Calculus AB and Calculus BC

Answers Explained

Multiple-Choice

Part A

1. (C) Use the Rational Function Theorem.

2. (D)

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)

5. (E)

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

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

8. (B)

9. (C) Let 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,

10. (D) P ′(x) = 2g(x) · g ′(x)

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

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

13. (E) Since

y ′ = 2(x − 3)⁻² and =

y ″ is positive when x < 3.

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

15. (B)

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:

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

f ′(1) exists and is equal to 2.

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

20. (A)

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

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

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)

25. (A)

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

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)

28. (C) Note that 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) −4(x − 3) + 8. f (3.02) −4(0.02) + 8.

30. (C)

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

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)

On the curve of f (x) − e^x − x², the two points labeled are (0,1) and (1, e − 1).

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

34. (B)

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

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

^* 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:

36. (C) so v = 3t³ + c.

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

Now so s = t³ + t + c.

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

37. (A) Let u = x². Then

38. (C)

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:

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

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

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

Thus,

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

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

Hence,

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

Then

43. (D)

−v(0) = −π + 3; so v(0) = π − 3.

44. (C)

45. (D) Let y = (x³ − 4x² + 8)e^cos(x2). The equation of the tangent at point (2, y (2)) is y − y(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) (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 and the cone’s volume,

Then

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

(Review Chapter 4)

(d) Use one of these Riemann sums:

(Review Chapter 6)

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

(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 + e^x)(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

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.

(Review Chapter 4)

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

(b)

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

(Review Chapter 7)

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

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

Solving for y gives the solution:

Note that is defined only if

only if the numerator and denominator have the same sign.

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

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

AB/BC 6. (a)

(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 an improper integral.

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

So y → e⁰ or 1.

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

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

16. (E) Separating variables yields 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 =

17. (B) y ′ = x + 2x ln x and

18. (D)

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

where so the next step produces

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

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

We send region S about the x-axis. Using washers, Δ V = π(R² − r²) Δx. Symmetry allows us to double the volume generated by the first-quadrant portion of S. So for V we have

A calculator yields 108.177.

31. (A) Use the Ratio Test:

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

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 x π, the maximum error (numerically) occurs when x = π. Since

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

34. (B)

Graph x = 4 − t² and y = 2^t 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

35. (E)

36. (B)

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

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

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

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 (3 sin 2θ)² d ≈ 3.53.

Free-Response

Part A

1. (a) Use the Ratio Test:

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

Since

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

(b) Because 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 at n = 5 terms.

(c) 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), draw a segment at (−1, −1) that decreases steeply. Repeat this process at each of the other points. The result is shown below.

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

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

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

(c) The differential equation is separable:

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

The solution is

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

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

(b)

(c) Use disks. Then ΔV = πx² Δy,

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 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

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 the sine function varies from −1 to 1 as the argument varies from

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 Then

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 1. Thus − ∞ < ln cos x 0.

20. (E) implies Then and 3 = b^1/2. So 3² = b.

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

Solve for y:

22. (D) Since f (1) = 0, x − 1 is a factor of f. Since f (x) divided by x − 1 yields x² − x − 2, f (x) = (x − 1) (x + 1) (x − 2); the roots are x = 1,−1, and 2.

23. (B) If then − ∞< tan x < ∞ and 0 < e^tanx < ∞.

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

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

26. (A) arccos

27. (A) Interchange x and y: x = 2e^−y

Solve for y:

Thus

28. (C) The function in (C) is not one-to-one since, for each y between (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

33. (D) 2 cos 5 = 0 when

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

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

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 the denominator approaches ∞

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

11. (D)

12. (B)

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

14. (C) Since (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

16. (C) As x → 0, takes on varying finite values as it increases. Since the sine function repeats, 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 both x = 2 and are vertical asymptotes. Also, is a horizontal asymptote.

18. (B) Use the Rational Function Theorem.

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

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

21. (A) As x → π, (π − x) → 0.

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

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

24. (B) Only x = 1 and x = 2 need be checked. Since for x ≠ 1, 2, and = −3 = f (1), f is continuous at x = 1. Since 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 → ∞, 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

28. (D) See Figure N2–1.

