AP Calculus AB · Mock Exam 2

CalcIQ — Mock Exam 2: Full AP Simulation (All 8 Units)

This is a full-length practice simulation of the AP Calculus AB exam. It is independent study material and is not endorsed by or affiliated with the College Board. "AP" is a registered trademark of the College Board.

Student Instructions — Read Before You Begin

This mock exam reproduces the full AP Calculus AB format and timing. Work it in one sitting if you can, using a clock.

| Section | Part | Questions | Time | Calculator |

|---|---|---|---|---|

| I (Multiple Choice) | A | 30 MC | 60 min | NO calculator |

| I (Multiple Choice) | B | 15 MC | 45 min | Calculator permitted |

| II (Free Response) | A | 2 FRQ | 30 min | Calculator permitted |

| II (Free Response) | B | 4 FRQ | 60 min | NO calculator |

When you finish Section I, check your work against the Section I Answer Key, then continue to Section II.

SECTION I, PART A

30 questions · 60 minutes · NO CALCULATOR

Calculators are not permitted on this part. Where a numerical answer is impractical by hand, exact (symbolic) answers are given as choices.

1NO CALC
The graph of a function f is shown.
y = f(x) on [−1, 5] × [−1, 5
Which statement is true at x = 2?
2NO CALC
lim_{x→0} sin(3x)/(5x) =
3NO CALC
lim_{x→∞} (3x² − 5)/(2x² + x) =
Question 4NO CALC
Let 
f(x) = (x² − 4)/(x − 2),  x ≠ 2
       a,                 x = 2
For what value of a is f continuous at x = 2?
5NO CALC
The function f(x) = x³ − x − 1 is continuous on [1, 2] with f(1) = −1 and f(2) = 5. Which theorem guarantees a c in (1, 2) with f(c) = 0, and why?
6NO CALC
Using the limit definition, the derivative of f(x) = x² is f'(x) =
7NO CALC
If f(x) = x cos x, then f'(x) =
8NO CALC
If g(x) = x/(x² + 1), then g'(x) =
9NO CALC
If f(x) = tan x, then f'(π/4) =
10NO CALC
If y = √(3x + 1), then dy/dx =
11NO CALC
Given x² + xy + y² = 7, find dy/dx at the point (1, 2).
12NO CALC
If f(x) = ln(x² + 1), then f'(x) =
13NO CALC
Let h(x) = f(g(x)) where g(2) = 5, g'(2) = 3, f'(5) = −4. Then h'(2) =
14NO CALC
A particle moves along a line with position s(t) = t³ − 6t² + 9t. The particle is moving to the left (velocity negative) when
15NO CALC
The graph of f' (the derivative of f) is shown.
y = f'(x) on [0, 6] × [−3, 3
At which x does f have a relative maximum?
16NO CALC
The graph of f' (the derivative of f) is shown below.
y = f'(x) on [0, 6] × [−3, 3
On which interval is the graph of f concave up?
17NO CALC
lim_{x→0} (1 − cos x)/x² =
18NO CALC
The line tangent to the curve y = x³ at the point (1, 1) is used to approximate the value of the function near x = 1. The linear approximation of (1.1)³ is
19NO CALC
The function f(x) = x² satisfies the hypotheses of the Mean Value Theorem on [1, 3]. The value of c in (1, 3) guaranteed by the MVT is
20NO CALC
Let f(x) = x⁴ − 6x². The x-coordinates of the points of inflection of f are
21NO CALC
A right circular cylinder is inscribed so that a rectangle of width 2x and height 4 − x² sits under the parabola y = 4 − x² (for 0 ≤ x ≤ 2), with its base on the x-axis. The width 2x that maximizes the rectangle's area A = 2x(4 − x²) occurs at
22NO CALC
∫ (6x² + 4x) dx =
23NO CALC
∫_0^{π/2} cos x dx =
24NO CALC
∫ x e^{x²} dx =
25NO CALC
Let g(x) = ∫_0^x f(t) dt, where the graph of f is shown.
y = f(t) on [0, 5] × [−2, 3
What is g(4) = ∫_0^4 f(t) dt?
26NO CALC
Using g(x) = ∫_0^x f(t) dt from Question 25, at x = 4 the function g has
27NO CALC
∫_1^e (1/x) dx =
28NO CALC
If F(x) = ∫_2^x √(t³ + 1) dt, then F'(x) =
29NO CALC
The region bounded by y = √x, the x-axis, and x = 4 is revolved about the x-axis. The volume of the resulting solid is
30NO CALC
The solution to the differential equation dy/dx = x/y with initial condition y(0) = 3 is

End of Section I, Part A.

SECTION I, PART B

15 questions · 45 minutes · CALCULATOR PERMITTED

A graphing calculator is permitted on this part. Several questions are intended to be solved using nDeriv, fnInt, or a numeric equation solver. Unless otherwise specified, round to three decimal places only at the end, and choose the closest answer.

