = f (x + k);

Image

Or let u = x + k. Then dx = du, when x = 0, u = k; when x = 3, u = 3 + k.

68. (E) See the figure. The equation of the generating circle is (x − 3)2 + y2 = 1, which yields Image

Image

Image

69. (D) Note that f (g(u)) = tan−1 (e2u); then the derivative is Image

70. (D) Let Image Then cos (xy) [xy′ + y] = y . Solve for y .

71. (E) Image

Image

72. (C) About the x-axis; see the figure. Washer.

Image

73. (C) By the Mean Value Theorem, there is a number c in [1, 2] such that

Image

74. (D) The enclosed region, S, is bounded by y = sec x, the y-axis, and y = 4. It is to be rotated about the y-axis.

Image

Use disks; then ΔV = πR2 H = π(arc sec y)2 Δy. Using the calculator, we find that

Image

75. (C) If Q is the amount at time t. then Q = 40ekt. Since Q = 20 when t = 2, k = −0.3466. Now find Q when t = 3, from Q = 40e−(0.3466) 3, getting Q = 14 to the nearest gram.

76. (A) The velocity v(t) is an antiderivative of a(t), where Image So Image arctan t + C. Since v (1) = 0, C = − π.

Image

77. (D) Graph y = tan x and y = 2 − x in [−1, 3] × [−1, 3]. Note that

Image

The limits are y = 0 and y = b, where b is the ordinate of the intersection of the curve and the line. Using the calculator, solve

arctan y = 2 − y

and store the answer in memory as B. Evaluate the desired area:

Image

Image

78. (E) Center the ellipse at the origin and let (x, y) be the coordinates of the vertex of the inscribed rectangle in the first quadrant, as shown in the figure.

Image

To maximize the rectangle’s area A = 4xy, solve the equation of the ellipse, getting

Image

So Image Graph Image in the window [0,5] × [0,150],

The calculator shows that the maximum area (the y-coordinate) equals 100.

Image

79. (B) Image

80. (B) When f ′ is positive, f increases. By the Fundamental Theorem of Calculus, f′ (x) = 1 − 2 (cos x)3. Graph f ′ in [0, 2π] × [−2, 4]. It is clear that f ′ > 0 on the interval a < x < b. Using the calculator to solve 1 − 2(cos x)3 = 0 yields a = 0.654 and b = 5.629.

Image

81. (C) Image

82. (B) The volume is composed of elements of the form ΔV = (2x)2 Δy. If h is the depth, in feet, then, after t hr,

Image

Image

83. (B) Separating variables yields

Image

P(0) = 300 gives c = 700. P(5) = 500 yields 500 = 1000 − 700e−5k, so k Image + 0.0673. Now P(10) = 1000 − 700e−0.673 Image 643.

84. (C) Image dx = 4 arctan 1 = π. H′ (1) = f (1) = 2.

The equation of the tangent line is y − π = 2(x − 1).

85. (C) Using midpoint diameters to determine cylinders, estimate the volume to be

V Image π · 82 · 25 + π · 62 · 25 + π · 42 · 25 + π · 32 · 25.

86. (A) Image

87. (C) H′ (3) = f′ (g(3)) · g′ (3) = f′ (2) · g′ (3).

88. (E) M′ (3) = f (3) · g′ (3) + g(3) · f′ (3) = 4 · 3 + 2 · 2.

89. (E) Image

90. (C) Image

91. (D) Here are the pertinent curves, with d denoting the depth of the water:

Image

92. (B) Use areas; then Image Thus, f (7) − f (1) = 7.

93. (B) The region x units from the stage can be approximated by the semicircular ring shown; its area is then the product of its circumference and its width.

Image

The number of people standing in the region is the product of the area and the density:

Image

To find the total number of people, evaluate

Image

94. (B) Image is positive, but decreasing; hence Image

95. (C) Image

96. (E) On 2 t 5, the object moved Image ft to the right; then on 5 t 8, it moved only Image ft to the left.

97. (B) Image

98. (D) Evaluate Image

99. (A) Image

Use, also, the facts that the speed is given by Image and that the point moves counterclockwise; then Image = 4, yielding Image and Image at the given point. The velocity vector, v, at (3, 4) must therefore be Image

100. (A) Image

101. (B) The formula for length of arc is

Image

Since y = 2x, we find

Image

102. (D) a(t) = (0, et); the acceleration is always upward.

103. (A) At (0, 1), Image so Euler’s method yields (0.1, 1 + 0.1(4)) = (0.1, 1.4). Image has particular solution y = e4x; the error is e4(0.1) − 1.4.

104. (D) Image Note that the series converges by the Alternating Series Test. Since the first term dropped in the estimate is Image the estimate is too high, but within 0.1 of the true sum.

105. (C) Image which equals a constant times the harmonic series.

106. (D) We seek x such that

Image

Then |x| < e and the radius of convergence is e.

107. (B) The error is less than the maximum value of Image

This maximum occurs at c = x = Image

108. (D) Image

Note that the curve is traced exactly once by the parametric equations from t = 0 to t = 1.

Answers Explained

Part A

1. (a) Image

(b) It appears that the rate of change of f, while negative, is increasing. This implies that the graph of f is concave upward.

(c) L = 7.6(0.7) + 5.7(0.3) + 4.2(0.5) + 3.1(0.6) + 2.2(0.4) = 11.87.

(d) Using disks ΔV = πr2 Δx. One possible answer uses the left endpoints of the subintervals as values of r:

V ≈ π(7.6)2 (0.7) + π(5.7)2 (0.3) + π(4.2)2 (0.5) + π(3.1)2 (0.6) + π(2.2)2 (0.4)

2. (a) 12y0 + 0.3 = 24 yields y0 Image 1.975.

(b) Replace x by 0.3 in the equation of the curve:

Image

The calculator’s solution to three decimal places is y0 = 1.990.

(c) Since the true value of y0 at x = 0.3 exceeds the approximation, conclude that the given curve is concave up near x = 0. (Therefore, it is above the line tangent at x = 0.)

3. Graph f ′(x) = 2x sin xe(−x2) + 1 in [−7, 7] × [−10, 10].

Image

(a) Since f ′ is even and f contains (0, 0), f is odd and its graph is symmetric about the origin.

(b) Since f is decreasing when f ′ < 0, f decreases on the intervals (a, c) and (j, 1). Use the calculator to solve f′ (x) = 0. Conclude that f decreases on −6.202 < x < −3.294 and (symmetrically) on 3.294 < x < 6.202.

(c) f has a relative maximum at x = q if f ′(q) = 0 and if f changes from increasing (f ′ > 0) to decreasing (f ′ < 0) at q. There are two relative maxima here:
at x = a = −6.202 and at x = j = 3.294.

(d) f has a point of inflection when the graph of f changes its concavity; that is, when f′ changes from increasing to decreasing, as it does at points d and h, or when f′ changes from decreasing to increasing, as it does at points b, g, and k. So there are five points of inflection altogether.

4. In the graph below, C is the piece of the curve lying in the first quadrant. S is the region bounded by the curve C and the coordinate axes.

Image

(a) Graph Image in [0,3] × [0,5]. Since you want dy/dx, the slope of the tangent, where y = 1, use the calculator to solve

Image

(storing the answer at B). Then evaluate the slope of the tangent to C at y = 1:

f ′(B) ≈ −21.182.

(b) Since ΔA = yΔx, A = Image

(c) When S is rotated about the x-axis, its volume can be obtained using disks:

Image

5. See the figure, where R is the point (a, b), and seek a such that

Image

Image

6. Graph y = sin 2x in [−1, 3.2] × [−1, 1]. Note that y = f .

Image

(a) The graph of f is concave downward where f ″ is negative, namely, on (b, d). Use the calculator to solve sin 2x = 0, obtaining b = 1.651 and d = 2.651. The answer to (a) is therefore 1.651 < x < 2.651.

(b) f ′ has a relative minimum at x = d, because f ″ equals 0 at d, is less than 0 on (b, d), and is greater than 0 on (d, g). Thus f ′ has a relative minimum (from part a) at x = 2.651.

(c) The graph of f ′ has a point of inflection wherever its second derivative f ″′ changes from positive to negative or vice versa. This is equivalent to f ″ changing from increasing to decreasing (as at a and g) or vice versa (as at c). Therefore, the graph of f ′ has three points of inflection on [− 1, 3.2].

7. Graph f (x) = cos x and g(x) = x2 − 1 in [−2, 2] × [−2,2], Here, y1 = f and y2 = g.

Image

(a) Solve cos x = x2 − 1 to find the two points of intersection: (1.177,0.384) and (−1.177,0.384).

(b) Since ΔA = (y1 − y2) Δx = [f (x) − g(x)] Δx, the area A bounded by the two curves is

Image

8. (a) Use the Trapezoid Rule, with h = 60 min:

Image

(b) Draw a horizontal line at y = 20 (as shown on the graph below), representing the rate at which letters are processed then.

Image

(i) Letters began to pile up when they arrived at a rate greater than that at which they were being processed, that is, at t = 10 A.M.

(ii) The pile was largest when the letters stopped piling up, at t = 2 P.M.

(iii) The number of letters in the pile is represented by the area of the small trapezoid above the horizontal line: Image

(iv) The pile began to diminish after 2 P.M., when letters were processed at a rate faster than they arrived, and vanished when the area of the shaded triangle represented 1500 letters. At 5 P.M. this area is Image letters, so the pile vanished shortly after 5 P.M.

9. Draw a vertical element of area as shown below.

Image

(a) Let a represent the x-value of the positive point of intersection of y = x4 and y = sin x. Solving a4 = sin a with the calculator, we find a = 0.9496.

Image

(b) Elements of volume are triangular prisms with height h = 3 and base b = (sinxx4), as shown below.

Image

(c) When R is rotated around the x-axis, the element generates washers. If r1, and r2 are the radii of the larger and smaller disks, respectively, then

Image

10. The figure above shows an elliptical cross section of the tank. Its equation is

Image

(a) The volume of the tank, using disks, is Image where the ellipse’s symmetry about the x-axis has been exploited. The equation of the ellipse is equivalent to x2 = 6.25(100 − y2), so

Image

Use the calculator to evaluate this integral, storing the answer as V to have it available for part (b).

The capacity of the tank is 7.48V, or 196,000 gal of water, rounded to the nearest 1000 gal.

(b) Let k be the y-coordinate of the water level when the tank is one-fourth full. Then

Image

and the depth of the water is k + 10.

11. (a) Let h represent the depth of the water, as shown.

Image

Then h is the altitude of an equilateral triangle, and the base Image

The volume of water is

Image

Now Image and it is given that Image Thus, when h = 4,

Image

(b) Let x represent the length of one of the sides, as shown.

The bases of the trapezoid are 24 − 2x and 24 − 2x + Image and the height is Image

Image

The volume of the trough (in in.3) is given by

Image

Since Image the maximum volume is attained by folding the metal 8 inches from tne edges.

12. (a) Both π/4 and the expression in brackets yield 0.7853981634, which is accurate to ten decimal places.

(b) Image

(c) Image this agrees with the value of Image to four decimal places.

(d) The series

Image

converges very slowly. Example 56 evaluated the sum of 60 terms of the series for π (which equals 4 tan−1 1). To four decimal places, we get π = 3.1249, which yields 0.7812 for π/4—not accurate even to two decimal places.

13. (a) The given series is alternating. Since Image

Since ln x is an increasing function,

Image

The series therefore converges.

(b) Since the series converges by the Alternating Series Test, the error in using the first n terms for the sum of the whole series is less than the absolute value of the (n + 1)st term. Thus the error is less than Image Solve for n using Image

Image

The given series converges very slowly!

(c) The series Image is conditionally convergent. The given alternating series converges since the nth term approaches 0 and Image However, the nonnegative series diverges by the Integral Test, since

Image

14. (a) Solve by separation of variables:

Image

Let c = e−10c; then

Image

Now use initial condition y = 2 at t = 0:

Image

and the other condition, y = 5 at t = 2, gives

Image

(b) Since c = 4 and Image

Solving for y yields Image

(c) Image means 1 + 4 · 2t = 1.25, so t = 4.

(d) Image so the value of y approaches 10.

15. (a) Since Image Since y = 18 − 2 · 22 = 10, P is at (2,10).

(b) Since Image Since Image Therefore

Image

(c) Let D = the object’s distance from the origin. Then

Image

(d) The object hits the x-axis when y = 18 − 2x2 = 0, or x = 3. Since Image

(e) The length of the arc of y = 18 − 2x2 for 0 ≤ x ≤ 3 is given by

Image

16. (a) See graph.

Image

(b) You want to maximize Image

Image

See signs analysis.

Image

The maximum y occurs when t = 1, because y changes from increasing to decreasing there.

(c) Since x(1) = 4arctan 1 = π and Image the coordinates of the highest point are (π,6).

Since Image This vector is shown on the graph.

(d) Image Thus the particle approaches the point (2π,0).

17. (a) To find the smallest rectangle with sides parallel to the x- and y-axes, you need a rectangle formed by vertical and horizontal tangents as shown in the figure. The vertical tangents are at the x-intercepts, x = ±3. The horizontal tangents are at the points where y (not r) is a maximum. You need, therefore, to maximize

Image

Use the calculator to find that Image when θ = 0.7854. Therefore, y = 1.414, so the desired rectangle has dimensions 6 × 2.828.

(b) Since the polar formula for the area is Image the area of R (enclosed by r) is Image which is 14.137.

Part B

18. The graph shown below satisfies all five conditions. So do many others!

Image

19. (a) f′ is defined for all x in the interval. Since f is therefore differentiable, it must also be continuous.

(b) Because f′ (2) = 0 and f′ changes from negative to positive there, f has a relative minimum at x = 2. To the left of x = 9, f′ is negative, so f is decreasing as it approaches that endpoint and reaches another relative minimum there.

(c) Because f′ is negative to the right of x = −3, f decreases from its left end-point, indicating a relative max there. Also, f′ (2) = 0 and f′ changes from positive to negative there, so f has a relative minimum at x = 7.

(d) Note that f (7) − f (−3) = Image Since there is more area above the x-axis than below the x-axis on [−3,7], the integral is positive and f (7) − f (−3) > 0. This implies that f (7) > f (−3), and that the absolute maximum occurs at x = 7.

(e) At x = 2 and also at x = 6, f′ changes from increasing to decreasing, indicating that f changes from concave upward to concave downward at each. At x = 4, f′ changes from decreasing to increasing, indicating that f changes from concave downward to concave upward there. Hence the graph of f has points of inflection at x = 2, 4, and 6.

20. Draw a sketch of the region bounded above by y1 = 8 − 2x2 and below by y2 = x2 − 4, and inscribe a rectangle in this region as described in the question. If (x, y1) and (x, y2) are the vertices of the rectangle in quadrants I and IV, respectively, then the area

A = 2x (y1y2) = 2x(12 − 3x2), or A(x) = 24x − 6x3.

Then A′ (x) = 24 − 18x2 = 6(4 − 3x2), which equals 0 when Image Check to verify that A (x) < 0 at this point. This assures that this value of x yields maximum area, which is given by Image

21. The graph of f ′(x) is shown here.

Image

22. The rate of change in volume when the surface area is 54 ft3 is Image ft3 /sec.

23. See the figure. The equation of the circle is x2 + y2 = a2; the equation of RS is y = ax. If y2 is an ordinate of the circle and y1 of the line, then,

Image

24. (a) The region is sketched in the figure. The pertinent points of intersection are labeled.

Image

(b) The required area consists of two parts. The area of the triangle is represented by Image and is equal to 1, while the area of the region bounded at the left by x = 1, above by y = x + 4, and at the right by the parabola is represented by Image This equals

Image

The required area, thus, equals Image

25. (a) 1975 to 1976 and 1978 to 1980.

(b) 1975 to 1977 and 1979 to 1981.

(c) 1976 to 1977 and 1980 to 1981.

26. (a) Since Image then, separating variables, Image Integrating gives

Image

and, since v = 20 when t = 0, C = ln 20. Then (1) becomes ln Imageor, solving for v,

Image

(b) Note that v > 0 for all t. Let s be the required distance traveled (as v decreases from 20 to 5); then

Image

where, when v = 20, t = 0. Also, when v = 5, use (2) to get Image or − ln 4 = −2t. So t = ln 2. Evaluating s in (3) gives

Image

27. Let (x,y) be the point in the first quadrant where the line parallel to the x-axis meets the parabola. The area of the triangle is given by

A = xy = x(27 − x2) = 27xx3 for 0 ≤ xImage

Then A′ (x) = 27 − 3x2 = 3(3 + x)(3 − x), and A′ (x) = 0 at x = 3.

Since A′ changes from positive to negative at x = 3, the area reaches its maximum there.