29. (E) Note, from Figure N2–1, that

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

31. (A) Note that if x ≠ 0 and that

32. (A)

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 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)

37. (E) As x → 0⁻, arctan

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

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

41. (B)

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

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

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

5. (E)

6. (D) Rewrite:

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

8. (D) Use the Quotient Rule:

9. (C) Since

10. (E) Use formula (18):

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

12. (D) By the Quotient Rule,

13. (D) Since ln (x² + 1)

14. (C)

15. (A) Since (−csc 2x cot 2x · 2).

16. (A) y ′ = e^−x (−2 sin 2x) + cos 2x(−e^−x).

17. (C) y ′ = (2 sec x)(sec x tan x).

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

19. (B)

20. (C)

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

22. (A) 1 − sin (x + y)(1 + y ′) = 0;

23. (D) cos x + sin y · y′ = 0;

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

25. (A)

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

27. (E) f ′(x) = 8x^−1/2;

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

29. (D) 2x + 2yy ′ = 0;

30. (E) Replace t by 1.

31. (D)

32. (D) y ′ = e^x · 1 + e^x (x − 1) = xe^x;

y ″ = xe^x + e^x and y ″(0) = 0 · 1 + 1 = 1.

33. (E) When simplified,

34. (B) Since (if sin t ≠ 0)

= −2 sin t = −4 sin t cos t and

then Thus:

BC ONLY

NOTE: Since each of the limits in Questions 35–39 yields an indeterminate form of the type 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) =

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) but f (1) = 4.

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

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

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

42. (A) Note that f (0) = = 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) = is h(x) = then h ′(x) = Replace x by 3.

44. (D) After 50(!) applications of L’Hôpital’s Rule we get which “equals” ∞. A perfunctory examination of the limit, however, shows immediately that the answer is ∞. In fact, 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);

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

46. (B)

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

47. (C)

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

48. (E)

[We can replace 1 − cos x by getting

49. (D)

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

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

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

53. (A)

54. (D)

55. (E)

56. (C) Since

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)

60. (D)

61. (C)

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

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

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 = and AC has m = − The tangent line is perpendicular to the radius.

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 There appear to be 3 points at which tangent lines are parallel to

74. (C)

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

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).

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.

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

81. (A)

82. (A)

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

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 also, f (x), or 2x³ − 3x, equals −1, by observation, for x = 1. So h ′(−1) or (when x = 1) equals

86. (D)

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

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)

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

93. (E) Now

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

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

96. (E) Since f (x) = 3^x − x³, then f ′(x) = 3^x 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 Solving 3^x ln 3 − 3x² = 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).

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) =

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

103. (D)

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

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 The equation is

4. (B) Since y ′ = e^−x (1 − x) and e^−x > 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, At (4, 2), y ′ = 2. The equation of the tangent at (4, 2) is y − 2 = 2(x − 4).

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

8. (D) Since 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 The linear approximation for the true value of y when x changes from 3 to 3.04 is

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

or

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

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

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

14. (A) The domain of y is {x | x 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 = 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 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.

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

28. (A) Note that

29. (B)

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

31. (C) Since

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 and that

Replace h by 8.

33. (D) 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 = 2xe^−x2. Thus which is 0 when the positive value of x is 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.

The area A of the triangle is given by

Then 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 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 Since

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

40. (A) Since f ′(x) = e^−x (1 − x), f ′(0) > 0.

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

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

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

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 therefore, at t = 1, Also, x = 3 and y = 2.

47. (A) Let f (x) = x^1/3, and find the slope of the tangent line at (64, 4). Since If we move one unit to the left of 64, the tangent line will drop approximately unit.

48. (D)

49. (E) e^kh e^k·0 + ke^k·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 the slope of the normal is

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

y − 1 = −1(x + 2) or x + y + 1 = 0.

53. (C)

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

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

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

66. (E)

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

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

70. (B)

71. (A) Since 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 the acceleration is always directed downward. Its magnitude, is 2 for all t.

73. (D) Using the notations v_x, v_y, a_x, and a_y, we are given that where k is a constant. Then

74. (E)

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

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

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

81. (E) The equation of the tangent is y = −2x + 5. Its intercepts are 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 Using s² = x² + y², we get

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

83. (B) (from Question 82) is zero when 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) 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 at Q, R, and T. At Q, at T,

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) 4x + 6 and f (6.02) 30.08.

92. (C)

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.

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

4. (D) Rewrite:

5. (E) Rewrite:

Use (3).