31CALC
Let f(x) = x³ − 3x + 1. The value of f'(2) is

> TI-84: MATH → 8:nDeriv( → nDeriv(X³−3X+1, X, 2)

32CALC
For f(x) = ln(x² + 1), the value of f'(1) is closest to
33CALC
The graphs of y = sin x and y = x/2 enclose a region in the first quadrant between x = 0 and their first positive intersection. The area of that region is closest to
34CALC
∫_0^1 cos(x²) dx is closest to
35CALC
The average value of f(x) = e^{−x²} on the interval [0, 2] is closest to
36CALC
A particle moves along the x-axis with velocity v(t) = t² − 4t + 3 for 0 ≤ t ≤ 4. The total distance traveled by the particle is
37CALC
Using the velocity v(t) = t² − 4t + 3 from Question 36, the displacement (net change in position) of the particle from t = 0 to t = 4 is
38CALC
The equation e^x = 3x has two solutions. The smaller solution is closest to

> Solve e^X − 3X = 0 numerically, or graph Y₁ = e^X, Y₂ = 3X and find both intersections.

39CALC
— Rate table. Water is pumped into a reservoir at a rate R(t), in gallons per hour, recorded every 2 hours: | t (hours) | 0 | 2 | 4 | 6 | 8 | |---|---|---|---|---|---| | R(t) (gal/hr) | 10 | 14 | 18 | 16 | 12 | Using a trapezoidal sum with the four subintervals shown, the approximate total amount of water pumped in over 0 ≤ t ≤ 8 is
40CALC
— Rate table. Using the same table as Question 39, a left Riemann sum with the four subintervals approximates ∫_0^8 R(t) dt as
41CALC
A spherical balloon is being inflated. When the radius is r = 3 cm, it is increasing at dr/dt = 2 cm/sec. The rate at which the volume V = (4/3)πr³ is increasing at that instant is
42CALC
The region bounded by y = 4 − x² and y = x + 2 has area closest to
43CALC
A population grows according to dP/dt = kP. The population doubles every 5 years. By what factor has the population grown after 15 years?
44CALC
Let g(x) = ∫_0^x f(t) dt where f(t) = cos(t²). The value of g(1.5) is closest to
45CALC
For f(x) = 2^x, the instantaneous rate of change of f at x = 3, i.e. f'(3), is closest to

> TI-84: nDeriv(2^X, X, 3). Note f'(x) = 2^x ln 2.

End of Section I.

SECTION I — ANSWER KEY

(Each correct answer was independently recomputed and symbolically verified.)

Part A (No Calculator)

1. (D) the limit does not exist. The left-hand limit is 4 and the right-hand limit is 1; since they differ, lim_{x→2} f(x) does not exist. (C)/(B) report a one-sided value or the function value f(2)=3; (A) is false because the limit fails to exist (and even the limit ≠ f(2)).

2. (B) 3/5. sin(3x)/(5x) = (3/5)·sin(3x)/(3x) → 3/5. (C) 1 forgets the coefficients; (A) inverts the ratio.

3. (A) 3/2. Ratio of leading coefficients of equal-degree polynomials, 3/2. (C) treats it as bottom-heavy; (D) as top-heavy.

4. (C) 4. lim_{x→2}(x²−4)/(x−2) = lim (x+2) = 4, so a = 4 removes the hole. (D) wrongly assumes the discontinuity is non-removable.

5. (B) IVT. f is continuous on [1, 2] and 0 lies between f(1) = −1 and f(2) = 5, so the IVT guarantees a root. (A)/(D)/(C) misname the theorem or its hypothesis; (C) is false since f(1) ≠ f(2).

6. (D) 2x. lim_{h→0}[(x+h)² − x²]/h = lim (2x + h) = 2x. (A) 2x+h forgets to take the limit.

7. (A) cos x − x sin x. Product rule: (1)(cos x) + x(−sin x). (C) has the wrong sign on the second term — the classic product-rule sign slip.

8. (C) (1 − x²)/(x² + 1)². Quotient rule: [(1)(x²+1) − x(2x)]/(x²+1)² = (1 − x²)/(x²+1)². (D) reverses the numerator sign.

9. (D) 2. d/dx tan x = sec²x; at π/4, sec(π/4) = √2, so (√2)² = 2. (C) √2 forgot to square the secant.

10. (A) 3/(2√(3x+1)). Chain rule: (1/(2√(3x+1)))·3. (B) drops the chain-rule factor of 3 — the most common error.

11. (B) −4/5. Implicit: 2x + y + x y' + 2y y' = 0 → y' = −(2x + y)/(x + 2y). At (1,2): −(2+2)/(1+4) = −4/5. (D) inverts; (C) drops a sign.

12. (D) 2x/(x² + 1). Chain rule: (1/(x²+1))·2x. (C) forgets the chain-rule factor 2x.

13. (A) −12. h'(2) = f'(g(2))·g'(2) = f'(5)·3 = (−4)(3) = −12. (C) 15 ignores the sign; (D) −20 multiplies the wrong pair.

14. (B) 1 < t < 3. v(t) = 3t² − 12t + 9 = 3(t−1)(t−3) < 0 on (1, 3). (D) uses the wrong factorization; (C)/(A) take only one side of the sign chart.