The maximum area is A(3) = 3(27 − 32) = 54.

28. Image

(b) Using the Ratio Test, you know that the series converges when Image that is, when |x| < 1, or −1 < x < 1. Thus, the radius of convergence is 1.

(c) Image

(d) Since the series converges by the Alternating Series Test, the error in the answer for (c) is less absolutely than Image

29. From the equations for x and y,

dx = (1 − cos θ) dθ and dy = sin θ dθ.

(a) The slope at any point is given by Image which here is Image When Image

(b) When Image The equation of the tangent is Image

30. Both curves are circles with centers at, respectively, (2,0) and Image the circles intersect at Image The common area is given by

Image

The answer is 2(π − 2).

31. (a) For f (x) = cos x, f′ (x) = −sin x, f (x) = −cos x, f ′″(x) = sin x, f(4) (x) = cos x, f(5) (x) = −sin x, f(6) (x) = −cos x. The Taylor polynomial of order 4 about 0 is

Image

Note that the next term of the alternating Maclaurin series for cos x is Image

(b) Image

(c) The error in (b), convergent by the Alternating Series Test, is less absolutely than the first term dropped:

Image

32. (a) Since Image y = 2t + 1 and x = 4t3 + 6t + 3t.

(b) Since Image then, when t = 1, |a| = 36.

33. See the figure. The required area A is twice the sum of the following areas: that of the limaçon from 0 to Image and that of the circle from Image Thus

Image

Answers Explained

Multiple-Choice

Part A

1. (C) Use the Rational Function Theorem.

2. (C) Note that Image where f (x) = ln x.

3. (B) Since y ′ = −2xe−x2, therefore y ″ = −2(x · e x2 · (−2x) + e x2). Replace x by 0.

4. (B) Image

5. (B) h ′(3) = g ′(f (3)) · f ′(3) = g ′(4) · Image

6. (E) Since f ′(x) exists for all x, it must equal 0 for any x0 for which f is a relative maximum, and it must also change sign from positive to negative as x increases through x0. For the given derivative, no x satisfies both of these conditions.

7. (E) By the Quotient Rule (formula (6)),

Image

8. (A) Here, f ′(x) is e x (1 − x); f has maximum value when x = 1.

9. (A) Note that (1) on a horizontal line the slope segments are all parallel, so the slopes there are all the same and Image must depend only on y; (2) along the x-axis (where y = 0) the slopes are infinite; and (3) as y increases, the slope decreases.

10. (E) Acceleration is the derivative (the slope) of velocity v; v is largest on 8 < t < 9.

11. (C) Velocity v is the derivative of position; because v > 0 until t = 6 and v < 0 thereafter, the position increases until t = 6 and then decreases; since the area bounded by the curve above the axis is larger than the area below the axis, the object is farthest from its starting point at t = 6.

12. (D) From t = 5 to t = 8, the displacement (the integral of velocity) can be found by evaluating definite integrals based on the areas of two triangles:

Image Thus, if K is the object’s position at t = 5, then K − 3 = 10 at t = 8.

13. (A) The integral is of the form Image evaluate Image

14. (E) Image

15. (D) Image

16. (A) f (x) = e x is decreasing and concave upward.

17. (B) Implicit differentiation yields 2yy ′ = 1; so Image At a vertical tangent, Image is undefined; y must therefore equal 0 and the numerator be non-zero. The original equation with y = 0 is 0 = xx3, which has three solutions.

18. (B) Let t = x − 1; then t = −1 when x = 0, t = 5 when x = 6, and dt = dx.

19. (B) The required area, A, is given by the integral

Image

20. (B) The average value is Image The definite integral represents the sum of the areas of a trapezoid and a rectangle: Image (8 + 3)(6) = 4(7) = 61.

21. (A) Solve the differential equation Image by separation of variables: Image yields y = ce2x. The initial condition yields 1 = ce2 · 2; so c = e −4 and y = e2x−4.

22. (C) Changes in values of f ″ show that f ″′ is constant; hence f ″ is linear, f ′ is quadratic, and f must be cubic.

23. (B) By implicit differentiation, Image so the equation of the tangent line at (3,0) is y = −9(x−3).

24. (A) Image

25. (D) The graph shown has the x-axis as a horizontal asymptote.

26. (B) Since Image to render f (x) continuous at x = 1 f (1) must be defined to be 1.

27. (B) f ′(x) = 15x4 − 30x2; f ″(x) = 60x3 − 60x = 60x(x + 1)(x − 1); this equals 0 when x = −1, 0, or 1. Here are the signs within the intervals:

Image

The graph of f changes concavity at x = −1, 0, and 1.

28. (C) Note that Image so f has a critical value at x = −4. As x passes through −4, the sign of f ′ changes from − to +, so f has a local minimum at x = −4.

Part B

29. (B) We are given that (1) f ′(a) > 0; (2)f ″(a) < 0; and (3) G ′(a) < 0. Since G ′(x) = 2 f (x) · f ′(x), therefore G ′(a) = 2f (a) · f (a). Conditions (1) and (3) imply that (4)f (a) < 0. Since G ″(x) = 2[f (x) · f ″ (x) + (f ′(x))2], therefore G″(a) = 2[f (a)f ″ (a) + (f′ (a))2]. Then the sign of G ″(a) is 2[(−) · (−) + (+)] or positive, where the minus signs in the parentheses follow from conditions (4) and (2).

30. (E) Since Image it equals 0 for Image When x = 3, c = 9; this yields a minimum since f ″(3) > 0.

Image

31. (E) Use your calculator to graph velocity against time. Speed is the absolute value of velocity. The greatest deviation from v = 0 is at t = c. With a calculator, c = 9.538.

32. (D) Because f ′ changes from increasing to decreasing, f ″ changes from positive to negative and thus the graph of f changes concavity.

33. (D) Image We evaluate this definite integral by finding the area of a trapezoid (negative) and a triangle: Image so the tangent line passes through the point (3,−2). The slope of the line is H ′(3) = f (3) = 2, so an equation of the line is y − (−2) = 2(x − 3).

34. (D) The distance is approximately 14(6) + 8(2) + 3(4).

35. (D) Image R(x)dx = 166.396.

36. (A) Selecting an answer for this question from your calculator graph is unwise. In some windows the graph may appear continuous; in others there may seem to be cusps, or a vertical asymptote. Put the calculator aside. Find

Image

These limits indicate the presence of a jump discontinuity in the function at x = 1.

37. (D) Image

Image

38. (E) Image

Image

39. (E) In the figure above, S is the region bounded by y = sec x, the y axis, and y = 4. Send region S about the x-axis. Use washers; then ΔV = π(R2r2) Δx. Symmetry allows you to double the volume generated by the first quadrant of S, so V is

Image

A calculator yields 108.177.

40. (A) The curve falls when f ′(x) < 0 and is concave up when f ″(x) > 0.

41. (B) Image To find g ′(0), find x such that f (x) = 0. By inspection, Image

Image

42. (D) It is given that Image you want Image where Image

Image

Since y2 = 25 −x2, it follows that Image and, when x = 3, y = 4

Image

Image

The function f (x) = 2 sin x + sin 4x is graphed above.

43. (E) Since f (0) = f (π) and f is both continuous and differentiable, Rolle’s Theorem predicts at least one c in the interval such that f ′(c) = 0.

There are four points in [0,π] of the calculator graph above where the tangent is horizontal.

44. (B) Since Image a positive constant, Image where c is a positive constant. Then Image which is also positive.

45. (C) If Q(t) is the amount of contaminant in the tank at time t and Q0 is the initial amount, then

Image = kQ and Q(t) = Q0 ekt.

Since Q(1) = 0.8Q0, 0.8Q0 = Q0ek · 1, 0.8 = ek, and

Q(t) = Q0(0.8)t.

We seek t when Q(t) = 0.02Q0. Thus,

0.02Q0 = Q0(0.8)t

and

t Image 17.53 min.

Free-Response

Part A

AB/BC 1. (a) This is the graph of f ′(x).

Image

(b) f is increasing when f ′(x) > 0. The graph shows this to be true in the interval a < x < b. Use the calculator to find a and b (where ex − 2 cos 3x = 0); then a = 0.283 < x < 3.760 = b.

(c) See signs analysis.

Image

Since f decreases to the right of endpoint x = 0, f has a local maximum at x = 0. There is another local maximum at x = 3.760, because f changes from increasing to decreasing there.

(d) See signs analysis.

Image

Since the graph of f changes concavity at p, q, and r, there are three points of inflection.

AB2. (a) Since April 1 is 3 months from January 1 and June 30 is 3 months later, we form the sum for the interval [3,6]:

Image

We estimate the company sold 1051 software units during the second quarter.

(b) S(t) = 1.2(2)t/3

(c) Image The model’s estimate of 1039 sales is slightly lower, but the two are in close agreement.

(d) Image the model predicts an average sales rate of 649.2 units per month from January 1, 2012, through December 31, 2012.

Part B

AB 3. (a) At (2,5), Image so the tangent line is y − 5 = 4(x − 2).

Solving for y yields f (x) ≈ 5 + 4(x − 2).

(b) f (2.1) ≈ 5 + 4(2.1 − 2) = 5.4.

(c) The differential equation Image is separable:

Image

Since f passes through (2,5), it must be true that Image

Thus c = 9, and the positive root is used.

The solution is Image

(d) Image

AB 4. (a) Draw a vertical element of area, as shown.

Image

Image

(b) (i) Use washers; then

Image

(ii) See the figure above.

Image

AB/BC 5. Image

f is decreasing where f ′(x) < 0, which occurs for −2 < x < 1.

(b) f is decreasing on the interval −2 < x < 1, so there is a minimum at (1, −e2). Note that, as x approaches ±∞, f (x) = e2x(x2 − 2) is always positive. Hence (1,−e2) is the global minimum.

(c) As x approaches +∞, f (x) = e2x(x2 − 2) also approaches +∞. There is no global maximum.

AB/BC 6. (a) S = 4πr2, so Image Substitute given values; then

Image

Since Image therefore Image Substituting known values

gives Image

(b) Regions of consistent density are concentric spherical shells. The volume of each shell is approximated by its surface area (4πx2) times its thickness (Δx). The weight of each shell is its density times its volume (g/cm3 · cm3). If, when the snowball is 12 cm in diameter, ΔG is the weight of a spherical shell x cm from the center, then Image and the integral to find the weight of the snowball is

Image

Answers Explained

Multiple-Choice

Part A

1. (E) Image

2. (A) Divide both numerator and denominator by Image

3. (E) Since eln u = u, y = 1.

4. (D) f (0) = 3, and, Image

5. (B) Image

6. (D) Here y ′ = 3 sin2 (1 − 2x) cos (1 − 2x) · (−2).

7. (B) Image

8. (B) Let s be the distance from the origin: then

Image

Since Image

9. (B) For Image this limit represents f ′(25).

10. (B) Image This definite integral represents the area of a quadrant of the circle x2 + y2 = 1, hence Image

11. (C) Image

12. (A) The integral is rewritten as

Image

13. (B) Image

14. (D) Note:

Image

15. (B) The winning times are positive, decreasing, and concave upward.

16. (E) G(x) = H(x) + Image represents the area of a trapezoid.

17. (C) f ′(x) = 0 for x = 1 and f ″(1) > 0.

Image

18. (B) Solution curves appear to represent odd functions with a horizontal asymptote. In the figure above, the curve in (B) of the question has been superimposed on the slope field.

19. (B) Note that

Image

20. (C) v is not differentiable at t = 3 or t = 5.

Image

21. (B) Speed is the magnitude of velocity; its graph is shown in the answer explanation for question 20.

22. (B) The average rate of change of velocity is Image

23. (E) The curve has vertical asymptotes at x = 2 and x = −2 and a horizontal asymptote at y = −2.

24. (E) The function is not defined at x = −2; Image Defining f (−2) = 4 will make f continuous at x = −2, but f will still be discontinuous at x = 1.

25. (B) Image

26. (A) Image

27. (E) ln (4 + x2) = ln (4 + (−x)2); y Image

28. (A) f (x) = Image (x sin πx) = πx cos πx + sin πx.

Part B

29. (B) See the figure below. A = Image

Image

30. (C) See the figure below. About the x-axis: Washer. ΔV = π(y2 − 12) Δx,

Image

31. (E) We solve the differential equation Image by separation:

Image

If s = 1 when t = 0, we have C = 1; hence, Image when t = 1.

32. (D) Image

The roots of f (x) = x2 − 4x − 5 = (x − 5)(x + 1) are x = −1 and 5. Since areas A and B are equal, therefore Image Thus,

Image

Solving on a calculator gives k (or x) equal to 8.

33. (D) If N is the number of bacteria at time t, then N = 200ekt. It is given that 3 = e10k. When t = 24, N = 200e24k. Therefore N = 200(e10k)2.4 = 200(3)2.4 Image 2793 bacteria.

34. (C) Since Image For x = 2t − 1, t = 3 yields x = 5 and t = 5 yields x = 9.

35. (C) Using implicit differentiation on the equation

x3 + xyy2 = 10

yields

Image

The tangent is vertical when Image is undefined; that is, when 2yx = 0.

Replacing y by Image in (1) gives

Image

or

4x3 + x2 = 40.

Let y1 = 4x3 + x2 − 40. Inspection of the equation y1 = f (x) = 0 reveals that there is a root near x = 2. Solving on a calculator yields x = 2.074.

36. (D) G ′(x) = f (3x − 1) · 3.

37. (B) Since f changes from positive to negative at t = 3, G ′ does also where 3x − 1 = 3.

38. (D) Using your calculator, evaluate y ′(2).

Image

39. (E) 2(3) + 2(0) + 2(4) + 2(1).

See the figure above.

40. (E) Image

At x = 3, the answer is 2[2(−2) + 52] = 42.

41. (A) The object is at rest when v(t) = ln(2 − t2) = 0; that occurs when 2−t2 = 1, so t = 1. The acceleration is Image

42. (D) Image

implies that x = 3.

43. (C) Image represents the rate of change of the surface area; if y is inversely proportional to x, then, Image

Image

44. (E) The velocity functions are

v1 = −2t sin (t2 + 1)

and

Image

Graph both functions in [0, 3] × [−5, 5]. The graphs intersect four times during the first 3 sec, as shown in the figure above.

45. (B) Image

Free-Response

Part A

AB/BC1. Image

(d) To work with g(x) = f −1(x), interchange x and y:

x

7.6

5.7

4.2

3.8

2.2

1.6

g(x)

2.5

3.2

3.5

4.0

4.5

5.0

Image

AB 2. Let M = the temperature of the milk at time t. Then

Image

The differential equation is separable:

Image

where c = e C.

Find c, using the fact that M = 40° when t = 0:

40 = 68−ce0 means c = 28.

Find k, using the fact that M = 43° when t = 3:

Image

Hence Image

Now find t when M = 60°:

Image

Since the phone rang at t = 3, you have 30 min to solve the problem.

Part B

AB 3. (a) Image

See the figure.

(b) The average value of a function on an interval is the area under the graph of the function divided by the interval width, here Image

(c) From part (a) you know that the area of the region is given by ImageSince Image as k increases the area of the region approaches 3π.

AB/BC 4. (a) The rectangular slices have base y, height 5y, and thickness along the x-axis:

Image

(b) The disks have radius x and thickness along the y-axis:

Image

Now we solve for x in terms of y:

Image

(NOTE: Although the shells method is not a required AP topic, another correct integral for this volume is Image

AB/BC 5. (a) The volume of the cord is V = πr2h. Differentiate with respect to time, then substitute known values. (Be sure to use consistent units; here, all measurements have been converted to inches.)

Image

(b) Let Image represent the angle of elevation and h the height, as shown.

Image

When h = 80, your distance to the jumper is 100 ft, as shown.

Then

Image

Image

AB/BC 6. (a) Image the negative of the area of the shaded rectangle in the figure. Hence F(−2) = − (3)(2) = −6.

Image

Image is represented by the shaded triangles in the figure.

Image

(b) Image so F(x) = 0 at x = 1. Image because the regions above and below the x-axis have the same area. Hence F(x) = 0 at x = 3.

Image

(c) F is increasing where F ′ = f is positive: −2 ≤ x ≤ 2.

(d) The maximum value of F occurs at x = 2, where F ′ = f changes from positive to negative. Image

The minimum value of F must occur at one of the endpoints. Since F(−2) = − 6 and F(6) = −3, the minimum is at x = −2.

(e) F has points of inflection where F ″ changes sign, as occurs where F ′ = f goes from decreasing to increasing, at x = 3.

Answers Explained

Multiple-Choice

Part A

1. (E) Here, Image

2. (C) The given limit equals Image where f (x) = sin x.