6. (B) Rewrite:

Using (3) yields

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

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

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

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

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

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

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

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

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

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

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

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

17. (E)

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

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

19. (A) Integrate term by term.

20. (D) Division yields

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

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

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

24. (C) Use formula (6) with

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

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

27. (E) Use formula (6):

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

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

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

Use formula (3).

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

32. (E) Use formula (13) with

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

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

35. (E) Use formula (7).

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

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

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

39. (C)

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

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

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

Then x − 1 = A(x − 2) + Bx.

Set x = 0: −1 = −2A and

Set x = 2: 1 = 2B and

So the given integral equals

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

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

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

46. (B) Use formula (15) with

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

48. (C) See Example 44.

49. (D) The integral is of the form

50. (A) The integral has the form Use formula (18), with u = e^x, du = e^x dx.

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

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

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

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

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 and The Parts Formula yields

57. (E) The integral has the form where

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

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

60. (D) Multiply to get

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

62. (E) The integral equals it is equivalent to 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 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

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

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

66. (E) Rewrite:

67. (B) Hint: Divide, getting

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

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

The Parts Formula yields

70. (B) Hint: Note that

Or multiply the integrand by recognizing that the correct answer is equivalent to −ln|e^−x − 1|.

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

72. (C) Hint: Observe that e^{2 ln u} = u².

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

74. (B) Hint: Expand and note that

Use formulas (9) and (7).

75. (E) Multiply by 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, Replacing x by 1 and y by −3 yields

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 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 Use partial fractions to get

Answers Explained

1. (C) The integral is equal to

2. (B) Rewrite as This equals

3. (E) Rewrite as

4. (B) This integral equals

5. (D)

6. (A) Rewrite as

7. (D)

8. (A)

9. (C)

10. (D) You get = −(e⁻¹ − 1).

11. (B)

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

13. (E)

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

when x = 2, The integral is equivalent to

15. (D) Evaluate This equals

16. (A)

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

The result is

18. (E)

19. (A)

20. (E)

21. (C)

22. (A)

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

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

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

26. (D) 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)

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

for R(2) use the inscribed rectangles:

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:

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:

33. (C) Because 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

34. (C) The average value is equal to

35. (C) The average value is equal to

36. (E) The average value is The integral represents the area of a trapezoid: (5 + 3) · 10 = 40. The average value is (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, Thus, the average value of f on [0,6] is There are points on f with y-values of in the intervals [0,2] and [2,4],

40. (B)

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

42. (C) (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

47. (E) Let and u = x²; then

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

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

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

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

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

53. (B)

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)

2. (C)

3. (A)

4. (D)

5. (D)

6. (C)

7. (E)

8. (A)

9. (A)

10. (D)

11. (D)

12. (C)

13. (D)

14. (A)

15. (B)

16. (B)

17. (C)

VOLUME

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

18. (E)

19. (D)

20. (A)

21. (C)

22. (D)

23. (B)

24. (A)

25. (D)

26. (B)

27. (E)

ARC LENGTH

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

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

IMPROPER INTEGRALS

30. (A) The integral equals

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

32. (B) Redefine as

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

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

35. (C) Evaluate

36. (A)

37. (C)

38. (E)

39. (B)

40. (D)

41. (B)

42. (A)

AREA

43. (C)

44. (D)

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

46. (B) 2θ dθ = 1.571, using a graphing calculator.

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

VOLUME

48. (D)

49. (B)

50. (C)

51. (B)

52. (C)

53. (A)

54. (D)

ARC LENGTH

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

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

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

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 and changes sign both when t = 1 and when t = 3.

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

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

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

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

sin t dt = 1 − cos 2.

6. (D) .

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

8. (D) Since R(0) = 〈0,1〉,

c₁ = 0 and c₂ = 1.

9. (A) a = v′(t) = 〈1,1〉 for all t.

10. (B)

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

If R is perpendicular to v, then so

and x² + y² = k (k > 0).

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

x² + y² = 25.

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

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

14. (D)

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

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

16. (C)

17. (A)

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

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 =

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 mi long and its area is The strip’s population is approximately 2(12 − 2r) Δr. The total population of the entire city is twice the integral as it includes both halves of the city.

22. (C)

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 2πr_k Δr or 4πr_k. The population in the ring is

(4πr_k)· (density at r_k) = 4πr_k · f (r_k).

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) 6912 hundred people—or about 691,200 people.

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

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

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 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

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.