15. (C) x = 2. f has a relative max where f' changes from positive to negative, which occurs at x = 2. At x = 5, f' changes − to +, giving a minimum (distractor C). (B)/(D) are endpoints, not interior sign changes.

16. (D) (3, 6). f is concave up where f'' > 0, i.e. where f' is increasing. From the graph, f' increases on (3, 6). (A) is where f' decreases (concave down — confusing the sign of f' with the sign of f''); (C) ignores the turning point; (B) is false.

17. (A) 1/2. Standard limit (L'Hôpital twice, or Taylor): (1 − cos x)/x² → 1/2. (C) 1 is the value of sin x / x, a confusion.

18. (B) 1.3. Tangent at (1,1): y = 1 + 3(x − 1); at x = 1.1, y = 1 + 0.3 = 1.3. (A) 1.331 is the exact value, not the linear approximation. (D) 1.1 forgets the slope 3.

19. (C) c = 2. MVT: f'(c) = [f(3) − f(1)]/(3 − 1) = (9 − 1)/2 = 4. Since f'(x) = 2x, solve 2c = 4 → c = 2, which lies in (1, 3). (A) c = 4 confuses the slope value with c; (B)/(D) are setup slips.

20. (D) x = ±1. f'' = 12x² − 12 = 0 → x = ±1, and concavity changes there. (B) ±√3 are the x-intercepts of f; (A) ±√2 are the critical points' neighbors — both common confusions.

21. (A) x = 2/√3. A = 8x − 2x³, A' = 8 − 6x² = 0 → x² = 4/3 → x = 2/√3. (B) x=1 and (D) √2 are plausible-looking but don't satisfy A' = 0.

22. (B) 2x³ + 2x² + C. ∫6x² dx = 2x³ and ∫4x dx = 2x². (D) 3x³ divides 6 by 2 incorrectly (should multiply 6 by 1/3); (C) forgets to divide by the new power; (A) differentiates instead of integrating.

23. (C) 1. ∫_0^{π/2} cos x dx = [sin x]_0^{π/2} = 1 − 0 = 1. (B) −1 reverses the limits.

24. (D) (1/2)e^{x²} + C. u-sub u = x², du = 2x dx: (1/2)∫e^u du = (1/2)e^{x²} + C. (C) forgets the 1/2; (A) multiplies by 2 instead of dividing.

25. (A) 6. ∫_0^4 f = (rectangle area 0→2) + (triangle area 2→4) = (2·2) + (½·2·2) = 4 + 2 = 6. (B) 4 counts only the rectangle; (C) 8 mis-areas the triangle.

26. (B) relative maximum. g has a relative max where g'(x) = f(x) changes from positive to negative; from the graph f crosses from + to − at t = 4. (A) reverses the sign change; (D)/(C) miss the extremum.

27. (C) 1. ∫_1^e (1/x) dx = [ln x]_1^e = 1 − 0 = 1. (B) e − 1 integrates 1 instead of 1/x.

28. (D) √(x³ + 1). By the FTC Part 1, F'(x) = √(x³ + 1). (C) needlessly applies the chain rule (upper limit is just x); (B) subtracts the lower-limit value (not how FTC1 works); (A) antidifferentiates incorrectly.

29. (A) 8π. Disc: V = π∫_0^4 (√x)² dx = π∫_0^4 x dx = π·8 = 8π. (C) 8 drops the π; (D) integrates √x rather than its square; (B) uses x=4 as a radius.

30. (B) y = √(x² + 9). Separate: y dy = x dx → y²/2 = x²/2 + C → y² = x² + C'. With y(0) = 3: 9 = C', so y = √(x² + 9). Check: dy/dx = x/√(x²+9) = x/y. ✓ (A)/(D) drop the square root; (C) solves dy/dx = 2xy.

Part B (Calculator)

31. (C) 9. f'(x) = 3x² − 3; f'(2) = 12 − 3 = 9. (B) 3 is f'(0)'s magnitude piece; (A)/(D) are arithmetic slips.

32. (D) 1.000. f'(x) = 2x/(x²+1); f'(1) = 2/2 = 1. (A) 0.5 uses 1/(x²+1); (B) 0.693 = ln 2 (value, not slope).

33. (A) 0.421. First positive crossing of sin x = x/2 at x ≈ 1.895; ∫_0^{1.895}(sin x − x/2)dx ≈ 0.421. (C) 0.841 = sin 1; (B) halves the region.

34. (B) 0.905. fnInt(cos(X²), X, 0, 1) ≈ 0.9045. (D) 0.841 = sin 1; (C) is a wrong-bound slip.

35. (C) 0.441. (1/2)∫_0^2 e^{−x²}dx ≈ (1/2)(0.8821) ≈ 0.441. (D) 0.882 forgets the 1/(b−a); (B) divides by 4.

36. (D) 4. v = (t−1)(t−3), zero at t=1,3. Total distance = ∫_0^4 |v| dt = 4. (B) 4/3 is the displacement; (C)/(A) are partial integrals.

37. (A) 4/3. Displacement = ∫_0^4 (t²−4t+3) dt = 4/3. (B) 0 wrongly assumes symmetry; (C) 4 is total distance.