3. (C) Since f (x) = x ln x,

f ′(x) = 1 + ln x, Image

4. (A) Differentiate implicitly to get Image Substitute (−1, 1) to find Image the slope at this point, and write the equation of the tangent: y − 1 = −1(x + 1).

5. (C) f ′(x) = 4x3 − 12x2 + 8x = 4x(x − 1)(x − 2). To determine the signs of f ′(x), inspect the sign at any point in each of the intervals x < 0, 0 < x < 1, 1 < x < 2, and x > 2. The function increases whenever f ′(x) > 0.

6. (C) The integral is equivalent to Image where u = 4 + 2sinx.

7. (D) Here Image which is zero for x = e. Since the sign of y ′ changes from positive to negative as x increases through e, this critical value yields a relative maximum. Note that Image

8. (E) Since Image is always positive, there are no reversals in motion along the line.

Image

9. (E) The slope field suggests the curve shown above as a particular solution.

10. (E) Since Image is discontinuous at x = 1; the domain of F is therefore x > 1. On [2, 5] f (x) < 0, so Image which is positive for x > 1.

Image

11. (B) In the graph above, W(t), the water level at time t, is a cosine function with amplitude 6 ft and period 12hr:

Image

12. (B) Solve the differential equation Image Use x = −1,

y = 2 to determine Image

13. (D) Image

14. (A) Image

15. (C) G ′(2) = 4, so G (x) Image 4(x − 2) + 5.

16. (E) Image See the figure below.

Image

17. (E) Note that Image

18. (B) Note that (0, 0) is on the graph, as are (1, 2) and (−1, −2). So only (B) and (E) are possible. Since Image only (B) is correct.

19. (D) See the figure.

Image

20. (E) Image

21. (D) In (D), f (x) is not defined at x = 0. Verify that each of the other functions satisfies both conditions of the Mean Value Theorem.

22. (E) The signs within the intervals bounded by the critical points are given below.

Image

Since f changes from increasing to decreasing at x = −3, f has a local maximum at −3. Also, f has a local minimum at x = 0, because it is decreasing to the left of zero and increasing to the right.

23. (B) Since ln Image then

Image

24. (D) Image

25. (B) As seen from the figure, Image where y = 2r,

Image

26. (C) From the figure below, Image

Image

27. (A) Since the degrees of numerator and denominator are the same, the limit as x→∞ is the ratio of the coefficients of the terms of highest degree: Image

28. (A) We see from the figure that ΔA = (y2y1) Δx;

Image

Image

Part B

29. (A) Let u = x2 + 2. Then

Image

and

Image

30. (E) Image will increase above the half-full level (that is, the height of the water will rise more rapidly) as the area of the cross section diminishes.

31. (B) Since Image

32. (C) Image

The initial condition H(0) = 120 shows c = 50. Evaluate H(10).

33. (B) Let P be the amount after t years. It is given that Image The solution of this differential equation is P = 4000e0.08t + C, where P (0) = 4000 yields C = 0. The answer is 4000e(0.08) ·10.

34. (C) The inverse of y = 1 + ex is x = 1 + ey or y = ln (x − 1); (x − 1) must be positive.

35. (E) Speed is the magnitude of velocity: |v(8)| = 4.

36. (E) For 3 < t < 6 the object travels to the right Image At t = 7 it has returned 1 unit to the left; by t = 8, 4 units to the left.

Image

37. (C) Use disks: ΔV = πr2Δy = πx2Δy, where x = ln y. Use your calculator to evaluate Image

38. (B) If Image then u2 = x − 2, x = u2 + 2, dx = 2u du. When x = 3, u = 1; when x = 6, u = 2.

39. (A) The tangent line passes through points (8,1) and (0,3). Its slope, Image is f ′(8).

40. (A) Graph f ″ in [0,6] × [−5,10]. The sign of f ″ changes only at x = a, as seen in the figure.

Image

41. (C) In the graph below, the first rectangle shows 2 tickets sold per minute for 15 min, or 30 tickets. Similarly, the total is 2(15) + 6(15) + 10(15) + 4(15).

Image

42. (B) Graph both functions in [−8,8] × [−5,5]. At point of intersection Q, both are decreasing. Tracing reveals x Image −4 at Q. If you zoom in on the curves at x = T, you will note that they do not actually intersect there.

Image

43. (D) Counterexamples are, respectively, for (A), f (x) = |x|, c = 0; for (B), f (x) = x3, c = 0; for (C), f (x) = x4, c = 0; for (E), f (x) = x2 on (−1, 1).

44. (E) f ′(x) > 0; the curve shows that f ′ is defined for all a < x < b, so f is differentiable and therefore continuous.

45. (D) Consider the blast area as a set of concentric rings; one is shown in the figure. The area of this ring, which represents the region x meters from the center of the blast, may be approximated by the area of the rectangle shown. Since the number of particles in the ring is the area times the density, ΔP = 2πx · Δx · N(x). To find the total number of fragments within 20 m of the point of the explosion, integrate: Image

Image

Free-Response

Part A

AB1. (a) Since x2y − 3y2 = 48,

Image

(b) At (5,3), Image so the equation of the tangent line is

Image

(c) Image

(d) Horizontal tangent lines have Image This could happen only if

2xy = 0, which means that x = 0 or y = 0.

If x = 0, 0y − 3y2 = 48, which has no real solutions.

If y = 0, x2 · 0 − 3 · 02 = 48, which is impossible. Therefore, there are no horizontal tangents.

AB/BC2. (a) Image

(b) The average value of a function is the integral across the given interval divided by the interval width. Here Image Estimate the value of the integral using trapezoid rule T with values from the table and Δt = 4:

Image

Hence

Avg(W) Image 35.167 ft.

(c) For Image use your calculator to evaluate F ′(16) ≈ −0.749. After 16 hr, the river depth is dropping at the rate of 0.749 ft/hr.

Image

Part B

AB/BC 3. (a) Image

(b) Let h = the hypotenuse of an isosceles right triangle, as shown in the figure. Then each leg of the triangle is Image and its area is Image

Image

An element of volume is

Image

and thus Image

Image

AB/BC 4. (a) Image so v(0) = 0 and v(10) = 48.

The average acceleration is Image

Acceleration Image

Image

(b) Since Q’s acceleration, for all t in 0 ≤ t ≤ 5, is the slope of its velocity graph, Image

(c) Find the distance each auto has traveled. For P, the distance is

Image

For auto Q, the distance is the total area of the triangle and trapezoid under the velocity graph shown below, namely,

Image

Auto P won the race.

AB5. (a) Using the differential equation, evaluate the derivative at each point, then sketch a short segment having that slope. For example, at (−1, −1), Image = 2(−1)((−1)2 + 1) − 4; draw a steeply decreasing segment at (−1, −1). Repeat this process at each of the other points. The result follows.

Image

(b) The differential equation Image is separable.

Image

It is given that f passes through (0,1), so 1 = tan (02 + c) and Image

The solution is f(x) = tan Image

The particular solution must be differentiable on an interval containing the initial point (0,1). The tangent function has vertical asymptotes at Image hence:

Image (Since x2 ≥ 0, we ignore the left inequality.)

Image

AB 6. (a) Image f (t)dt = F(2) = 4.

(b) One estimate might be Image f (t)dt F(7) − F(2) = 2 − 4 = −2.

(c) f (x) = F′ (x); F′ (x) = 0 at = 4.

(d) f ′(x) = F ″(x). F ″ is negative when F is concave downward, which is true for the entire interval 0 < x < 8.

(e) Image

Then the graph of G is the graph of F translated downward 4 units.

Image

Answers Explained

The explanations for questions not given below will be found in the answer section for AB Practice Examination 1. Identical questions in Section I of Practice Examinations AB1 and BC1 have the same number. For example, explanations of the answers for Question 1, not given below, will be found in Section I of Calculus AB Practice Examination 1, Answer 1.

Multiple-Choice

Part A

3. (E) Here,

Image

6. (D) Image

12. (B) Note that, when x = 2 sin θ, x2 = 4 sin2 θ, dx = 2 cos dθ, and Image 2 cos θ. Also,

Image

13. (C) The given integral is equivalent to Image

Image

The figure shows the graph of f (x) = 1 − |x| on [−1,1].

The area of triangle PQR is equal to Image

14. (B) Let y = x1/x; then take logarithms. ln Image As x → ∞ the fraction is of the form ∞/∞. Image So ye0 or 1.

17. (E) Separating variables yields Image so ln y = −ln cos x + C. With y = 3 when x = 0, C = ln 3. The general solution is therefore (cos x) y = 3. When Image

Image and y = 6.

20. (A) Represent the coordinates parametrically as (r cos θ, r sin θ). Then

Image

Note that Image sin 2θ, and evaluate Image (Alternatively, write x = cos 2θ cos θ and y = cos 2θ sin θ to find Image)

21. (D) Note that v is negative from t = 0 to t = 1, but positive from t = 1 to t = 2. Thus the distance traveled is given by

Image

22. (E) Separating variables yields y dy = (1 − 2x) dx. Integrating gives

Image

24. (A) Use parts; then u = x, dv = cos x dx; du = dx,v = sin x. Thus,

Image

25. (C) Using the above figure, consider a thin slice of the oil, and the work Δw done in raising it to the top of the tank:

Image

26. (A) By Taylor’s Theorem, the coefficient is Image For Image and Image making the coefficient Image

27. (D) Here,

Image

28. (D) Evaluate

Image

Part B

29. (E) The vertical component of velocity is

Image = 4 cos t + 12 cos 12t.

Evaluate at t = 1.

30. (B) At t = 0, we know H = 120, so 120 = 70 + ke −0.4(0), and thus k = 50. The average temperature for the first 10 minutes is Image

36. (B)

Image

Image

The required volume is 0.592.

38. (C) The Maclaurin expansion is

Image

The Lagrange remainder R, after n terms, for some c in the interval |x| Image 2, is

Image

Since R is greatest when c = 2, n needs to satisfy the inequality

Image

Using a calculator to evaluate Image successively at various integral values of x gives y(8) > 0.01, y(9) > 0.002, y(10) < 3.8 × 10 −4 < 0.0004. Thus we achieve the desired accuracy with a Taylor polynomial at 0 of degree at least 10.

Image

39. (E) On your calculator, graph one arch of the cycloid for t in [0, 2π] and (x,y) in [0, 7] × [−1, 3]. Use disks; then the desired volume is

Image

40. (C) In the first quadrant, both x and y must be positive; x(t) = et is positive for all t, but y(t) = 1 − t2 is positive only for −1 < t < 1. The arc length is

Image

41. (D) Chapter 10.

44. (E) Each is essentially a p-series, Image Such a series converges only if p > 1.

Free-Response

Part A

1. See solution for AB-1, Free-Response in AB Practice Exam One.

2. See solution for AB2, Free-Response in AB Practice Exam One.

Part B

3. (a) Because Image which never equals zero, the object is never at rest.

(b) Image so the object’s speed is

Image

Position is the antiderivative of velocity Image

Image

Since P(0) = (0,0), arcsin Image and thus c = 0.

Image

Since P(0) = (0,0), Image and thus c = 2.

Then

Image

(c) Solving Image for t yields t = 2 sinx. Therefore

Image

Since 0 ≤ t ≤ 1 means Image then cos x > 0, so y = 2 − 2cos x.

4. (a) To write the Maclaurin series for f (x) = ln(e+x), use Taylor’s theorem at x = 0.

Image

Image

(b) By the Ratio Test, the series converges when

Image

Thus, the radius of convergence is e.

Image

5. (a) To find the maximum rate of growth, first find the derivative of

Image

A signs analysis shows that Image changes from positive to negative there, confirming that Image is at its maximum when there are 300 trout.

(b) The differential equation Image is separable.

Image

To integrate the left side of this equation, use the method of partial fractions.

Image

(c) In (a) the population was found to be growing the fastest when F = 300. Then:

Image

6. See solution for AB/BC 6, Free-Response in AB Practice Exam One.

Answers Explained

The explanations for questions not given below will be found in the answer section for AB Practice Exam 2. Identical questions in Section I of Practice Examinations AB2 and BC2 have the same number. For example, explanations of the answers for Questions 4 and 5, not given below, will be found in Section I of Calculus AB Practice Exam 2, Answers 4 and 5.

Multiple-Choice

Part A

1. (B) Since Image to render f (x) continuous at x = 1, define f (1) to be 1.

2. (C) Note that

Image

where you let Image

3. (C) Obtain the first few terms of the Maclaurin series generated by Image

Image

6. (C) Here,

Image

and the magnitude of the acceleration, |a|, is given by

Image

7. (B) Image

By the Ratio Test the series converges when

Image

Checking the endpoints, we find:

Image is the alternating harmonic series, which converges.

Image is the harmonic series, which diverges.

Hence the interval of convergence is 0 ≤ x < 2.

10. (C) Image

11. (D) We integrate by parts using u = ln x, dv = dx; then Image v = x, and

Image

13. (C) Use the method of partial fractions, letting

Image

Letting x = 0, we find A = 2, and letting x = 3 yields B = −1.

Now Image

19. (E) At (2,1), Image Use Δx = 0.1; then Euler’s method moves to (2.1, 1 + 3(0.1)).

At (2.1, 1.3), Image so the next point is (2.2, 1.3 + 3.4(0.1)).

22. (D) Separate variables to get Image and integrate to get ln y = ln x + C.

Since y = 3 when x = 1, C = ln 3. Then y = e(ln x + ln 3) = eln x · eln 3 = 3x.

23. (D) The generating circle has equation x2 + y2 = 4. Using disks, the volume, V, is given by

Image

24. (C) Image The integrals in (A), (B), and (D) all diverge to infinity.

26. (D) Using separation of variables:

Image

Given initial point (0,0), we have 0 = tan(0 + C); hence C = 0 and the particular solution is y = tan(x).

Because this function has vertical asymptotes at Image and the particular solution must be differentiable in an interval containing the initial point x = 0, the domain is Image

28. (D) Image

Part B

30. (C) Since the equation of the spiral is r = lnImage, use the polar mode. The formula for area in polar coordinates is

Image

Therefore, calculate

Image

The result is 3.743.

31. (D) When y = 2t = 4, we have t = 2, so the line passes through point F(2) = (5,4).

Also Image so at t = 2 the slope of the tangent line is

Image

An equation for the tangent line is y − 4 = ln 2(x − 5).

32. (A) I. Image converges by the Ratio Test: Image

II. Image diverges by the nth Term Test: Image

III. Image diverges by the Comparison Test: Image diverges.

38. (E) If Q0 is the initial amount of the substance and Q is the amount at time t, then

Image

and Image Thus k = 0.08664. Using a calculator, find t when Image

Image so t ≈ 12.68. Don’t round off k too quickly.

40. (A) See figure below.

Image

Image

41. (D) See figure below.

Image

42. (E) Image

43. (C) The endpoints of the arc are Image and (e,1). The arc length is given by

Image

44. (B) Find k such that cos x will differ from Image by less than 0.001 at x = k.

Solve

Image

which yields x or k = 0.394.

Free-Response

Part A

1. See solution for AB-1, Free-Response in AB Practice Exam Two.

2. (a) Position is the antiderivative of Image

Image

To find c, substitute the initial condition that x = 1 when t = 0:

Image

At t = 2, x = 2 arctan Image and the position of the object is Image

Image

(c) The distance traveled is the length of the arc of y = x2 + 2 in the interval Image

Image

Part B

3. (a)

Image

(b) By the Ratio Test, the series converges when

Image

Thus, the radius of convergence is 2.

(c) Image is an alternating series. Since Image and Image it converges by the Alternating Series Test. Therefore the error is less than the magnitude of the first omitted term:

Image

4. See solution for AB/BC 4, Free-Response in AB Practice Exam Two.

5. See solution for AB-5, Free-Response in AB Practice Exam Two.

6. See solution for AB-6, Free-Response in AB Practice Exam Two.

Answers Explained

The explanations for questions not given below will be found in the answer section for AB Practice Exam 3. Identical questions in Section I of Practice Examinations AB3 and BC3 have the same number. For example, explanations of the answers for Questions 1 and 2, not given below, will be found in Section I of Calculus AB Practice Exam 3, Answers 1 and 2.

Multiple-Choice

Part A

3. (B) The series Image is geometric with Image it converges

to Image

5. (D) Image is the sum of an infinite geometric series with first term 1 and common ratio −2x. The series is 1 − 2x + 4x2 − 8x3 + 16x4 − ….

6. (A) Assume that

Image

Then

2x2x + 4 = A(x − 1)(x − 2) + Bx(x − 2) + Cx(x − 1).

Since you are looking for B, let x = 1:

2(1) − 1 + 4 = 0 + B(−1) + 0; B = −5.