Using the calculator, we get

27. (E) The FTC yields total change:

28. (C) The total change (increase) in population during the second hour is given by 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 using a Riemann sum.

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)

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

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

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

Answers Explained

1. (C) v(t) = 2t² − t + C; v(1) = 3; so C = 2.

2. (B) If a(t) = 20t³ − 6t, then

v(t) = 5t⁴ − 3t² + C₁,

s(t) = t⁵ − t³ + 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) = t⁵ − t³ + 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

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

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

8. (A)

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

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

11. (A) Integrating yields + 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), x² − y² = 3. Rewriting the d.e. y dy = x dx as reveals that the derivative does not exist when y = 0, which occurs at Since the particular solution must be differentiable in an interval containing x = −2, the domain is

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

14. (C) We separate variables. The particular solution is −e^−y = x − 2.

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

16. (E) Since it follows that

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

so y = kx.

17. (E) hence the general solution is y = ke^x, k ≠ 0.

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

19. (C) We are given that 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. 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.

23. (B)

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:

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” e^−x2: 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 = ce^x. 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

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. yields and initial condition y(1) = 5 means that or C = 4.5. Hence + 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 We integrate:

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

36. (B) Since 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

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

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

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

40. (C) The general solution of is ln y = x + C or y = ce^x. For a solution to pass through (2, 3), we have 3 = ce² and c = 3/e² 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 = e^x − x − 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) = e^−x2, we know from the FTC, since F(0) = 0, that

To approximate 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 where L is the upper limit. The graph shows L = 1000, so the differential equation must be Only equation C is of this form (k = 0.003).

44. (D) We start with x = 3 and y = 100. At x = 3, moving us to x = 3 + 2 = 5 and y = 100 + 5 = 105. From there 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:

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

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

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

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

We let Q(0) = Q₀. Then

We now find t when Q = 0.1Q₀:

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(a − y).

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

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

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:

BC ONLY

Answers Explained

1. (B) converges to −1.

2. (C) Note that

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

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

5. (C)

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

7. (B)

8. (A)

BC ONLY

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

10. (D) 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)

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

14. (C) unless x = 3.

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

16. (E)

17. (A)

18. (E) The series satisfies the Alternating Series Test, so the error is less than the first term dropped, namely, (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 then the first omitted term, is negative. Hence P₇ (x) exceeds tan⁻¹ x.

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

21. (A) If converges, so does 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 is 1, so this test fails; note for (E) that

25. (B) does not equal 0.

26. (E) Note the following counterexamples:

(A)

(B)

(C)

(D)

27. (C) Since 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) 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

30. (B) See Example 52.

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

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

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

Or use the series for e^x and let x = −0.1.

BC ONLY

35. (C)

Or generate the Maclaurin series for e^{sin x}.

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

37. (A)

38. (C)

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

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

40. (D)

Answers Explained

Part A

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

thus this function is continuous at 0. In (A), 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)

4. (D) The line x + 3y + 3 = 0 has slope 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) is f′ (1), where Or simplify the given fraction to

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)

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

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:

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

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

11. (D) Evaluate

12. (A) From the equation given, y = e^{sin x}.

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

14. (C) If y = e^x ln x, then 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

17. (C)

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

For y = 2, Note that 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 yields the minimum distance at

21. (D)

22. (D) See the figure. Since the area, A, of the ring equals π (y₂² − y₁²),

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

and = π (72x − 36x²) = 36πx (2 − x).

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

23. (A) This is of type

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

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

or

y² = 2x − 2x² + k

or

2x² + y² − 2x = k.

26. (E) 2(5) +

27. (E)

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

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

30. (E) The parameter k determines the amplitude of the sine curve. For f = k sin x and g = e^x 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 = e^q and k cos q = e^q,

and therefore

sin q = cos q.

The figure shows

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

At (3, 1), The linearization is

32. (C)

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

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

35. (E) 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

where The answer is

37. (B)

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

39. (C) The general solution is y = 3 ln|x² − 4| + C. The differential equation 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

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

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

giving

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

then the required area is given by

(See polar graph 63 in the Appendix.)

44. (C)

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

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

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

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

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

Then

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

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

49. (D)

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

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

See Example 26.

52. (C) Let Then ln y = −x ln x and

Now apply L’Hôpital’s Rule:

So, if ln y = 0, then

53. (D) The speed, |v|, equals and since x = 3y − y²,

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

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

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

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,

56. (B) Since

Then

Note that so the integral is proper.