38. (B) 0.619. Solutions of e^x = 3x are x ≈ 0.619 and x ≈ 1.512; the smaller is 0.619. (A) is the larger root; (C) 1/3 and (D) ln 3 are setup confusions.

39. (C) 118. Trapezoidal with Δt = 2: (2/2)[10 + 2(14) + 2(18) + 2(16) + 12] = 10 + 28 + 36 + 32 + 12 = 118. (B) 116 is the left sum; (D) 120 is the right sum; (A) 60 forgets the width.

40. (D) 116. Left sum: 2(10 + 14 + 18 + 16) = 2(58) = 116. (B) 120 is the right sum; (C) 232 forgets to drop the last point / doubles.

41. (A) 72π. dV/dt = 4πr²·dr/dt = 4π(9)(2) = 72π. (D) 36π forgets to double dr/dt; (C) 24π uses 2πr instead of 4πr²-style factor.

42. (B) 4.500. Curves meet at x = −2, 1; area = ∫_{−2}^1 [(4−x²) − (x+2)] dx = 9/2 = 4.5. (D) 9 forgets the 1/.../doubles; (C) halves it.

43. (C) 8. Doubling every 5 years → after 15 years (three doublings) the factor is 2³ = 8. (A) 3 counts the number of doublings; (D) 32 = 2⁵ uses the wrong exponent.

44. (C) 0.778. g(1.5) = ∫_0^{1.5} cos(t²) dt ≈ 0.7782. (D) 0.905 is ∫_0^1; (B) is a wrong-bound slip.

45. (D) 5.545. f'(x) = 2^x ln 2; f'(3) = 8 ln 2 ≈ 5.545. (C) 8 forgets the ln 2 factor; (B) 12 confuses with 3·2² = 12 (power-rule misuse).

(Section II — Free Response — continues below.)

SECTION II — FREE RESPONSE

6 questions · 90 minutes total · 50% of exam score

This section has two parts. You may not return to Part A once you begin Part B, and you may not use a calculator on Part B.

Directions for all free-response questions:

SECTION II, PART A

Questions 1–2 · 30 minutes · GRAPHING CALCULATOR PERMITTED

FRQ 1 — Rate In / Rate Out (Units 6, 8) · [CALC]

(9 points)

A storage tank contains water. At time t = 0 hours the tank contains 50 liters of water. For 0 ≤ t ≤ 8, water is pumped into the tank at a rate

D(t) = 2 + 5 sin(t/4)   liters per hour,

and water is drained out of the tank at a rate

E(t) = 5 − 5 cos(t/3)   liters per hour.

Both D and E are continuous, and t is measured in hours.

(a) How many liters of water are pumped into the tank during the 8-hour interval 0 ≤ t ≤ 8? (2 points)

(b) Let W(t) be the number of liters of water in the tank at time t. Find W′(3), and use it to determine whether the amount of water in the tank is increasing or decreasing at time t = 3. Show the computation that leads to your answer, and indicate units of measure. (2 points)

(c) Find the value of W(8). Using correct units, interpret the meaning of W(8) in the context of the problem. (2 points)

(d) For 0 ≤ t ≤ 8, at what time t is the amount of water in the tank greatest? Justify your answer. (3 points)

FRQ 1 — Model Solution

Setup. Because D and E are the rates in and out, the amount of water is

W(t) = 50 + ∫₀ᵗ [D(s) − E(s)] ds,     so     W′(t) = D(t) − E(t).

(a) The water pumped in is the integral of the in-rate:

∫₀⁸ D(t) dt = ∫₀⁸ [2 + 5 sin(t/4)] dt ≈ 44.323 liters.
TI-84: MATH → 9:fnInt( → fnInt(2 + 5sin(X/4), X, 0, 8) → 44.32294

During the 8 hours, approximately 44.323 liters of water are pumped into the tank.

(b) W′(t) = D(t) − E(t). At t = 3:

D(3) = 2 + 5 sin(3/4) ≈ 5.408     liters/hour
E(3) = 5 − 5 cos(3/3) = 5 − 5 cos 1 ≈ 2.298     liters/hour
W′(3) = D(3) − E(3) ≈ 5.408 − 2.298 = 3.110     liters/hour

Since W′(3) ≈ 3.110 > 0, the amount of water in the tank is increasing at time t = 3 (water is entering faster than it is leaving).

(c) By the net-change (accumulation) interpretation,

W(8) = 50 + ∫₀⁸ [D(t) − E(t)] dt ≈ 50 + 11.182 = 61.182 liters.
TI-84: fnInt(2 + 5sin(X/4) − (5 − 5cos(X/3)), X, 0, 8) → 11.18203

W(8) ≈ 61.182 means that after 8 hours the tank contains approximately 61.182 liters of water.

(d) The interior critical points occur where W′(t) = D(t) − E(t) = 0. Solving numerically on (0, 8):

TI-84: graph Y₁ = D(X) − E(X) and use 2:zero,  or solve D(X) − E(X) = 0
→ the only sign change occurs at t ≈ 5.935.