8. (B) Since ex Image 1 + x,e x2 Image 1 − x2. So

Image

12. (A) Image

17. (E) Image

18. (D) See the figure below, which shows that the length of a semicircle of radius 2 is needed here. The answer can, of course, be found by using the formula for arc length:

Image

20. (E) Image using long division, Image

22. (C) Using parts we let u = x2, dv = exdx; then du = 2x dx, v = ex, and

Image

We use parts again with u = x, dv = exdx; then du = dx, v = ex, and

Image

Now Image

23. (E) Image will increase above the half-full level (that is, the height of the water will rise more rapidly) as the area of the cross section diminishes.

27. (A) The required area is lined in the figure below.

Image

28. (D) Note that f (x) = x + 6 if x ≠ 6, that f (6) = 12, and that Image So f is defined and continuous at x = 6.

Part B

29. (E) The velocity functions are

Image

When these functions are graphed on a calculator, it is clear that they intersect four times during the first 3 sec, as shown below.

Image

30. (C) Changes in values of f show that f ″′ may be constant. Hence f may be linear, so f′ could be quadratic and thus f cubic.

31. (C) Expressed parametrically, x = sin3θ cosθ, y = sin3θ sin θ. Image is undefined where Image = −sin3θ sin θ + 3 cos 3θ cos θ = 0.

Use your calculator to solve for θ.

34. (D) See the figure below.

Image

The roots of f (x) = x2 − 4x − 5 = (x − 5)(x + 1) are x = −1 and 5. Since areas A and B are equal, therefore, Image Thus,

Image

A calculator yields k = 8.

37. (C) It is given that Image An antiderivative is Image

Since Image the constants are c1 = c2 = 1. The object’s speed is Image

Use a calculator to find that the object’s maximum speed is 2.217.

Image

39. (D) Use the Ratio Test:

Image

which is less than 1 if −3 < x < 3. When x = −3, the convergent alternating harmonic series is obtained.

40. (A) Since Image and since v(0) = 1, C = 1. Then Image yields s = t3 + t + C ′, and you can let s(0) = 0. Then you want s(3).

41. (A) Arc length is given by Image Here the integrand Image implies that Image hence, y = 3 ln x + C. Since the curve contains (1,2), 2 = 3 ln 1 + C, which yields C = 2.

42. (E) Image

At x = 3; the answer is 2[2(−2) + 52] = 42.

43. (D) Counterexamples are, respectively, for (A), f (x) = |x|, c = 0; for (B), f (x) = x3, c = 0; for (C), f (x) = x4, c = 0; for (E), f (x) = x2 on (−1, 1).

Free-Response

Part A

1. Image

(a) The following table shows x- and y-components of acceleration, velocity, and position:

Image

The last line in the table is the answer to part (a).

(b) To determine how far above the ground the ball is when it hits the wall, find out when x = 315, and evaluate y at that time.

Image

(c) The ball’s speed at the moment of impact in part (b) is |v(t)| evaluated at Image

Image

2. See solution for AB-2, Free-Response in AB Practice Exam Three.

Part B

3. See solution for AB/BC 3, Free-Response in AB Practice Exam Three.

4. See solution for AB-4, pages Free-Response in AB Practice Exam Three.

5. (a) The table below is constructed from the information given in Question 5 in BC Practice Exam Three.

n

f (n) (5)

Image

0

2

2

1

−2

−2

2

−1

Image

3

6

1

Image

Image

(c) Use Taylor’s theorem around x = 0.

n

g(n)(x)

g(n)(0)

Image

0

f (2x + 5)

f (5) = 2

2

1

2f′ (2x + 5)

2f′ (5) = 2(−2) = −4

−4

2

4f (2x + 5)

4f (5) = 4(−1) = −4

−2

3

8f′″(2x + 5)

8f′″(5) = 8(6) = 48

8

g(x) ≈ 2 − 4x − 2x2 + 8x3.

6. (a) At (−1,8), Image so the tangent line is

y − 8 = 5(x − (−1)). Therefore f (x) ≈ 8 + 5(x + 1).

(b) f (3) ≈ 8 + 5(0 + 1) = 13.

(c) At (−1,8), Image For Δx = 0.5, Δy = 0.5(5) = 2.5, so move to

(−1 + 0.5, 8 + 2.5) = (−0.5,10.5).

At (−0.5,10.5), Image For Δx = 0.5, Δy = 0.5(8) = 4, so move to (−0.5 + 0.5, 10.5 + 4).

Thus f (0) ≈ 14.5.

Image

Answers Explained

1. (D) Velocity Image and changes sign both when t = 1 and when t = 3.

2. (A) Since v > 0 for 0 Image t Image 2, the distance is equal to Image

3. (E) The answer is 8. Since the particle reverses direction when t = 2, and v > 0 for t > 2 but v < 0 for t < 2, therefore, the total distance is

Image

4. (E) Image so there is no change in position.

5. (B) Since v = sin t is positive on 0 < t Image 2, the distance covered is

Imagesin t dt = 1 − cos 2.

6. (D) Image.

7. (D) The velocity v of the car is linear since its acceleration is constant:

Image

8. (D) Image Since R(0) = 〈0,1〉,

c1 = 0 and c2 = 1.

9. (A) a = v(t) = 〈1,1〉 for all t.

10. (B) Image

11. (B) Since R = 〈x, y〉, its slope is Image since Image its slope is Image

If R is perpendicular to v, then Image so

Image and x2 + y2 = k (k > 0).

Since (4, 3) is on the curve, the equation must be

x2 + y2 = 25.

12. (D) Image since Image and e° + c2 = 0; hence c1 = 2 and c2 = −1.

13. (B) The object’s position is given by Image Since the object was at the origin at Image and 4 · 1 + c2 = 0, making the position Image When t = 0, x(0) = −2, y(0) = −4.

14. (D) Image

15. (B) We want the accumulated number of people to be 100:

Image

This occurs at h = 2 hours after 8 A.M.

16. (C) Image

17. (A) Image

18. (A) The number of new people who hear the rumor during the second week is

Image

Be careful with the units! The answer is the total change, of course, in F(t) from t = 7 to t = 14 days, where F (t) = t2 + 10t.

19. (B) Total gallons = Image

20. (A) Be careful! The number of cars is to be measured over a distance of x (not 20) mi. The answer to the question is a function, not a number. Note that choice (C) gives the total number of cars on the entire 20-mi stretch.

21. (C) Since the strip of the city shown in the figure is at a distance r mi from the highway, it is Image mi long and its area is Image The strip’s population is approximately 2(12 − 2r) Image Δr. The total population of the entire city is twice the integral Image as it includes both halves of the city.

Image

22. (C)

Image

The population equals ∑ (area · density). We partition the interval [0,10] along a radius from the center of town into 5 equal subintervals each of width Δr = 2 mi. We will divide Winnipeg into 5 rings. Each has area equal to (circumference × width), so the area is 2πrk Δr or 4πrk. The population in the ring is

(4πrk)· (density at rk) = 4πrk · f (rk).

A Riemann sum, using left-hand values, is 4π · 0 · 50 + 4π · 2 · 45 + 4π · 4 · 40 + 4π · 6 · 30 + 4π · 8 · 15 = 4π(90 + 160 + 180 + 120) Image 6912 hundred people—or about 691,200 people.

23. (E) The total amount dumped after 7 weeks is

Image

24. (B) The total change in temperature of the roast 20 min after it is put in the refrigerator is

Image

Since its temperature was 160°F when placed in the refrigerator, then 20 min later it is (160 − 89.7)°F or about 70°F. Note that the temperature of the refrigerator (45°F) is not used in answering the question because it is already “built into” the cooling rate.

25. (A) Let T be the number of weeks required to release 9 tons. We can use parts to integrate Image then substitute the limits. We must then set the resulting expression equal to 9 and solve for T. A faster, less painful alternative is to use a graphing calculator to solve the equation

Image

The answer is about 10.2 weeks.

26. (D) Note that the curve is above the x-axis on [0, 1], but below on [1, 3], and that the areas for x < 0 and x > 3 are unbounded.

Image

Using the calculator, we get

Image

27. (E) The FTC yields total change:

Image

28. (C) The total change (increase) in population during the second hour is given by Image The answer is 1046.

29. (C) Call the time in hours t and the function for visitors/hour R(t). Then the area under the curve represents the number of visitors V. We will estimate the time k when Image using a Riemann sum.

The table shows one
approach, based on
the Midpoint Rule.

Hour

Visitors/hour (midpoint est.)

Total visitors since noon

Noon–1 P.M.

5

5

1−2 P.M.

25

30

2−3 P.M.

70

100

3−4 P.M.

120

220

We estimate that there had been about 100 visitors by 3 P.M. and 220 by 4 P.M., so the 200th visitor arrived between 3 and 4 P.M.

30. (B) Image

31. (C) We partition [0, 2] into n equal subintervals each of time Δt hr. Since the 18-wheeler gets (4 + 0.01v) mi/gal of diesel, it uses Image

Since it covers v · Δt mi during Δt hr, it uses Image

Since Image we see that the diesel fuel used in the first 2 hr is

Image

Answers Explained

1. (C) v(t) = 2t2t + C; v(1) = 3; so C = 2.

2. (B) If a(t) = 20t3 − 6t, then

v(t) = 5t4 − 3t2 + C1,

s(t) = t5t3 + C1 t + C2,

Since

s(−1) = −1 + 1 − C1 + C2 = 2

and

s(1) = 1 − 1 + C1 + C2 = 4,

therefore

2C2 = 6, C2, = 3,

C1 = 1.

So

v(t) = 5t4 − 3t2 + 1.

3. (D) From Answer 2, s(t) = t5t3 + t + 3, so s(0) = C2 = 3.

4. (B) Since a(t) = −32, v(t) = −32t + 40, and the height of the stone s(t) = −16t2 + 40t + C. When the stone hits the ground, 4 sec later, s(t) = 0, so

Image

5. (C) From Answer 4

s(t) = −16t2 + 40t + 96.

Then

s ′(t) = −32t + 40,

which is zero if t = 5/4, and that yields maximum height, since s ″(t) = −32.

6. (E) The velocity v(t) of the car is linear, since its acceleration is constant and

Image

7. (B) Since v = 100 − 20t, s = 100t − 10t2 + C with s(0) = 0. So s(1) = 100 − 10 = 90 ft.

8. (A) Image

9. (A) The odometer measures the total trip distance from time t = a to t = b (whether the car moves forward or backward or reverses its direction one or more times from t = a to t = b). This total distance is given exactly by Image

10. (E) (A), (B), (C), and (D) are all true. (E) is false: see Answer 9.

11. (A) Integrating yields Image + C or y2 = x2 + 2C or y2 = x2 + C , where we have replaced the arbitrary constant 2C by C ′.

12. (C) For initial point (−2,1), x2y2 = 3. Rewriting the d.e. y dy = x dx as Image reveals that the derivative does not exist when y = 0, which occurs at Image Since the particular solution must be differentiable in an interval containing x = −2, the domain is Image

13. (E) We separate variables. Image The initial point yields ln Image hence c = −2. With y > 0, the particular solution is ln Image

14. (C) We separate variables. Image The particular solution is −ey = x − 2.

15. (B) The general solution is Image when x = 4 yields C = 0.

16. (E) Since Image it follows that

ln y = ln x + C or ln y = ln x + ln k;

so y = kx.

17. (E) Image hence the general solution is y = kex, k ≠ 0.

18. (A) We rewrite and separate variables, getting Image The general solution is

Image

19. (C) We are given that Image The general solution is ln |y| = 3 ln |x| + C.

Thus, |y| = c |x3 |; y = ±c x3. Since y = 1 when x = 1, we get c = 1.

20. (E) The d.e. Image reveals that the derivative does not exist when x = 0. Since the particular solution must be differentiable in an interval containing initial value x = 1, the domain is x > 0.

21. (E) The general solution is y = k ln |x| + C, and the particular solution is y = 2 ln |x| + 2.

22. (D) We carefully(!) draw a curve for a solution to the d.e. represented by the slope field. It will be the graph of a member of the family y = sin x + C. At the right we have superimposed the graph of the particular solution y = sin x − 0.5.

Image

23. (B)

Image

It’s easy to see that the answer must be choice (A), (B), or (C), because the slope field depends only on x: all the slope segments for a given x are parallel. Also, the solution curves in the slope field are all concave up, as they are only for choices (A) and (B). Finally, the solution curves all have a minimum at x = 2, which is true only for differential equation (B).

24. (E) The solution curve is y = tan x, which we can obtain from the differential equation y ′ = 1 + y2 with the condition y(0) = 0 as follows:

Image

Since y(0) = 0, C = 0. Verify that (A) through (D) are incorrect.

NOTE: In matching slope fields and differential equations in Questions 25–29, keep in mind that if the slope segments along a vertical line are all parallel, signifying equal slopes for a fixed x, then the differential equation can be written as y ′ = f (x). Replace “vertical” by “horizontal” and “x” by “ y” in the preceding sentence to obtain a differential equation of the form y ′ = g(y).

25. (B) The slope field for y ′ = y must by II; it is the only one whose slopes are equal along a horizontal line.

26. (D) Of the four remaining slope fields, IV is the only one whose slopes are not equal along either a vertical or a horizontal line (the segments are not parallel). Its d.e. therefore cannot be either of type y ′ = f (x) or y ′ = g(y). The d.e. must be implicitly defined—that is, of the form y ′ = F(x,y). So the answer here is IV.

27. (C) The remaining slope fields, I, III, and V, all have d.e.’s of the type y ′ = f (x). The curves “lurking” in III are trigonometric curves—not so in I and V.

28. (A) Given y ′ = 2x, we immediately obtain the general solution, a family of parabolas, y = x2 + C. (Trace the parabola in I through (0, 0), for example.)

29. (E) V is the only slope field still unassigned! Furthermore, the slopes “match” ex2: the slopes are equal “about” the y-axis; slopes are very small when x is close to −2 and 2; and e−x2 is a maximum at x = 0.

30. (A) From Answer 25, we know that the d.e. for slope field II is y ′ = y. The general solution is y = cex. For a solution curve to pass through point (0, −1), we have −1 = ce0; and c = −1.

31. (C) Euler’s method for y ′ = x, starting at (1, 5), with Δx = 0.1, yields

x

y

(SLOPE) *· Δx

= Δ y

1

5

1 · (0.1)

= 0.1

*The slope is x.

1.1

5.1

(1.1) · (0.1)

= 0.11

1.2

5.21

32. (B) We want to compare the true value of y(1.2) to the estimated value of 5.21 obtained using Euler’s method in Solution 31. Solving the d.e. Image yields Image and initial condition y(1) = 5 means that Image or C = 4.5. Hence Image + 4.5 = 5.22. The error is 5.22−5.21 = 0.01.

33. (A) Slopes depend only on the value of y, and the slope field suggests that y ′ = 0 whenever y = 0 or y = −2.

34. (D) The slope field suggests that the solution function increases (or decreases) without bound as x increases, but approaches y = 1 as a horizontal asymptote as x decreases.

35. (D) We separate variables to get Image We integrate:

Image With t = 0 and s = 1, C = 0. When Image we get Image

36. (B) Since Image and ln R = ct + C. When t = 0, R = R0; so ln R0 = C or ln R = ct + ln R0. Thus

ln R − ln R0 = ct; ln Image

37. (D) The question gives rise to the differential equation Image where P = 2P0 when t = 50. We seek Image for t = 75. We get ln Image with ln 2 = 50k; then

Image

38. (A) We let S equal the amount present at time t; using S = 40 when t = 0 yields ln Image Since, when t = 2, S = 10, we get

Image

39. (A) We replace g(x) by y and then solve the equation Image We use the constraints given to find the particular solution Image

40. (C) The general solution of Image is ln y = x + C or y = cex. For a solution to pass through (2, 3), we have 3 = ce2 and c = 3/e2 Image 0.406.

41. (C) At a point of intersection, y ′ = x + y and x + y = 0. So y ′ = 0, which implies that y has a critical point at the intersection. Since y ″ = 1 + y ′ = 1 + (x + y) = 1 + 0 = 1, y ″ > 0 and the function has a local minimum at the point of intersection. [See Figure N9–5, showing the slope field for y ′= x + y and the curve y = exx − 1 that has a local minimum at (0, 0).]

42. (A) Although there is no elementary function (one made up of polynomial, trigonometric, or exponential functions or their inverses) that is an anti-derivative of F ′(x) = ex2, we know from the FTC, since F(0) = 0, that

Image

To approximate Image use your graphing calculator

For upper limits of integration x = 50 and x = 60, answers are identical to 10 decimal places. Rounding to three decimal places yields 0.886.