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

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

61. (D) represents the area of the same region as 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 Use your calculator to solve the equation for c (in radians).

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

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 hence f has two points of inflection.

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

The nth derivative at

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);

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

69. (D) Note that f (g(u)) = tan⁻¹ (e^2u); then the derivative is

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

71. (E)

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

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

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.

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

75. (C) If Q is the amount at time t. then Q = 40e^−kt. 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 So arctan t + C. Since v (1) = 0, C = − π.

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

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:

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.

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

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

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

79. (B)

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.

81. (C)

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,

83. (B) Separating variables yields

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

84. (C) 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 π · 8² · 25 + π · 6² · 25 + π · 4² · 25 + π · 3² · 25.

86. (A)

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)

90. (C)

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

92. (B) Use areas; then 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.

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

To find the total number of people, evaluate

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

95. (C)

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

97. (B)

98. (D) Evaluate

99. (A)

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

100. (A)

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

Since y = 2^x, we find

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

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

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

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

106. (D) We seek x such that

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

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

This maximum occurs at c = x =

108. (D)

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

Answers Explained

Part A

1. (a)

(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 1.975.

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

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 x − e^(−x2) + 1 in [−7, 7] × [−10, 10].

(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.

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

(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 =

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

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

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

(a) The graph of f is concave downward where f ″ is negative, namely, on (b, d). Use the calculator to solve sin 2^x = 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.

(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

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

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

(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:

(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 letters, so the pile vanished shortly after 5 P.M.

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

(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.

(b) Elements of volume are triangular prisms with height h = 3 and base b = (sinx − x⁴), as shown below.

(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

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

(a) The volume of the tank, using disks, is 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

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

and the depth of the water is k + 10.

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

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

The volume of water is

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

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

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

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

Since 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)

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

(d) The series

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

Since ln x is an increasing function,

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 Solve for n using

The given series converges very slowly!

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

14. (a) Solve by separation of variables:

Let c = e^−10c; then

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

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

(b) Since c = 4 and

Solving for y yields

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

(d) so the value of y approaches 10.

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

(b) Since Since Therefore

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

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

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

16. (a) See graph.

(b) You want to maximize

See signs analysis.

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

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

Since This vector is shown on the graph.

(d) 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

Use the calculator to find that 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 the area of R (enclosed by r) is which is 14.137.

Part B

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

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) = 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 (y₁ − y₂) = 2x(12 − 3x²), or A(x) = 24x − 6x³.

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

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

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

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

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

(b) The required area consists of two parts. The area of the triangle is represented by 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 This equals

The required area, thus, equals

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 then, separating variables, Integrating gives

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

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

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

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) = 27x − x³ for 0 ≤ x ≤

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.

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

(c)

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

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 which here is When

(b) When The equation of the tangent is

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

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

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

(b)

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

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

(b) Since 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 and that of the circle from Thus

Answers Explained

Multiple-Choice

Part A

1. (C) Use the Rational Function Theorem.

2. (C) Note that 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)

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

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)),

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 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:

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 evaluate

14. (E)

15. (D)

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

17. (B) Implicit differentiation yields 2yy ′ = 1; so At a vertical tangent, is undefined; y must therefore equal 0 and the numerator be non-zero. The original equation with y = 0 is 0 = x − x³, 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

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

21. (A) Solve the differential equation by separation of variables: yields y = ce^2x. The initial condition yields 1 = ce^{2 · 2}; so c = e ⁻⁴ and y = e^2x−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, so the equation of the tangent line at (3,0) is y = −9(x−3).

24. (A)

25. (D) The graph shown has the x-axis as a horizontal asymptote.

26. (B) Since 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:

The graph of f changes concavity at x = −1, 0, and 1.

28. (C) Note that 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 it equals 0 for When x = 3, c = 9; this yields a minimum since f ″(3) > 0.

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) We evaluate this definite integral by finding the area of a trapezoid (negative) and a triangle: 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) 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

These limits indicate the presence of a jump discontinuity in the function at x = 1.

37. (D)

38. (E)

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(R² − r²) Δx. Symmetry allows you to double the volume generated by the first quadrant of S, so V is

A calculator yields 108.177.

40. (A) The curve falls when f ′(x) < 0 and is concave up when f ″(x) > 0.