At t ≈ 5.935, W′ changes from positive to negative (water stops accumulating and begins to drop), so t ≈ 5.935 gives a relative maximum of W. Compare the candidate with the endpoints, since W is continuous on the closed interval [0, 8] (Extreme Value Theorem):

W(0)      = 50
W(5.935)  = 50 + ∫₀^{5.935} [D − E] dt ≈ 64.228
W(8)      ≈ 61.182

The amount of water in the tank is greatest at t ≈ 5.935 hours, where W ≈ 64.228 liters. Justification: W′ = D − E changes from positive to negative only at t ≈ 5.935, so this is the only candidate for a maximum besides the endpoints; comparing values, W(5.935) ≈ 64.228 exceeds W(0) = 50 and W(8) ≈ 61.182.

FRQ 1 — Rubric (9 points)

| Part | Pts | Earned for |

|---|---|---|

| (a) | 2 | Integral ∫₀⁸ D(t) dt of the in-rate (1); answer ≈ 44.323 liters (1) |

| (b) | 2 | W′(3) = D(3) − E(3) ≈ 3.110 with the computation shown (1); "increasing" with correct units/reasoning (1) |

| (c) | 2 | W(8) = 50 + ∫₀⁸ (D − E) dt ≈ 61.182 (1); interpretation with units — liters in tank at t = 8 (1) |

| (d) | 3 | Sets W′ = D − E = 0, finds t ≈ 5.935 (1); justification that W′ changes + → − (or candidate test set up) (1); compares with endpoints and concludes t ≈ 5.935 (1) |

Where students lose points. In (a), integrating D − E instead of D alone (the question asks only how much is pumped in). In (b), reporting D(3) or E(3) alone, omitting units, or computing in degrees rather than radians (the calculator must be in radian mode). In (c), giving a number with no contextual interpretation, or forgetting the initial 50. In (d), naming the time without the endpoint comparison, or stopping at "W′ = 0" without the sign-change justification.

FRQ 2 — Particle Motion (Units 4, 8) · [CALC]

(9 points)

A particle moves along the x-axis so that its velocity at time t (for 0 ≤ t ≤ 6) is given by

v(t) = e^(t/4) · sin t .

At time t = 0, the particle is at position x(0) = 2. The velocity v(t) is measured in meters per second and t in seconds.

(a) Find the acceleration of the particle at time t = 2. Is the speed of the particle increasing or decreasing at t = 2? Give a reason for your answer. (2 points)

(b) Find the displacement of the particle over the interval 0 ≤ t ≤ 6. Then find the position x(6). (2 points)

(c) Find the total distance traveled by the particle over the interval 0 ≤ t ≤ 6. (3 points)

(d) Find the average velocity of the particle over 0 ≤ t ≤ 6, and explain, with units, what this value represents in the context of the problem. (2 points)

FRQ 2 — Model Solution

(a) Acceleration is a(t) = v′(t). By calculator,

TI-84: MATH → 8:nDeriv( → nDeriv(e^(X/4)sin(X), X, 2) → −0.31132

so a(2) ≈ −0.311 m/s². Now check velocity: v(2) = e^(0.5) sin 2 ≈ 1.499 > 0.

Since v(2) > 0 and a(2) < 0, the velocity and acceleration have opposite signs, so the speed is decreasing at t = 2 (the particle is slowing down).

(b) Displacement is the integral of velocity:

∫₀⁶ v(t) dt = ∫₀⁶ e^(t/4) sin t dt ≈ −3.404 meters.
TI-84: fnInt(e^(X/4)sin(X), X, 0, 6) → −3.40353

The displacement is ≈ −3.404 m. Then

x(6) = x(0) + ∫₀⁶ v(t) dt = 2 + (−3.404) ≈ −1.404 meters.

(c) Total distance is the integral of speed, ∫₀⁶ |v(t)| dt. On [0, 6], v(t) = e^(t/4) sin t = 0 where sin t = 0, i.e. at t = π ≈ 3.142 (the only interior zero, since 2π ≈ 6.283 > 6). On (0, π), v > 0; on (π, 6), v < 0. So

Distance = ∫₀^π v dt − ∫_π^6 v dt
         ≈ 3.005 − (−6.409)
         = 3.005 + 6.409
         ≈ 9.414 meters.
TI-84: fnInt(abs(e^(X/4)sin(X)), X, 0, 6) → 9.41441

The total distance traveled is ≈ 9.414 m.

(d) Average velocity over [0, 6] is

(1/(6 − 0)) ∫₀⁶ v(t) dt ≈ (1/6)(−3.404) ≈ −0.567 m/s.

The average velocity is ≈ −0.567 m/s. This means that over the 6-second interval the particle's net displacement (≈ −3.404 m) corresponds to an average rate of motion of about 0.567 meters per second in the negative x-direction.