43. (C) Logistic growth is modeled by equations of the form Image where L is the upper limit. The graph shows L = 1000, so the differential equation must be Image Only equation C is of this form (k = 0.003).

44. (D) We start with x = 3 and y = 100. At x = 3, Image moving us to x = 3 + 2 = 5 and y = 100 + 5 = 105. From there Image so when x = 5 + 2 = 7 we estimate y = 105 + (−1) = 104.

45. (C) We separate the variables in the given d.e., then solve:

Image

Since y(0) = 180, ln 112 = c. Then

Image

When t = 10, y = 68 + 112e−1.1 Image 105°F.

46. (E) The solution of the d.e. in Question 45, where y is the temperature of the coffee at time t, is

Image

47. (D) If Q is the concentration at time t, then Image We separate variables and integrate:

Image

We let Q(0) = Q0. Then

Image

We now find t when Q = 0.1Q0:

Image

48. (E) Please read the sections on Applications for Restricted Growth and Applications for Logistic Growth in this chapter for the characteristics of the logistic model.

49. (D) (A), (B), (C), and (E) are all of the form y ′ = ky(ay).

50. (B) The rate of growth, Image is greatest when its derivative is 0 and the curve of Image is concave down. Since

Image

which is equal to 0 if Image or 500, animals. The curve of y ′ is concave down for all P, since

Image

so P = 500 is the maximum population.

51. (A) The description of temperature change here is an example of Case II: the rate of change is proportional to the amount or magnitude of the quantity present (i.e., the temperature of the corpse) minus a fixed constant (the temperature of the mortuary).

52. (C) Since (A) is the correct answer to Question 51, we solve the d.e. in (A) given the initial condition T(0) = 32:

Image

BC ONLY

Answers Explained

1. (B) Image converges to −1.

2. (C) Note that Image

3. (D) The sine function varies continuously between −1 and 1 inclusive.

4. (E) Note that Image is a sequence of the type sn = rn with |r| < 1; also that Image by repeated application of L’Hôpital’s rule.

5. (C) Image

6. (E) The harmonic series Image is a counterexample for (A), (B), and (C). Image shows that (D) does not follow.

7. (B) Image

8. (A) Image

BC ONLY

9. (B) Find counterexamples for statements (A), (C), and (D).

10. (D) Image the general term of a divergent series.

11. (D) (A), (B), (C), and (E) all converge; (D) is the divergent geometric series with r = −1.1.

12. (D) Image

13. (A) If Image then f (0) is not defined.

14. (C) Image unless x = 3.

15. (B) The integrated series is Image See Question 27.

16. (E) Image

17. (A) Image

18. (E) The series satisfies the Alternating Series Test, so the error is less than the first term dropped, namely, Image (see (5)), in the chart for Common Maclaurin Series in Chapter 10 of this eBook. So n 500.

19. (D) Note that the Taylor series for tan−1 x satisfies the Alternating Series Test and that Image then the first omitted term, Image is negative. Hence P7 (x) exceeds tan−1 x.

20. (E) Now the first omitted term, Image is positive for x < 0. Hence P9 (x) is less than tan−1 x.

21. (A) If Image converges, so does Image where m is any positive integer; but their sums are probably different.

BC ONLY

22. (E) Each series given is essentially a p-series. Only in (E) is p > 1.

23. (C) Use the Integral Test.

24. (C) The limit of the ratio for the series Image is 1, so this test fails; note for (E) that

Image

25. (B) Image does not equal 0.

26. (E) Note the following counterexamples:

(A) Image

(B) Image

(C) Image

(D) Image

27. (C) Since Image the series converges if |x| < 1. We must test the end-points: when x = 1, we get the divergent harmonic series; x = −1 yields the convergent alternating harmonic series.

28. (A) Image for all x ≠ −1; since the given series converges to 0 if x = −1, it therefore converges for all x.

29. (B) The differentiated series is Image

Image

30. (B) See Example 52.

31. (D) Note that every derivative of ex is e at x = 1. The Taylor series is in powers of (x − 1) with coefficients of the form Image

32. (D) For f (x) = cos x around Image

33. (C) Note that ln q is defined only if q > 0, and that the derivatives must exist at x = a in the formula for the Taylor series.

34. (A) Use

Image

Image Or use the series for ex and let x = −0.1.

BC ONLY

35. (C) Image

Or generate the Maclaurin series for esin x.

36. (E) (A), (B), (C), and (D) are all true statements.

37. (A) Image

38. (C) Image

Since the series converges when Image that is, when Image the radius of convergence is Image

39. (E) The Maclaurin series sin x = Image converges by the Alternating Series Test, so the error |R4 | is less than the first omitted term. For x = 1, we have Image

40. (D) Image

Answers Explained

Part A

1. (D) If f (x) = Image for x ≠ 0 and f (0) = 0 then,

Image

thus this function is continuous at 0. In (A), Image does not exist; in (B), f has a jump discontinuity; in (C), f has a removable discontinuity; and in (E), f has an infinite discontinuity.

2. (C) To find the y-intercept, let x = 0; y = − 1.

3. (A) Image

4. (D) The line x + 3y + 3 = 0 has slope Image a line perpendicular to it has slope 3.

The slope of the tangent to y = x2 − 2x + 3 at any point is the derivative 2x − 2. Set 2x − 2 equal to 3.

5. (A) Image is f′ (1), where Image Or simplify the given fraction to Image

6. (E) Because p ″(2) < 0 and p ″(5) > 0, p changes concavity somewhere in the interval [2,5], but we cannot be sure p ″ changes sign at x = 4.

7. (C) Image

Save time by finding the area under y = |x − 4| from a sketch!

Image

8. (A) Since the degrees of numerator and denominator are the same, the limit as x→∞ is the ratio of the coefficients of the terms of highest degree: Image

9. (D) On the interval [1, 4], f′ (x) = 0 only for x = 3. Since f (3) is a relative minimum, check the endpoints to find that f (4) = 6 is the absolute maximum of the function.

10. (C) To find lim f as x → 5 (if it exists), multiply f by Image

Image

and if x ≠ 5 this equals Image So lim f (x) as x → 5 is Image For f to be continuous at x = 5, f (5) or c must also equal Image

11. (D) Evaluate Image

12. (A) Image From the equation given, y = esin x.

13. (D) If f (x) = x cos x, then f ′(x) = −x sin x + cos x, and

Image

14. (C) If y = ex ln x, then Image which equals e when x = 1. Since also y = 0 when x = 1, the equation of the tangent is y = e(x − 1).

15. (B) v = 4(t − 2)3 and changes sign exactly once, when t = 2.

16. (C) Evaluate Image

17. (C) Image

18. (C) Since v = 3t2 + 3, it is always positive, while a = 6t and is positive for t > 0 but negative for t < 0. The speed therefore increases for t > 0 but decreases for t < 0.

19. (A) Note from the figure that the area, A, of a typical rectangle is

Image

Image

For y = 2, Image Note that Image is always negative.

20. (B) If S represents the square of the distance from (3, 0) to a point (x, y) on the curve, then S = (3 − x)2 + y2 = (3 − x)2 + (x2 − 1). Setting Image yields the minimum distance at Image

21. (D) Image

22. (D) See the figure. Since the area, A, of the ring equals π (y22y12),

A = π [(6xx2)2x4] = π [36x2 − 12x3 + x4x4 ]

and Image = π (72x − 36x2) = 36πx (2 − x).

It can be verified that x = 2 produces the maximum area.

Image

23. (A) This is of type Image

Image

24. (A) About the y-axis; see the figure. Washer.

Image

25. (E) Separating variables, we get y dy = (1 − 2x) dx. Integrating gives

Image

or

y2 = 2x − 2x2 + k

or

2x2 + y2 − 2x = k.

26. (E) 2(5) + Image

Image

27. (E) Image

28. (D) Use L’Hôpital’s Rule or rewrite the expression as Image

29. (D) For f (x) = tan x, this is Image

30. (E) The parameter k determines the amplitude of the sine curve. For f = k sin x and g = ex to have a common point of tangency, say at x = q, the curves must both go through (q, y) and their slopes must be equal at q. Thus, we must have

k sin q = eq and k cos q = eq,

and therefore

sin q = cos q.

Image

The figure shows Image

Image

31. (D) We differentiate implicitly to find the slope Image

Image

At (3, 1), Image The linearization is Image

32. (C) Image

33. (A) About the x-axis. Disk.

Image

34. (C) Let f (x) = ax; then Image ln a = ln a.

35. (E) Image is a function of x alone; curves appear to be asymptotic to the y-axis and to increase more slowly as |x| increases.

36. (D) The given limit is equivalent to

Image

where Image The answer is Image

37. (B) Image

38. (C) In the figure, the curve for y = ex has been superimposed on the slope field.

Image

39. (C) The general solution is y = 3 ln|x2 − 4| + C. The differential equation Image reveals that the derivative does not exist for x = ±2. The particular solution must be differentiable in an interval containing the initial value x = −1, so the domain is −2 < x < 2.

40. (A) The solution curve shown is y = ln x, so the differential equation is Image

Image

41. (D) Image = sec θ; dx = sec2 θ; 0 x 1; so 0 ≤ θ ≤ Image

42. (C) The equations may be rewritten as Image = sin u and y = 1 − 2 sin2 u,

giving Image

43. (D) Use the formula for area in polar coordinates,

Image

then the required area is given by

Image

(See polar graph 63 in the Appendix.)

44. (C) Image

45. (A) The first three derivatives of Image The first four terms of the Maclaurin series (about x = 0) are 1, + 2x, Image

Note also that Image represents the sum of an infinite geometric series with first term 1 and common ratio 2x. Hence,

Image

46. (D) We use parts, first letting u = x2, dv = e−x dx; then du = 2x dx, v = −e−x and

Image

Now we use parts again, letting u = x, dv = ex dx; then du = dx, v = −ex and

Image

Alternatively, we could use the Tic-Tac-Toe Method:

Image

Then Image

47. (E) Use formula (20) in the Appendix to rewrite the integral as

Image

48. (E) The area, A, is represented by Image

49. (D) Image

50. (C) Check to verify that each of the other improper integrals converges.

51. (D) Note that the integral is improper.

Image

See Example 26.

52. (C) Let Image Then ln y = −x ln x and

Image

Now apply L’Hôpital’s Rule:

Image

So, if Image ln y = 0, then Image

53. (D) The speed, |v|, equals Image and since x = 3yy2,

Image

Then |v| is evaluated, using y = 1, and equals Image

54. (A) This is an indeterminate form of type Image use L’Hôpital’s Rule:

Image

55. (E) We find A and B such that Image

After multiplying by the common denominator, we have

3x + 2 = A(x − 4) + B(x + 3).

Substituting x = −3 yields A = 1, and x = 4 yields B = 2; hence,

Image

56. (B) Since Image

Then Image

Note that Image so the integral is proper.

57. (D) We represent the spiral as P(θ) = (θ cos θ, θ sin θ). So

Image

Part B

58. (D) Since h is increasing, h ′ ≥ 0. The graph of h is concave downward for x < 2 and upward for x > 2, so h ″ changes sign at x = 2, where it appears that h′ = 0 also.

59. (C) I is false since, for example, f ′(−2) = f ′(1) = 0 but neither g(−2) nor g(1) equals zero.

II is true. Note that f = 0 where g has relative extrema, and f is positive, negative, then positive on intervals where g increases, decreases, then increases.

III is also true. Check the concavity of g: when the curve is concave down, h < 0; when up, h > 0.

60. (A) If Image

61. (D) Image represents the area of the same region as Image translated one unit to the left.

62. (D) According to the Mean Value Theorem, there exists a number c in the interval [1,1.5] such that Image Use your calculator to solve the equation Image for c (in radians).

63. (E) Here are the relevant sign lines:

Image

We see that f ′ and f ″ are both positive only if x > 1.

64. (E) Note from the sign lines in Question 63 that f changes from decreasing to increasing at x = 1, so f has a local minimum there.

Also, the graph of f changes from concave up to concave down at x = 0, then back to concave up at Image hence f has two points of inflection.

65. (C) The derivatives of ln (x + 1) are Image

The nth derivative at Image

66. (C) The absolute-value function f (x) = |x| is continuous at x = 0, but f ′(0) does not exist.

67. (C) Let F ′(x) = f (x); then F ′(x + k) = f (x + k);

Image

Or let u = x + k. Then dx = du, when x = 0, u = k; when x = 3, u = 3 + k.

68. (E) See the figure. The equation of the generating circle is (x − 3)2 + y2 = 1, which yields Image

Image

Image

69. (D) Note that f (g(u)) = tan−1 (e2u); then the derivative is Image

70. (D) Let Image Then cos (xy) [xy′ + y] = y . Solve for y .

71. (E) Image

Image

72. (C) About the x-axis; see the figure. Washer.

Image

73. (C) By the Mean Value Theorem, there is a number c in [1, 2] such that

Image

74. (D) The enclosed region, S, is bounded by y = sec x, the y-axis, and y = 4. It is to be rotated about the y-axis.

Image

Use disks; then ΔV = πR2 H = π(arc sec y)2 Δy. Using the calculator, we find that

Image

75. (C) If Q is the amount at time t. then Q = 40ekt. Since Q = 20 when t = 2, k = −0.3466. Now find Q when t = 3, from Q = 40e−(0.3466) 3, getting Q = 14 to the nearest gram.

76. (A) The velocity v(t) is an antiderivative of a(t), where Image So Image arctan t + C. Since v (1) = 0, C = − π.

Image

77. (D) Graph y = tan x and y = 2 − x in [−1, 3] × [−1, 3]. Note that

Image

The limits are y = 0 and y = b, where b is the ordinate of the intersection of the curve and the line. Using the calculator, solve

arctan y = 2 − y

and store the answer in memory as B. Evaluate the desired area:

Image

Image

78. (E) Center the ellipse at the origin and let (x, y) be the coordinates of the vertex of the inscribed rectangle in the first quadrant, as shown in the figure.

Image

To maximize the rectangle’s area A = 4xy, solve the equation of the ellipse, getting

Image

So Image Graph Image in the window [0,5] × [0,150],

The calculator shows that the maximum area (the y-coordinate) equals 100.

Image

79. (B) Image

80. (B) When f ′ is positive, f increases. By the Fundamental Theorem of Calculus, f′ (x) = 1 − 2 (cos x)3. Graph f ′ in [0, 2π] × [−2, 4]. It is clear that f ′ > 0 on the interval a < x < b. Using the calculator to solve 1 − 2(cos x)3 = 0 yields a = 0.654 and b = 5.629.

Image

81. (C) Image

82. (B) The volume is composed of elements of the form ΔV = (2x)2 Δy. If h is the depth, in feet, then, after t hr,

Image

Image

83. (B) Separating variables yields

Image

P(0) = 300 gives c = 700. P(5) = 500 yields 500 = 1000 − 700e−5k, so k Image + 0.0673. Now P(10) = 1000 − 700e−0.673 Image 643.

84. (C) Image dx = 4 arctan 1 = π. H′ (1) = f (1) = 2.

The equation of the tangent line is y − π = 2(x − 1).

85. (C) Using midpoint diameters to determine cylinders, estimate the volume to be

V Image π · 82 · 25 + π · 62 · 25 + π · 42 · 25 + π · 32 · 25.

86. (A) Image

87. (C) H′ (3) = f′ (g(3)) · g′ (3) = f′ (2) · g′ (3).

88. (E) M′ (3) = f (3) · g′ (3) + g(3) · f′ (3) = 4 · 3 + 2 · 2.

89. (E) Image

90. (C) Image

91. (D) Here are the pertinent curves, with d denoting the depth of the water:

Image

92. (B) Use areas; then Image Thus, f (7) − f (1) = 7.

93. (B) The region x units from the stage can be approximated by the semicircular ring shown; its area is then the product of its circumference and its width.

Image

The number of people standing in the region is the product of the area and the density:

Image

To find the total number of people, evaluate

Image

94. (B) Image is positive, but decreasing; hence Image

95. (C) Image

96. (E) On 2 t 5, the object moved Image ft to the right; then on 5 t 8, it moved only Image ft to the left.

97. (B) Image

98. (D) Evaluate Image

99. (A) Image

Use, also, the facts that the speed is given by Image and that the point moves counterclockwise; then Image = 4, yielding Image and Image at the given point. The velocity vector, v, at (3, 4) must therefore be Image

100. (A) Image

101. (B) The formula for length of arc is

Image

Since y = 2x, we find

Image

102. (D) a(t) = (0, et); the acceleration is always upward.

103. (A) At (0, 1), Image so Euler’s method yields (0.1, 1 + 0.1(4)) = (0.1, 1.4). Image has particular solution y = e4x; the error is e4(0.1) − 1.4.

104. (D) Image Note that the series converges by the Alternating Series Test. Since the first term dropped in the estimate is Image the estimate is too high, but within 0.1 of the true sum.

105. (C) Image which equals a constant times the harmonic series.

106. (D) We seek x such that

Image

Then |x| < e and the radius of convergence is e.

107. (B) The error is less than the maximum value of Image

This maximum occurs at c = x = Image

108. (D) Image

Note that the curve is traced exactly once by the parametric equations from t = 0 to t = 1.

Answers Explained

Part A

1. (a) Image

(b) It appears that the rate of change of f, while negative, is increasing. This implies that the graph of f is concave upward.

(c) L = 7.6(0.7) + 5.7(0.3) + 4.2(0.5) + 3.1(0.6) + 2.2(0.4) = 11.87.

(d) Using disks ΔV = πr2 Δx. One possible answer uses the left endpoints of the subintervals as values of r:

V ≈ π(7.6)2 (0.7) + π(5.7)2 (0.3) + π(4.2)2 (0.5) + π(3.1)2 (0.6) + π(2.2)2 (0.4)

2. (a) 12y0 + 0.3 = 24 yields y0 Image 1.975.

(b) Replace x by 0.3 in the equation of the curve:

Image

The calculator’s solution to three decimal places is y0 = 1.990.

(c) Since the true value of y0 at x = 0.3 exceeds the approximation, conclude that the given curve is concave up near x = 0. (Therefore, it is above the line tangent at x = 0.)

3. Graph f ′(x) = 2x sin xe(−x2) + 1 in [−7, 7] × [−10, 10].

Image

(a) Since f ′ is even and f contains (0, 0), f is odd and its graph is symmetric about the origin.

(b) Since f is decreasing when f ′ < 0, f decreases on the intervals (a, c) and (j, 1). Use the calculator to solve f′ (x) = 0. Conclude that f decreases on −6.202 < x < −3.294 and (symmetrically) on 3.294 < x < 6.202.

(c) f has a relative maximum at x = q if f ′(q) = 0 and if f changes from increasing (f ′ > 0) to decreasing (f ′ < 0) at q. There are two relative maxima here:
at x = a = −6.202 and at x = j = 3.294.

(d) f has a point of inflection when the graph of f changes its concavity; that is, when f′ changes from increasing to decreasing, as it does at points d and h, or when f′ changes from decreasing to increasing, as it does at points b, g, and k. So there are five points of inflection altogether.

4. In the graph below, C is the piece of the curve lying in the first quadrant. S is the region bounded by the curve C and the coordinate axes.

Image

(a) Graph Image in [0,3] × [0,5]. Since you want dy/dx, the slope of the tangent, where y = 1, use the calculator to solve

Image

(storing the answer at B). Then evaluate the slope of the tangent to C at y = 1:

f ′(B) ≈ −21.182.

(b) Since ΔA = yΔx, A = Image

(c) When S is rotated about the x-axis, its volume can be obtained using disks:

Image

5. See the figure, where R is the point (a, b), and seek a such that

Image

Image

6. Graph y = sin 2x in [−1, 3.2] × [−1, 1]. Note that y = f .

Image

(a) The graph of f is concave downward where f ″ is negative, namely, on (b, d). Use the calculator to solve sin 2x = 0, obtaining b = 1.651 and d = 2.651. The answer to (a) is therefore 1.651 < x < 2.651.

(b) f ′ has a relative minimum at x = d, because f ″ equals 0 at d, is less than 0 on (b, d), and is greater than 0 on (d, g). Thus f ′ has a relative minimum (from part a) at x = 2.651.

(c) The graph of f ′ has a point of inflection wherever its second derivative f ″′ changes from positive to negative or vice versa. This is equivalent to f ″ changing from increasing to decreasing (as at a and g) or vice versa (as at c). Therefore, the graph of f ′ has three points of inflection on [− 1, 3.2].

7. Graph f (x) = cos x and g(x) = x2 − 1 in [−2, 2] × [−2,2], Here, y1 = f and y2 = g.

Image

(a) Solve cos x = x2 − 1 to find the two points of intersection: (1.177,0.384) and (−1.177,0.384).

(b) Since ΔA = (y1 − y2) Δx = [f (x) − g(x)] Δx, the area A bounded by the two curves is

Image

8. (a) Use the Trapezoid Rule, with h = 60 min:

Image

(b) Draw a horizontal line at y = 20 (as shown on the graph below), representing the rate at which letters are processed then.

Image

(i) Letters began to pile up when they arrived at a rate greater than that at which they were being processed, that is, at t = 10 A.M.

(ii) The pile was largest when the letters stopped piling up, at t = 2 P.M.

(iii) The number of letters in the pile is represented by the area of the small trapezoid above the horizontal line: Image

(iv) The pile began to diminish after 2 P.M., when letters were processed at a rate faster than they arrived, and vanished when the area of the shaded triangle represented 1500 letters. At 5 P.M. this area is Image letters, so the pile vanished shortly after 5 P.M.

9. Draw a vertical element of area as shown below.

Image

(a) Let a represent the x-value of the positive point of intersection of y = x4 and y = sin x. Solving a4 = sin a with the calculator, we find a = 0.9496.

Image

(b) Elements of volume are triangular prisms with height h = 3 and base b = (sinxx4), as shown below.

Image

(c) When R is rotated around the x-axis, the element generates washers. If r1, and r2 are the radii of the larger and smaller disks, respectively, then

Image

10. The figure above shows an elliptical cross section of the tank. Its equation is

Image

(a) The volume of the tank, using disks, is Image where the ellipse’s symmetry about the x-axis has been exploited. The equation of the ellipse is equivalent to x2 = 6.25(100 − y2), so

Image

Use the calculator to evaluate this integral, storing the answer as V to have it available for part (b).

The capacity of the tank is 7.48V, or 196,000 gal of water, rounded to the nearest 1000 gal.

(b) Let k be the y-coordinate of the water level when the tank is one-fourth full. Then

Image

and the depth of the water is k + 10.

11. (a) Let h represent the depth of the water, as shown.

Image

Then h is the altitude of an equilateral triangle, and the base Image

The volume of water is

Image

Now Image and it is given that Image Thus, when h = 4,

Image

(b) Let x represent the length of one of the sides, as shown.

The bases of the trapezoid are 24 − 2x and 24 − 2x + Image and the height is Image

Image

The volume of the trough (in in.3) is given by

Image

Since Image the maximum volume is attained by folding the metal 8 inches from tne edges.

12. (a) Both π/4 and the expression in brackets yield 0.7853981634, which is accurate to ten decimal places.

(b) Image

(c) Image this agrees with the value of Image to four decimal places.

(d) The series

Image

converges very slowly. Example 56 evaluated the sum of 60 terms of the series for π (which equals 4 tan−1 1). To four decimal places, we get π = 3.1249, which yields 0.7812 for π/4—not accurate even to two decimal places.

13. (a) The given series is alternating. Since Image

Since ln x is an increasing function,

Image

The series therefore converges.

(b) Since the series converges by the Alternating Series Test, the error in using the first n terms for the sum of the whole series is less than the absolute value of the (n + 1)st term. Thus the error is less than Image Solve for n using Image

Image

The given series converges very slowly!

(c) The series Image is conditionally convergent. The given alternating series converges since the nth term approaches 0 and Image However, the nonnegative series diverges by the Integral Test, since

Image

14. (a) Solve by separation of variables:

Image

Let c = e−10c; then

Image

Now use initial condition y = 2 at t = 0:

Image

and the other condition, y = 5 at t = 2, gives

Image

(b) Since c = 4 and Image

Solving for y yields Image

(c) Image means 1 + 4 · 2t = 1.25, so t = 4.

(d) Image so the value of y approaches 10.

15. (a) Since Image Since y = 18 − 2 · 22 = 10, P is at (2,10).

(b) Since Image Since Image Therefore

Image

(c) Let D = the object’s distance from the origin. Then

Image

(d) The object hits the x-axis when y = 18 − 2x2 = 0, or x = 3. Since Image

(e) The length of the arc of y = 18 − 2x2 for 0 ≤ x ≤ 3 is given by

Image

16. (a) See graph.

Image

(b) You want to maximize Image

Image

See signs analysis.

Image

The maximum y occurs when t = 1, because y changes from increasing to decreasing there.

(c) Since x(1) = 4arctan 1 = π and Image the coordinates of the highest point are (π,6).

Since Image This vector is shown on the graph.

(d) Image Thus the particle approaches the point (2π,0).

17. (a) To find the smallest rectangle with sides parallel to the x- and y-axes, you need a rectangle formed by vertical and horizontal tangents as shown in the figure. The vertical tangents are at the x-intercepts, x = ±3. The horizontal tangents are at the points where y (not r) is a maximum. You need, therefore, to maximize

Image

Use the calculator to find that Image when θ = 0.7854. Therefore, y = 1.414, so the desired rectangle has dimensions 6 × 2.828.

(b) Since the polar formula for the area is Image the area of R (enclosed by r) is Image which is 14.137.

Part B

18. The graph shown below satisfies all five conditions. So do many others!

Image

19. (a) f′ is defined for all x in the interval. Since f is therefore differentiable, it must also be continuous.

(b) Because f′ (2) = 0 and f′ changes from negative to positive there, f has a relative minimum at x = 2. To the left of x = 9, f′ is negative, so f is decreasing as it approaches that endpoint and reaches another relative minimum there.

(c) Because f′ is negative to the right of x = −3, f decreases from its left end-point, indicating a relative max there. Also, f′ (2) = 0 and f′ changes from positive to negative there, so f has a relative minimum at x = 7.

(d) Note that f (7) − f (−3) = Image Since there is more area above the x-axis than below the x-axis on [−3,7], the integral is positive and f (7) − f (−3) > 0. This implies that f (7) > f (−3), and that the absolute maximum occurs at x = 7.

(e) At x = 2 and also at x = 6, f′ changes from increasing to decreasing, indicating that f changes from concave upward to concave downward at each. At x = 4, f′ changes from decreasing to increasing, indicating that f changes from concave downward to concave upward there. Hence the graph of f has points of inflection at x = 2, 4, and 6.

20. Draw a sketch of the region bounded above by y1 = 8 − 2x2 and below by y2 = x2 − 4, and inscribe a rectangle in this region as described in the question. If (x, y1) and (x, y2) are the vertices of the rectangle in quadrants I and IV, respectively, then the area

A = 2x (y1y2) = 2x(12 − 3x2), or A(x) = 24x − 6x3.

Then A′ (x) = 24 − 18x2 = 6(4 − 3x2), which equals 0 when Image Check to verify that A (x) < 0 at this point. This assures that this value of x yields maximum area, which is given by Image

21. The graph of f ′(x) is shown here.

Image

22. The rate of change in volume when the surface area is 54 ft3 is Image ft3 /sec.

23. See the figure. The equation of the circle is x2 + y2 = a2; the equation of RS is y = ax. If y2 is an ordinate of the circle and y1 of the line, then,

Image

24. (a) The region is sketched in the figure. The pertinent points of intersection are labeled.

Image

(b) The required area consists of two parts. The area of the triangle is represented by Image and is equal to 1, while the area of the region bounded at the left by x = 1, above by y = x + 4, and at the right by the parabola is represented by Image This equals

Image

The required area, thus, equals Image

25. (a) 1975 to 1976 and 1978 to 1980.

(b) 1975 to 1977 and 1979 to 1981.

(c) 1976 to 1977 and 1980 to 1981.

26. (a) Since Image then, separating variables, Image Integrating gives

Image

and, since v = 20 when t = 0, C = ln 20. Then (1) becomes ln Imageor, solving for v,

Image

(b) Note that v > 0 for all t. Let s be the required distance traveled (as v decreases from 20 to 5); then

Image

where, when v = 20, t = 0. Also, when v = 5, use (2) to get Image or − ln 4 = −2t. So t = ln 2. Evaluating s in (3) gives

Image

27. Let (x,y) be the point in the first quadrant where the line parallel to the x-axis meets the parabola. The area of the triangle is given by

A = xy = x(27 − x2) = 27xx3 for 0 ≤ xImage

Then A′ (x) = 27 − 3x2 = 3(3 + x)(3 − x), and A′ (x) = 0 at x = 3.

Since A′ changes from positive to negative at x = 3, the area reaches its maximum there.

The maximum area is A(3) = 3(27 − 32) = 54.

28. Image

(b) Using the Ratio Test, you know that the series converges when Image that is, when |x| < 1, or −1 < x < 1. Thus, the radius of convergence is 1.

(c) Image

(d) Since the series converges by the Alternating Series Test, the error in the answer for (c) is less absolutely than Image

29. From the equations for x and y,

dx = (1 − cos θ) dθ and dy = sin θ dθ.

(a) The slope at any point is given by Image which here is Image When Image

(b) When Image The equation of the tangent is Image

30. Both curves are circles with centers at, respectively, (2,0) and Image the circles intersect at Image The common area is given by

Image

The answer is 2(π − 2).

31. (a) For f (x) = cos x, f′ (x) = −sin x, f (x) = −cos x, f ′″(x) = sin x, f(4) (x) = cos x, f(5) (x) = −sin x, f(6) (x) = −cos x. The Taylor polynomial of order 4 about 0 is

Image

Note that the next term of the alternating Maclaurin series for cos x is Image

(b) Image

(c) The error in (b), convergent by the Alternating Series Test, is less absolutely than the first term dropped:

Image

32. (a) Since Image y = 2t + 1 and x = 4t3 + 6t + 3t.

(b) Since Image then, when t = 1, |a| = 36.

33. See the figure. The required area A is twice the sum of the following areas: that of the limaçon from 0 to Image and that of the circle from Image Thus

Image

Answers Explained

Multiple-Choice

Part A

1. (C) Use the Rational Function Theorem.

2. (C) Note that Image where f (x) = ln x.

3. (B) Since y ′ = −2xe−x2, therefore y ″ = −2(x · e x2 · (−2x) + e x2). Replace x by 0.

4. (B) Image

5. (B) h ′(3) = g ′(f (3)) · f ′(3) = g ′(4) · Image

6. (E) Since f ′(x) exists for all x, it must equal 0 for any x0 for which f is a relative maximum, and it must also change sign from positive to negative as x increases through x0. For the given derivative, no x satisfies both of these conditions.

7. (E) By the Quotient Rule (formula (6)),

Image

8. (A) Here, f ′(x) is e x (1 − x); f has maximum value when x = 1.

9. (A) Note that (1) on a horizontal line the slope segments are all parallel, so the slopes there are all the same and Image must depend only on y; (2) along the x-axis (where y = 0) the slopes are infinite; and (3) as y increases, the slope decreases.

10. (E) Acceleration is the derivative (the slope) of velocity v; v is largest on 8 < t < 9.

11. (C) Velocity v is the derivative of position; because v > 0 until t = 6 and v < 0 thereafter, the position increases until t = 6 and then decreases; since the area bounded by the curve above the axis is larger than the area below the axis, the object is farthest from its starting point at t = 6.

12. (D) From t = 5 to t = 8, the displacement (the integral of velocity) can be found by evaluating definite integrals based on the areas of two triangles:

Image Thus, if K is the object’s position at t = 5, then K − 3 = 10 at t = 8.

13. (A) The integral is of the form Image evaluate Image

14. (E) Image

15. (D) Image

16. (A) f (x) = e x is decreasing and concave upward.

17. (B) Implicit differentiation yields 2yy ′ = 1; so Image At a vertical tangent, Image is undefined; y must therefore equal 0 and the numerator be non-zero. The original equation with y = 0 is 0 = xx3, which has three solutions.

18. (B) Let t = x − 1; then t = −1 when x = 0, t = 5 when x = 6, and dt = dx.

19. (B) The required area, A, is given by the integral

Image

20. (B) The average value is Image The definite integral represents the sum of the areas of a trapezoid and a rectangle: Image (8 + 3)(6) = 4(7) = 61.

21. (A) Solve the differential equation Image by separation of variables: Image yields y = ce2x. The initial condition yields 1 = ce2 · 2; so c = e −4 and y = e2x−4.

22. (C) Changes in values of f ″ show that f ″′ is constant; hence f ″ is linear, f ′ is quadratic, and f must be cubic.

23. (B) By implicit differentiation, Image so the equation of the tangent line at (3,0) is y = −9(x−3).

24. (A) Image

25. (D) The graph shown has the x-axis as a horizontal asymptote.

26. (B) Since Image to render f (x) continuous at x = 1 f (1) must be defined to be 1.

27. (B) f ′(x) = 15x4 − 30x2; f ″(x) = 60x3 − 60x = 60x(x + 1)(x − 1); this equals 0 when x = −1, 0, or 1. Here are the signs within the intervals:

Image

The graph of f changes concavity at x = −1, 0, and 1.