FRQ 2 — Rubric (9 points)

| Part | Pts | Earned for |

|---|---|---|

| (a) | 2 | a(2) = v′(2) ≈ −0.311 (1); "speed decreasing" justified by v(2) > 0 and a(2) < 0 having opposite signs (1) |

| (b) | 2 | Displacement ∫₀⁶ v dt ≈ −3.404 (1); x(6) = 2 + (−3.404) ≈ −1.404 (1) |

| (c) | 3 | Recognizes distance = ∫₀⁶ |v| dt, finds the sign change at t = π (1); correct split integral / absolute-value setup (1); answer ≈ 9.414 (1) |

| (d) | 2 | Average velocity = (1/6)∫₀⁶ v dt ≈ −0.567 (1); interpretation with units and direction (1) |

Where students lose points. In (a), concluding "speeding up" by looking only at the sign of a — you must compare the signs of v and a. In (b), confusing displacement with distance. In (c), the classic error: integrating v (giving displacement, −3.404) instead of |v|, or failing to locate the zero at t = π and split the integral. In (d), forgetting the 1/(b−a) factor (that gives displacement, not average velocity) or omitting units.

SECTION II, PART B

Questions 3–6 · 60 minutes · NO CALCULATOR

During the timed portion for Part B, you may continue to work on the problems in Part A without the use of a calculator.

FRQ 3 — Analysis from the Graph of f′ (Unit 5) · [NO CALC]

(9 points)

Let f be a function that is continuous on [0, 8] and differentiable on (0, 8), with f(0) = 5. The graph of f′, the derivative of f, consists of four line segments and is shown below.

y = f′(x) on [0, 8] × [−5, 5

(a) Find f(4) and f(8). Show the reasoning that leads to your answers. (2 points)

(b) On the open interval (0, 8), find the x-coordinate of each relative extremum of f, and classify each as a relative maximum or minimum. Justify your answer. (2 points)

(c) On the open interval (0, 8), find the x-coordinate of each inflection point of the graph of f. Justify your answer. (2 points)

(d) Write an equation of the line tangent to the graph of f at x = 2. Use it to find an approximation for f(2.1). (3 points)

FRQ 3 — Model Solution

Key principle. The given graph is f′. So:

(a) By the Fundamental Theorem of Calculus, f(b) − f(a) = ∫ₐᵇ f′(x) dx.

f(4) = f(0) + ∫₀⁴ f′(x) dx = 5 + (+8) = 13.

(On [0, 4], f′ ≥ 0; the region is two triangles of area 4 each, total +8.)

f(8) = f(4) + ∫₄⁸ f′(x) dx = 13 + (−8) = 5.

(On [4, 8], f′ ≤ 0; the region is two triangles of area 4 each, total −8.)

f(4) = 13 and f(8) = 5.

(b) Relative extrema of f occur where f′ changes sign. From the graph, f′ > 0 on (0, 4) and f′ < 0 on (4, 8), so f′ changes sign only at x = 4.

Justification: At x = 4, f′ changes from positive to negative, so f has a relative maximum at x = 4. There is no relative minimum on (0, 8), since f′ does not change from negative to positive anywhere on the interior. (At x = 0 and x = 8, f′ = 0 but does not change sign, so those are not extrema.)

(c) Inflection points of f occur where f″ = (f′)′ changes sign — that is, where the graph of f′ changes from increasing to decreasing or vice versa. From the graph, f′ is increasing on (0, 2), decreasing on (2, 6), and increasing on (6, 8).

Justification: f″ = (f′)′. At x = 2, f′ changes from increasing to decreasing, so f″ changes from positive to negative; at x = 6, f′ changes from decreasing to increasing, so f″ changes from negative to positive. Therefore the graph of f has inflection points at x = 2 and x = 6.

(d) From the graph, f′(2) = 4 (the peak of f′). The y-value:

f(2) = f(0) + ∫₀² f′(x) dx = 5 + 4 = 9

(the triangle on [0, 2] has base 2, height 4, area 4). The tangent line at x = 2 is

y = f(2) + f′(2)(x − 2) = 9 + 4(x − 2).

Then

f(2.1) ≈ 9 + 4(2.1 − 2) = 9 + 4(0.1) = 9.4.

Tangent line: y = 9 + 4(x − 2); f(2.1) ≈ 9.4.

FRQ 3 — Rubric (9 points)

| Part | Pts | Earned for |

|---|---|---|

| (a) | 2 | f(4) = 13 via 5 + ∫₀⁴ f′ and signed area +8 (1); f(8) = 5 via signed area −8 (1) |

| (b) | 2 | Relative max at x = 4 (1); justification that f′ changes from + to − there (and no min) (1) |

| (c) | 2 | Inflection at x = 2 and x = 6 (1); justification that f″ = (f′)′ changes sign (f′ changes increasing/decreasing) (1) |

| (d) | 3 | f(2) = 9 (1); tangent line y = 9 + 4(x − 2) using f′(2) = 4 (1); f(2.1) ≈ 9.4 (1) |

Where students lose points. In (a), counting the below-axis region on [4, 8] as positive (this is the most common error and gives f(8) = 21). In (b), saying "f has a max where f′ has a max" — false; f's extrema are where f′ changes sign (x = 4), not at f′'s peak (x = 2). In (c), naming x = 4 (an extremum of f, not an inflection) or omitting x = 6. In (d), using a value of f′ other than f′(2) = 4, or miscomputing f(2).