28. (C) Note that Image so f has a critical value at x = −4. As x passes through −4, the sign of f ′ changes from − to +, so f has a local minimum at x = −4.

Part B

29. (B) We are given that (1) f ′(a) > 0; (2)f ″(a) < 0; and (3) G ′(a) < 0. Since G ′(x) = 2 f (x) · f ′(x), therefore G ′(a) = 2f (a) · f (a). Conditions (1) and (3) imply that (4)f (a) < 0. Since G ″(x) = 2[f (x) · f ″ (x) + (f ′(x))2], therefore G″(a) = 2[f (a)f ″ (a) + (f′ (a))2]. Then the sign of G ″(a) is 2[(−) · (−) + (+)] or positive, where the minus signs in the parentheses follow from conditions (4) and (2).

30. (E) Since Image it equals 0 for Image When x = 3, c = 9; this yields a minimum since f ″(3) > 0.

Image

31. (E) Use your calculator to graph velocity against time. Speed is the absolute value of velocity. The greatest deviation from v = 0 is at t = c. With a calculator, c = 9.538.

32. (D) Because f ′ changes from increasing to decreasing, f ″ changes from positive to negative and thus the graph of f changes concavity.

33. (D) Image We evaluate this definite integral by finding the area of a trapezoid (negative) and a triangle: Image so the tangent line passes through the point (3,−2). The slope of the line is H ′(3) = f (3) = 2, so an equation of the line is y − (−2) = 2(x − 3).

34. (D) The distance is approximately 14(6) + 8(2) + 3(4).

35. (D) Image R(x)dx = 166.396.

36. (A) Selecting an answer for this question from your calculator graph is unwise. In some windows the graph may appear continuous; in others there may seem to be cusps, or a vertical asymptote. Put the calculator aside. Find

Image

These limits indicate the presence of a jump discontinuity in the function at x = 1.

37. (D) Image

Image

38. (E) Image

Image

39. (E) In the figure above, S is the region bounded by y = sec x, the y axis, and y = 4. Send region S about the x-axis. Use washers; then ΔV = π(R2r2) Δx. Symmetry allows you to double the volume generated by the first quadrant of S, so V is

Image

A calculator yields 108.177.

40. (A) The curve falls when f ′(x) < 0 and is concave up when f ″(x) > 0.

41. (B) Image To find g ′(0), find x such that f (x) = 0. By inspection, Image

Image

42. (D) It is given that Image you want Image where Image

Image

Since y2 = 25 −x2, it follows that Image and, when x = 3, y = 4

Image

Image

The function f (x) = 2 sin x + sin 4x is graphed above.

43. (E) Since f (0) = f (π) and f is both continuous and differentiable, Rolle’s Theorem predicts at least one c in the interval such that f ′(c) = 0.

There are four points in [0,π] of the calculator graph above where the tangent is horizontal.

44. (B) Since Image a positive constant, Image where c is a positive constant. Then Image which is also positive.

45. (C) If Q(t) is the amount of contaminant in the tank at time t and Q0 is the initial amount, then

Image = kQ and Q(t) = Q0 ekt.

Since Q(1) = 0.8Q0, 0.8Q0 = Q0ek · 1, 0.8 = ek, and

Q(t) = Q0(0.8)t.

We seek t when Q(t) = 0.02Q0. Thus,

0.02Q0 = Q0(0.8)t

and

t Image 17.53 min.

Free-Response

Part A

AB/BC 1. (a) This is the graph of f ′(x).

Image

(b) f is increasing when f ′(x) > 0. The graph shows this to be true in the interval a < x < b. Use the calculator to find a and b (where ex − 2 cos 3x = 0); then a = 0.283 < x < 3.760 = b.

(c) See signs analysis.

Image

Since f decreases to the right of endpoint x = 0, f has a local maximum at x = 0. There is another local maximum at x = 3.760, because f changes from increasing to decreasing there.

(d) See signs analysis.

Image

Since the graph of f changes concavity at p, q, and r, there are three points of inflection.

AB2. (a) Since April 1 is 3 months from January 1 and June 30 is 3 months later, we form the sum for the interval [3,6]:

Image

We estimate the company sold 1051 software units during the second quarter.

(b) S(t) = 1.2(2)t/3

(c) Image The model’s estimate of 1039 sales is slightly lower, but the two are in close agreement.

(d) Image the model predicts an average sales rate of 649.2 units per month from January 1, 2012, through December 31, 2012.

Part B

AB 3. (a) At (2,5), Image so the tangent line is y − 5 = 4(x − 2).

Solving for y yields f (x) ≈ 5 + 4(x − 2).

(b) f (2.1) ≈ 5 + 4(2.1 − 2) = 5.4.

(c) The differential equation Image is separable:

Image

Since f passes through (2,5), it must be true that Image

Thus c = 9, and the positive root is used.

The solution is Image

(d) Image

AB 4. (a) Draw a vertical element of area, as shown.

Image

Image

(b) (i) Use washers; then

Image

(ii) See the figure above.

Image

AB/BC 5. Image

f is decreasing where f ′(x) < 0, which occurs for −2 < x < 1.

(b) f is decreasing on the interval −2 < x < 1, so there is a minimum at (1, −e2). Note that, as x approaches ±∞, f (x) = e2x(x2 − 2) is always positive. Hence (1,−e2) is the global minimum.

(c) As x approaches +∞, f (x) = e2x(x2 − 2) also approaches +∞. There is no global maximum.

AB/BC 6. (a) S = 4πr2, so Image Substitute given values; then

Image

Since Image therefore Image Substituting known values

gives Image

(b) Regions of consistent density are concentric spherical shells. The volume of each shell is approximated by its surface area (4πx2) times its thickness (Δx). The weight of each shell is its density times its volume (g/cm3 · cm3). If, when the snowball is 12 cm in diameter, ΔG is the weight of a spherical shell x cm from the center, then Image and the integral to find the weight of the snowball is

Image

Answers Explained

Multiple-Choice

Part A

1. (E) Image

2. (A) Divide both numerator and denominator by Image

3. (E) Since eln u = u, y = 1.

4. (D) f (0) = 3, and, Image

5. (B) Image

6. (D) Here y ′ = 3 sin2 (1 − 2x) cos (1 − 2x) · (−2).

7. (B) Image

8. (B) Let s be the distance from the origin: then

Image

Since Image

9. (B) For Image this limit represents f ′(25).

10. (B) Image This definite integral represents the area of a quadrant of the circle x2 + y2 = 1, hence Image

11. (C) Image

12. (A) The integral is rewritten as

Image

13. (B) Image

14. (D) Note:

Image

15. (B) The winning times are positive, decreasing, and concave upward.

16. (E) G(x) = H(x) + Image represents the area of a trapezoid.

17. (C) f ′(x) = 0 for x = 1 and f ″(1) > 0.

Image

18. (B) Solution curves appear to represent odd functions with a horizontal asymptote. In the figure above, the curve in (B) of the question has been superimposed on the slope field.

19. (B) Note that

Image

20. (C) v is not differentiable at t = 3 or t = 5.

Image

21. (B) Speed is the magnitude of velocity; its graph is shown in the answer explanation for question 20.

22. (B) The average rate of change of velocity is Image

23. (E) The curve has vertical asymptotes at x = 2 and x = −2 and a horizontal asymptote at y = −2.

24. (E) The function is not defined at x = −2; Image Defining f (−2) = 4 will make f continuous at x = −2, but f will still be discontinuous at x = 1.

25. (B) Image

26. (A) Image

27. (E) ln (4 + x2) = ln (4 + (−x)2); y Image

28. (A) f (x) = Image (x sin πx) = πx cos πx + sin πx.

Part B

29. (B) See the figure below. A = Image

Image

30. (C) See the figure below. About the x-axis: Washer. ΔV = π(y2 − 12) Δx,

Image

31. (E) We solve the differential equation Image by separation:

Image

If s = 1 when t = 0, we have C = 1; hence, Image when t = 1.

32. (D) Image

The roots of f (x) = x2 − 4x − 5 = (x − 5)(x + 1) are x = −1 and 5. Since areas A and B are equal, therefore Image Thus,

Image

Solving on a calculator gives k (or x) equal to 8.

33. (D) If N is the number of bacteria at time t, then N = 200ekt. It is given that 3 = e10k. When t = 24, N = 200e24k. Therefore N = 200(e10k)2.4 = 200(3)2.4 Image 2793 bacteria.

34. (C) Since Image For x = 2t − 1, t = 3 yields x = 5 and t = 5 yields x = 9.

35. (C) Using implicit differentiation on the equation

x3 + xyy2 = 10

yields

Image

The tangent is vertical when Image is undefined; that is, when 2yx = 0.

Replacing y by Image in (1) gives

Image

or

4x3 + x2 = 40.

Let y1 = 4x3 + x2 − 40. Inspection of the equation y1 = f (x) = 0 reveals that there is a root near x = 2. Solving on a calculator yields x = 2.074.

36. (D) G ′(x) = f (3x − 1) · 3.

37. (B) Since f changes from positive to negative at t = 3, G ′ does also where 3x − 1 = 3.

38. (D) Using your calculator, evaluate y ′(2).

Image

39. (E) 2(3) + 2(0) + 2(4) + 2(1).

See the figure above.

40. (E) Image

At x = 3, the answer is 2[2(−2) + 52] = 42.

41. (A) The object is at rest when v(t) = ln(2 − t2) = 0; that occurs when 2−t2 = 1, so t = 1. The acceleration is Image

42. (D) Image

implies that x = 3.

43. (C) Image represents the rate of change of the surface area; if y is inversely proportional to x, then, Image

Image

44. (E) The velocity functions are

v1 = −2t sin (t2 + 1)

and

Image

Graph both functions in [0, 3] × [−5, 5]. The graphs intersect four times during the first 3 sec, as shown in the figure above.

45. (B) Image

Free-Response

Part A

AB/BC1. Image

(d) To work with g(x) = f −1(x), interchange x and y:

x

7.6

5.7

4.2

3.8

2.2

1.6

g(x)

2.5

3.2

3.5

4.0

4.5

5.0

Image

AB 2. Let M = the temperature of the milk at time t. Then

Image

The differential equation is separable:

Image

where c = e C.

Find c, using the fact that M = 40° when t = 0:

40 = 68−ce0 means c = 28.

Find k, using the fact that M = 43° when t = 3:

Image

Hence Image

Now find t when M = 60°:

Image

Since the phone rang at t = 3, you have 30 min to solve the problem.

Part B

AB 3. (a) Image

See the figure.

(b) The average value of a function on an interval is the area under the graph of the function divided by the interval width, here Image

(c) From part (a) you know that the area of the region is given by ImageSince Image as k increases the area of the region approaches 3π.

AB/BC 4. (a) The rectangular slices have base y, height 5y, and thickness along the x-axis:

Image

(b) The disks have radius x and thickness along the y-axis:

Image

Now we solve for x in terms of y:

Image

(NOTE: Although the shells method is not a required AP topic, another correct integral for this volume is Image

AB/BC 5. (a) The volume of the cord is V = πr2h. Differentiate with respect to time, then substitute known values. (Be sure to use consistent units; here, all measurements have been converted to inches.)

Image

(b) Let Image represent the angle of elevation and h the height, as shown.

Image

When h = 80, your distance to the jumper is 100 ft, as shown.

Then

Image

Image

AB/BC 6. (a) Image the negative of the area of the shaded rectangle in the figure. Hence F(−2) = − (3)(2) = −6.

Image

Image is represented by the shaded triangles in the figure.

Image

(b) Image so F(x) = 0 at x = 1. Image because the regions above and below the x-axis have the same area. Hence F(x) = 0 at x = 3.

Image

(c) F is increasing where F ′ = f is positive: −2 ≤ x ≤ 2.

(d) The maximum value of F occurs at x = 2, where F ′ = f changes from positive to negative. Image

The minimum value of F must occur at one of the endpoints. Since F(−2) = − 6 and F(6) = −3, the minimum is at x = −2.

(e) F has points of inflection where F ″ changes sign, as occurs where F ′ = f goes from decreasing to increasing, at x = 3.

Answers Explained

Multiple-Choice

Part A

1. (E) Here, Image

2. (C) The given limit equals Image where f (x) = sin x.

3. (C) Since f (x) = x ln x,

f ′(x) = 1 + ln x, Image

4. (A) Differentiate implicitly to get Image Substitute (−1, 1) to find Image the slope at this point, and write the equation of the tangent: y − 1 = −1(x + 1).

5. (C) f ′(x) = 4x3 − 12x2 + 8x = 4x(x − 1)(x − 2). To determine the signs of f ′(x), inspect the sign at any point in each of the intervals x < 0, 0 < x < 1, 1 < x < 2, and x > 2. The function increases whenever f ′(x) > 0.

6. (C) The integral is equivalent to Image where u = 4 + 2sinx.

7. (D) Here Image which is zero for x = e. Since the sign of y ′ changes from positive to negative as x increases through e, this critical value yields a relative maximum. Note that Image

8. (E) Since Image is always positive, there are no reversals in motion along the line.

Image

9. (E) The slope field suggests the curve shown above as a particular solution.

10. (E) Since Image is discontinuous at x = 1; the domain of F is therefore x > 1. On [2, 5] f (x) < 0, so Image which is positive for x > 1.

Image

11. (B) In the graph above, W(t), the water level at time t, is a cosine function with amplitude 6 ft and period 12hr:

Image

12. (B) Solve the differential equation Image Use x = −1,

y = 2 to determine Image

13. (D) Image

14. (A) Image

15. (C) G ′(2) = 4, so G (x) Image 4(x − 2) + 5.

16. (E) Image See the figure below.

Image

17. (E) Note that Image

18. (B) Note that (0, 0) is on the graph, as are (1, 2) and (−1, −2). So only (B) and (E) are possible. Since Image only (B) is correct.

19. (D) See the figure.

Image

20. (E) Image

21. (D) In (D), f (x) is not defined at x = 0. Verify that each of the other functions satisfies both conditions of the Mean Value Theorem.

22. (E) The signs within the intervals bounded by the critical points are given below.

Image

Since f changes from increasing to decreasing at x = −3, f has a local maximum at −3. Also, f has a local minimum at x = 0, because it is decreasing to the left of zero and increasing to the right.

23. (B) Since ln Image then

Image

24. (D) Image

25. (B) As seen from the figure, Image where y = 2r,

Image

26. (C) From the figure below, Image

Image

27. (A) Since the degrees of numerator and denominator are the same, the limit as x→∞ is the ratio of the coefficients of the terms of highest degree: Image

28. (A) We see from the figure that ΔA = (y2y1) Δx;

Image

Image

Part B

29. (A) Let u = x2 + 2. Then

Image

and

Image

30. (E) Image will increase above the half-full level (that is, the height of the water will rise more rapidly) as the area of the cross section diminishes.

31. (B) Since Image

32. (C) Image

The initial condition H(0) = 120 shows c = 50. Evaluate H(10).

33. (B) Let P be the amount after t years. It is given that Image The solution of this differential equation is P = 4000e0.08t + C, where P (0) = 4000 yields C = 0. The answer is 4000e(0.08) ·10.

34. (C) The inverse of y = 1 + ex is x = 1 + ey or y = ln (x − 1); (x − 1) must be positive.

35. (E) Speed is the magnitude of velocity: |v(8)| = 4.

36. (E) For 3 < t < 6 the object travels to the right Image At t = 7 it has returned 1 unit to the left; by t = 8, 4 units to the left.

Image

37. (C) Use disks: ΔV = πr2Δy = πx2Δy, where x = ln y. Use your calculator to evaluate Image

38. (B) If Image then u2 = x − 2, x = u2 + 2, dx = 2u du. When x = 3, u = 1; when x = 6, u = 2.

39. (A) The tangent line passes through points (8,1) and (0,3). Its slope, Image is f ′(8).

40. (A) Graph f ″ in [0,6] × [−5,10]. The sign of f ″ changes only at x = a, as seen in the figure.

Image

41. (C) In the graph below, the first rectangle shows 2 tickets sold per minute for 15 min, or 30 tickets. Similarly, the total is 2(15) + 6(15) + 10(15) + 4(15).

Image

42. (B) Graph both functions in [−8,8] × [−5,5]. At point of intersection Q, both are decreasing. Tracing reveals x Image −4 at Q. If you zoom in on the curves at x = T, you will note that they do not actually intersect there.

Image

43. (D) Counterexamples are, respectively, for (A), f (x) = |x|, c = 0; for (B), f (x) = x3, c = 0; for (C), f (x) = x4, c = 0; for (E), f (x) = x2 on (−1, 1).