FRQ 4 — Area and Volume (Unit 8) · [NO CALC]

(9 points)

Let R be the region in the xy-plane bounded by the graphs of

y = 4 − x²     and     y = 2 − x .

(a) Find the area of region R. (2 points)

(b) Region R is the base of a solid. For this solid, each cross-section perpendicular to the x-axis is a square. Find the volume of the solid. (2 points)

(c) Find the volume of the solid generated when R is revolved about the x-axis. (3 points)

(d) Find the average vertical distance between the two curves over the interval that bounds R, and explain what this value represents. (2 points)

FRQ 4 — Model Solution

Setup (used in every part). Find the intersections: 4 − x² = 2 − x → −x² + x + 2 = 0 → x² − x − 2 = 0 → (x − 2)(x + 1) = 0 → x = −1 and x = 2. Test x = 0: the parabola gives 4, the line gives 2, so on [−1, 2] the top curve is y = 4 − x² and the bottom curve is y = 2 − x. The vertical gap is

(top − bottom) = (4 − x²) − (2 − x) = −x² + x + 2 = 2 + x − x².

(a) Area:

A = ∫_{−1}^{2} [(4 − x²) − (2 − x)] dx = ∫_{−1}^{2} (2 + x − x²) dx
  = [2x + x²/2 − x³/3]_{−1}^{2}
  = (4 + 2 − 8/3) − (−2 + 1/2 + 1/3)
  = 10/3 − (−7/6)
  = 9/2.

Area = 9/2.

(b) Each square has side s = (top − bottom) = 2 + x − x², so A(x) = s² = (2 + x − x²)². (No π — the cross-sections are squares.)

V = ∫_{−1}^{2} (2 + x − x²)² dx = ∫_{−1}^{2} (x⁴ − 2x³ − 3x² + 4x + 4) dx
  = [x⁵/5 − x⁴/2 − x³ + 2x² + 4x]_{−1}^{2}

At x = 2: 32/5 − 8 − 8 + 8 + 8 = 32/5 + 0 = 32/5.

At x = −1: −1/5 − 1/2 + 1 + 2 − 4 = −1/5 − 1/2 − 1 = −17/10.

V = 32/5 − (−17/10) = 64/10 + 17/10 = 81/10.

Volume = 81/10.

(c) Revolving about the x-axis: both curves are nonnegative on [−1, 2] (the line 2 − x runs from 3 down to 0; the parabola from 3 down to 0), and the top curve is farther from the axis, so the solid is a washer with outer radius R = 4 − x² and inner radius r = 2 − x:

V = π ∫_{−1}^{2} ([4 − x²]² − [2 − x]²) dx
  = π ∫_{−1}^{2} (16 − 8x² + x⁴) − (4 − 4x + x²) dx
  = π ∫_{−1}^{2} (x⁴ − 9x² + 4x + 12) dx
  = π [x⁵/5 − 3x³ + 2x² + 12x]_{−1}^{2}

At x = 2: 32/5 − 24 + 8 + 24 = 32/5 + 8 = 72/5.

At x = −1: −1/5 + 3 + 2 − 12 = −1/5 − 7 = −36/5.

V = π (72/5 − (−36/5)) = π (108/5) = 108π/5.

Volume = 108π/5.

(d) The vertical distance between the curves is (2 + x − x²), and its average value over [−1, 2] is

avg = (1/(2 − (−1))) ∫_{−1}^{2} (2 + x − x²) dx = (1/3)(9/2) = 3/2.

The average vertical distance is 3/2. This means that if the gap between the two curves were constant across [−1, 2], it would equal 3/2; equivalently, a rectangle of width 3 (the interval length) and height 3/2 has the same area, 9/2, as region R.

FRQ 4 — Rubric (9 points)

| Part | Pts | Earned for |

|---|---|---|

| (a) | 2 | Integrand (4 − x²) − (2 − x) with limits −1, 2 (1); answer 9/2 (1) |

| (b) | 2 | A(x) = (2 + x − x²)² with no π (1); answer 81/10 (1) |

| (c) | 3 | Washer form π ∫ (R² − r²) with the constant π (1); correct radii R = 4 − x², r = 2 − x squared separately (1); answer 108π/5 (1) |

| (d) | 2 | Average-value setup (1/3) ∫_{−1}^{2} (top − bottom) dx with the 1/(b−a) factor (1); answer 3/2 with interpretation (1) |

Where students lose points. In (a), wrong intersection limits (e.g. using 0 and 2) or reversed order of subtraction. In (b), carrying a π into a non-revolution cross-section. In (c), writing (R − r)² instead of R² − r² — these are different numbers and lose two points; also dropping the π loses a point even if all else is right. In (d), omitting the 1/(b−a) factor (that gives 9/2, the area, not the average), or giving the number with no interpretation.

FRQ 5 — Differential Equation (Unit 7) · [NO CALC]

(9 points)

Consider the differential equation

dy/dx = 2x y .