44. (E) f ′(x) > 0; the curve shows that f ′ is defined for all a < x < b, so f is differentiable and therefore continuous.

45. (D) Consider the blast area as a set of concentric rings; one is shown in the figure. The area of this ring, which represents the region x meters from the center of the blast, may be approximated by the area of the rectangle shown. Since the number of particles in the ring is the area times the density, ΔP = 2πx · Δx · N(x). To find the total number of fragments within 20 m of the point of the explosion, integrate: Image

Image

Free-Response

Part A

AB1. (a) Since x2y − 3y2 = 48,

Image

(b) At (5,3), Image so the equation of the tangent line is

Image

(c) Image

(d) Horizontal tangent lines have Image This could happen only if

2xy = 0, which means that x = 0 or y = 0.

If x = 0, 0y − 3y2 = 48, which has no real solutions.

If y = 0, x2 · 0 − 3 · 02 = 48, which is impossible. Therefore, there are no horizontal tangents.

AB/BC2. (a) Image

(b) The average value of a function is the integral across the given interval divided by the interval width. Here Image Estimate the value of the integral using trapezoid rule T with values from the table and Δt = 4:

Image

Hence

Avg(W) Image 35.167 ft.

(c) For Image use your calculator to evaluate F ′(16) ≈ −0.749. After 16 hr, the river depth is dropping at the rate of 0.749 ft/hr.

Image

Part B

AB/BC 3. (a) Image

(b) Let h = the hypotenuse of an isosceles right triangle, as shown in the figure. Then each leg of the triangle is Image and its area is Image

Image

An element of volume is

Image

and thus Image

Image

AB/BC 4. (a) Image so v(0) = 0 and v(10) = 48.

The average acceleration is Image

Acceleration Image

Image

(b) Since Q’s acceleration, for all t in 0 ≤ t ≤ 5, is the slope of its velocity graph, Image

(c) Find the distance each auto has traveled. For P, the distance is

Image

For auto Q, the distance is the total area of the triangle and trapezoid under the velocity graph shown below, namely,

Image

Auto P won the race.

AB5. (a) Using the differential equation, evaluate the derivative at each point, then sketch a short segment having that slope. For example, at (−1, −1), Image = 2(−1)((−1)2 + 1) − 4; draw a steeply decreasing segment at (−1, −1). Repeat this process at each of the other points. The result follows.

Image

(b) The differential equation Image is separable.

Image

It is given that f passes through (0,1), so 1 = tan (02 + c) and Image

The solution is f(x) = tan Image

The particular solution must be differentiable on an interval containing the initial point (0,1). The tangent function has vertical asymptotes at Image hence:

Image (Since x2 ≥ 0, we ignore the left inequality.)

Image

AB 6. (a) Image f (t)dt = F(2) = 4.

(b) One estimate might be Image f (t)dt F(7) − F(2) = 2 − 4 = −2.

(c) f (x) = F′ (x); F′ (x) = 0 at = 4.

(d) f ′(x) = F ″(x). F ″ is negative when F is concave downward, which is true for the entire interval 0 < x < 8.

(e) Image

Then the graph of G is the graph of F translated downward 4 units.

Image

Answers Explained

The explanations for questions not given below will be found in the answer section for AB Practice Examination 1. Identical questions in Section I of Practice Examinations AB1 and BC1 have the same number. For example, explanations of the answers for Question 1, not given below, will be found in Section I of Calculus AB Practice Examination 1, Answer 1.

Multiple-Choice

Part A

3. (E) Here,

Image

6. (D) Image

12. (B) Note that, when x = 2 sin θ, x2 = 4 sin2 θ, dx = 2 cos dθ, and Image 2 cos θ. Also,

Image

13. (C) The given integral is equivalent to Image

Image

The figure shows the graph of f (x) = 1 − |x| on [−1,1].

The area of triangle PQR is equal to Image

14. (B) Let y = x1/x; then take logarithms. ln Image As x → ∞ the fraction is of the form ∞/∞. Image So ye0 or 1.

17. (E) Separating variables yields Image so ln y = −ln cos x + C. With y = 3 when x = 0, C = ln 3. The general solution is therefore (cos x) y = 3. When Image

Image and y = 6.

20. (A) Represent the coordinates parametrically as (r cos θ, r sin θ). Then

Image

Note that Image sin 2θ, and evaluate Image (Alternatively, write x = cos 2θ cos θ and y = cos 2θ sin θ to find Image)

21. (D) Note that v is negative from t = 0 to t = 1, but positive from t = 1 to t = 2. Thus the distance traveled is given by

Image

22. (E) Separating variables yields y dy = (1 − 2x) dx. Integrating gives

Image

24. (A) Use parts; then u = x, dv = cos x dx; du = dx,v = sin x. Thus,

Image

25. (C) Using the above figure, consider a thin slice of the oil, and the work Δw done in raising it to the top of the tank:

Image

26. (A) By Taylor’s Theorem, the coefficient is Image For Image and Image making the coefficient Image

27. (D) Here,

Image

28. (D) Evaluate

Image

Part B

29. (E) The vertical component of velocity is

Image = 4 cos t + 12 cos 12t.

Evaluate at t = 1.

30. (B) At t = 0, we know H = 120, so 120 = 70 + ke −0.4(0), and thus k = 50. The average temperature for the first 10 minutes is Image

36. (B)

Image

Image

The required volume is 0.592.

38. (C) The Maclaurin expansion is

Image

The Lagrange remainder R, after n terms, for some c in the interval |x| Image 2, is

Image

Since R is greatest when c = 2, n needs to satisfy the inequality

Image

Using a calculator to evaluate Image successively at various integral values of x gives y(8) > 0.01, y(9) > 0.002, y(10) < 3.8 × 10 −4 < 0.0004. Thus we achieve the desired accuracy with a Taylor polynomial at 0 of degree at least 10.

Image

39. (E) On your calculator, graph one arch of the cycloid for t in [0, 2π] and (x,y) in [0, 7] × [−1, 3]. Use disks; then the desired volume is

Image

40. (C) In the first quadrant, both x and y must be positive; x(t) = et is positive for all t, but y(t) = 1 − t2 is positive only for −1 < t < 1. The arc length is

Image

41. (D) Chapter 10.

44. (E) Each is essentially a p-series, Image Such a series converges only if p > 1.

Free-Response

Part A

1. See solution for AB-1, Free-Response in AB Practice Exam One.

2. See solution for AB2, Free-Response in AB Practice Exam One.

Part B

3. (a) Because Image which never equals zero, the object is never at rest.

(b) Image so the object’s speed is

Image

Position is the antiderivative of velocity Image

Image

Since P(0) = (0,0), arcsin Image and thus c = 0.

Image

Since P(0) = (0,0), Image and thus c = 2.

Then

Image

(c) Solving Image for t yields t = 2 sinx. Therefore

Image

Since 0 ≤ t ≤ 1 means Image then cos x > 0, so y = 2 − 2cos x.

4. (a) To write the Maclaurin series for f (x) = ln(e+x), use Taylor’s theorem at x = 0.

Image

Image

(b) By the Ratio Test, the series converges when

Image

Thus, the radius of convergence is e.

Image

5. (a) To find the maximum rate of growth, first find the derivative of

Image

A signs analysis shows that Image changes from positive to negative there, confirming that Image is at its maximum when there are 300 trout.

(b) The differential equation Image is separable.

Image

To integrate the left side of this equation, use the method of partial fractions.

Image

(c) In (a) the population was found to be growing the fastest when F = 300. Then:

Image

6. See solution for AB/BC 6, Free-Response in AB Practice Exam One.

Answers Explained

The explanations for questions not given below will be found in the answer section for AB Practice Exam 2. Identical questions in Section I of Practice Examinations AB2 and BC2 have the same number. For example, explanations of the answers for Questions 4 and 5, not given below, will be found in Section I of Calculus AB Practice Exam 2, Answers 4 and 5.

Multiple-Choice

Part A

1. (B) Since Image to render f (x) continuous at x = 1, define f (1) to be 1.

2. (C) Note that

Image

where you let Image

3. (C) Obtain the first few terms of the Maclaurin series generated by Image

Image

6. (C) Here,

Image

and the magnitude of the acceleration, |a|, is given by

Image

7. (B) Image

By the Ratio Test the series converges when

Image

Checking the endpoints, we find:

Image is the alternating harmonic series, which converges.

Image is the harmonic series, which diverges.

Hence the interval of convergence is 0 ≤ x < 2.

10. (C) Image

11. (D) We integrate by parts using u = ln x, dv = dx; then Image v = x, and

Image

13. (C) Use the method of partial fractions, letting

Image

Letting x = 0, we find A = 2, and letting x = 3 yields B = −1.

Now Image

19. (E) At (2,1), Image Use Δx = 0.1; then Euler’s method moves to (2.1, 1 + 3(0.1)).

At (2.1, 1.3), Image so the next point is (2.2, 1.3 + 3.4(0.1)).

22. (D) Separate variables to get Image and integrate to get ln y = ln x + C.

Since y = 3 when x = 1, C = ln 3. Then y = e(ln x + ln 3) = eln x · eln 3 = 3x.

23. (D) The generating circle has equation x2 + y2 = 4. Using disks, the volume, V, is given by

Image

24. (C) Image The integrals in (A), (B), and (D) all diverge to infinity.

26. (D) Using separation of variables:

Image

Given initial point (0,0), we have 0 = tan(0 + C); hence C = 0 and the particular solution is y = tan(x).

Because this function has vertical asymptotes at Image and the particular solution must be differentiable in an interval containing the initial point x = 0, the domain is Image

28. (D) Image

Part B

30. (C) Since the equation of the spiral is r = lnImage, use the polar mode. The formula for area in polar coordinates is

Image

Therefore, calculate

Image

The result is 3.743.

31. (D) When y = 2t = 4, we have t = 2, so the line passes through point F(2) = (5,4).

Also Image so at t = 2 the slope of the tangent line is

Image

An equation for the tangent line is y − 4 = ln 2(x − 5).

32. (A) I. Image converges by the Ratio Test: Image

II. Image diverges by the nth Term Test: Image

III. Image diverges by the Comparison Test: Image diverges.

38. (E) If Q0 is the initial amount of the substance and Q is the amount at time t, then

Image

and Image Thus k = 0.08664. Using a calculator, find t when Image

Image so t ≈ 12.68. Don’t round off k too quickly.

40. (A) See figure below.

Image

Image

41. (D) See figure below.

Image

42. (E) Image

43. (C) The endpoints of the arc are Image and (e,1). The arc length is given by

Image

44. (B) Find k such that cos x will differ from Image by less than 0.001 at x = k.

Solve

Image

which yields x or k = 0.394.

Free-Response

Part A

1. See solution for AB-1, Free-Response in AB Practice Exam Two.

2. (a) Position is the antiderivative of Image

Image

To find c, substitute the initial condition that x = 1 when t = 0:

Image

At t = 2, x = 2 arctan Image and the position of the object is Image

Image

(c) The distance traveled is the length of the arc of y = x2 + 2 in the interval Image

Image

Part B

3. (a)

Image

(b) By the Ratio Test, the series converges when

Image

Thus, the radius of convergence is 2.

(c) Image is an alternating series. Since Image and Image it converges by the Alternating Series Test. Therefore the error is less than the magnitude of the first omitted term:

Image

4. See solution for AB/BC 4, Free-Response in AB Practice Exam Two.

5. See solution for AB-5, Free-Response in AB Practice Exam Two.

6. See solution for AB-6, Free-Response in AB Practice Exam Two.

Answers Explained

The explanations for questions not given below will be found in the answer section for AB Practice Exam 3. Identical questions in Section I of Practice Examinations AB3 and BC3 have the same number. For example, explanations of the answers for Questions 1 and 2, not given below, will be found in Section I of Calculus AB Practice Exam 3, Answers 1 and 2.

Multiple-Choice

Part A

3. (B) The series Image is geometric with Image it converges

to Image

5. (D) Image is the sum of an infinite geometric series with first term 1 and common ratio −2x. The series is 1 − 2x + 4x2 − 8x3 + 16x4 − ….

6. (A) Assume that

Image

Then

2x2x + 4 = A(x − 1)(x − 2) + Bx(x − 2) + Cx(x − 1).

Since you are looking for B, let x = 1:

2(1) − 1 + 4 = 0 + B(−1) + 0; B = −5.

8. (B) Since ex Image 1 + x,e x2 Image 1 − x2. So

Image

12. (A) Image

17. (E) Image

18. (D) See the figure below, which shows that the length of a semicircle of radius 2 is needed here. The answer can, of course, be found by using the formula for arc length:

Image

20. (E) Image using long division, Image

22. (C) Using parts we let u = x2, dv = exdx; then du = 2x dx, v = ex, and

Image

We use parts again with u = x, dv = exdx; then du = dx, v = ex, and

Image

Now Image

23. (E) Image will increase above the half-full level (that is, the height of the water will rise more rapidly) as the area of the cross section diminishes.

27. (A) The required area is lined in the figure below.

Image

28. (D) Note that f (x) = x + 6 if x ≠ 6, that f (6) = 12, and that Image So f is defined and continuous at x = 6.

Part B

29. (E) The velocity functions are

Image

When these functions are graphed on a calculator, it is clear that they intersect four times during the first 3 sec, as shown below.

Image

30. (C) Changes in values of f show that f ″′ may be constant. Hence f may be linear, so f′ could be quadratic and thus f cubic.

31. (C) Expressed parametrically, x = sin3θ cosθ, y = sin3θ sin θ. Image is undefined where Image = −sin3θ sin θ + 3 cos 3θ cos θ = 0.

Use your calculator to solve for θ.

34. (D) See the figure below.

Image

The roots of f (x) = x2 − 4x − 5 = (x − 5)(x + 1) are x = −1 and 5. Since areas A and B are equal, therefore, Image Thus,

Image

A calculator yields k = 8.

37. (C) It is given that Image An antiderivative is Image

Since Image the constants are c1 = c2 = 1. The object’s speed is Image

Use a calculator to find that the object’s maximum speed is 2.217.

Image

39. (D) Use the Ratio Test:

Image

which is less than 1 if −3 < x < 3. When x = −3, the convergent alternating harmonic series is obtained.

40. (A) Since Image and since v(0) = 1, C = 1. Then Image yields s = t3 + t + C ′, and you can let s(0) = 0. Then you want s(3).

41. (A) Arc length is given by Image Here the integrand Image implies that Image hence, y = 3 ln x + C. Since the curve contains (1,2), 2 = 3 ln 1 + C, which yields C = 2.

42. (E) Image

At x = 3; the answer is 2[2(−2) + 52] = 42.

43. (D) Counterexamples are, respectively, for (A), f (x) = |x|, c = 0; for (B), f (x) = x3, c = 0; for (C), f (x) = x4, c = 0; for (E), f (x) = x2 on (−1, 1).

Free-Response

Part A

1. Image

(a) The following table shows x- and y-components of acceleration, velocity, and position:

Image

The last line in the table is the answer to part (a).

(b) To determine how far above the ground the ball is when it hits the wall, find out when x = 315, and evaluate y at that time.

Image

(c) The ball’s speed at the moment of impact in part (b) is |v(t)| evaluated at Image

Image

2. See solution for AB-2, Free-Response in AB Practice Exam Three.

Part B

3. See solution for AB/BC 3, Free-Response in AB Practice Exam Three.

4. See solution for AB-4, pages Free-Response in AB Practice Exam Three.

5. (a) The table below is constructed from the information given in Question 5 in BC Practice Exam Three.

n

f (n) (5)

Image

0

2

2

1

−2

−2

2

−1

Image

3

6

1

Image

Image

(c) Use Taylor’s theorem around x = 0.

n

g(n)(x)

g(n)(0)

Image

0

f (2x + 5)

f (5) = 2

2

1

2f′ (2x + 5)

2f′ (5) = 2(−2) = −4

−4

2

4f (2x + 5)

4f (5) = 4(−1) = −4

−2

3

8f′″(2x + 5)

8f′″(5) = 8(6) = 48

8

g(x) ≈ 2 − 4x − 2x2 + 8x3.

6. (a) At (−1,8), Image so the tangent line is

y − 8 = 5(x − (−1)). Therefore f (x) ≈ 8 + 5(x + 1).

(b) f (3) ≈ 8 + 5(0 + 1) = 13.

(c) At (−1,8), Image For Δx = 0.5, Δy = 0.5(5) = 2.5, so move to

(−1 + 0.5, 8 + 2.5) = (−0.5,10.5).

At (−0.5,10.5), Image For Δx = 0.5, Δy = 0.5(8) = 4, so move to (−0.5 + 0.5, 10.5 + 4).

Thus f (0) ≈ 14.5.

Image