(a) On the axes provided, a slope field for this differential equation is to be drawn at the twelve indicated points. Find the slope dy/dx at the points (1, 1), (−1, 1), (0, 2), and (2, 1), and describe the behavior of the slope field along the entire line x = 0. (2 points)

slope field for dy/dx = 2xy on [−2, 2] × [0, 3

(b) Describe all points in the plane at which dy/dx = 0, and explain what this tells you about the solution curves there. (2 points)

(c) Find the particular solution y = f(x) to the differential equation with the initial condition f(0) = 3. (4 points)

(d) For the particular solution in part (c), find f(1), and state whether f is concave up or concave down at x = 0. Justify your concavity statement. (1 point)

FRQ 5 — Model Solution

(a) dy/dx = 2xy:

at (1, 1):   2(1)(1) = 2
at (−1, 1):  2(−1)(1) = −2
at (0, 2):   2(0)(2) = 0
at (2, 1):   2(2)(1) = 4

Along the line x = 0, dy/dx = 2(0)(y) = 0 for every y, so every slope segment on the y-axis is horizontal (slope 0).

(b) dy/dx = 2xy = 0 exactly when x = 0 or y = 0. So the slope is zero at every point on the y-axis (x = 0) and at every point on the x-axis (y = 0). This means a solution curve has a horizontal tangent wherever it crosses the y-axis, and the line y = 0 is itself a (constant) solution.

(c) The equation is separable. Separate the variables (y ≠ 0) and integrate:

(1/y) dy = 2x dx
∫ (1/y) dy = ∫ 2x dx
ln|y| = x² + C

Exponentiate: |y| = e^{x² + C} = e^C · e^{x²}, so y = A e^{x²} where A = ±e^C is a nonzero constant. Apply the initial condition f(0) = 3:

3 = A e^{0} = A · 1   ⟹   A = 3.

Therefore

y = f(x) = 3 e^{x²}.

(d) f(1) = 3 e^{(1)²} = 3e. For concavity at x = 0: from dy/dx = 2xy, differentiate using the product rule,

d²y/dx² = 2y + 2x (dy/dx) = 2y + 2x(2xy) = 2y + 4x²y.

At x = 0, y = f(0) = 3: d²y/dx² = 2(3) + 0 = 6 > 0. Since f″(0) = 6 > 0, f is concave up at x = 0.

FRQ 5 — Rubric (9 points)

| Part | Pts | Earned for |

|---|---|---|

| (a) | 2 | Three or four of the four slopes correct: 2, −2, 0, 4 (1); states slopes along x = 0 are all 0 / horizontal (1) |

| (b) | 2 | dy/dx = 0 when x = 0 or y = 0 (1); interprets as horizontal tangents on the axes / y = 0 is a solution (1) |

| (c) | 4 | Separates variables (1/y) dy = 2x dx (1); antiderivatives ln|y| = x² + C (1); solves for y = A e^{x²} including the constant (1); applies f(0) = 3 to get y = 3e^{x²} (1) |

| (d) | 1 | f(1) = 3e and concave up at x = 0 with a supporting reason (f″(0) = 6 > 0) (1) |

Where students lose points. In (c), the four most common losses: (i) forgetting the constant of integration C, (ii) dropping the absolute value / mishandling the exponential so the constant A is lost, (iii) writing y = e^{x²} + C (adding the constant instead of multiplying — fatal, since e^{x²+C} = A e^{x²}), and (iv) not solving explicitly for y. An unjustified concavity claim in (d) earns no credit even if the conclusion is correct.

FRQ 6 — Table-Based Function: Tangent, MVT, IVT (Units 2–5) · [NO CALC]

(9 points)

The temperature of an object, H(t), is measured in degrees Celsius (°C) and recorded at selected times t, measured in hours, for 0 ≤ t ≤ 12. The function H is twice differentiable. Selected values are given in the table below. It is also known that H′(2) = 2 °C/hour.

| t (hours)   |  0 |  2 |  5 |  9 | 12 |
|-------------|----|----|----|----|----|
| H(t) (°C)   | 30 | 36 | 40 | 45 | 44 |

(a) Use the data in the table to approximate H′(4). Show the computation, and using correct units, interpret the meaning of H′(4) in the context of the problem. (2 points)

(b) Write an equation of the line tangent to the graph of H at t = 2, and use it to approximate H(2.5). (2 points)

(c) For 5 < t < 9, must there be a time t = c at which H′(c) = 1.25 °C/hour? Justify your answer. (2 points)

(d) Use a trapezoidal sum with the four subintervals indicated by the table to approximate the average temperature (1/12) ∫₀¹² H(t) dt. Indicate units of measure. (3 points)

FRQ 6 — Model Solution

(a) Approximate H′(4) with the average rate of change over the table interval [2, 5] that contains t = 4:

H′(4) ≈ (H(5) − H(2)) / (5 − 2) = (40 − 36) / (5 − 2) = 4/3 °C/hour.

Interpretation: H′(4) ≈ 4/3 °C/hour means that at time t = 4 hours, the temperature of the object is increasing at a rate of approximately 4/3 (≈ 1.33) degrees Celsius per hour.

(b) Using the given H′(2) = 2 and H(2) = 36, the tangent line at t = 2 is

y = H(2) + H′(2)(t − 2) = 36 + 2(t − 2).

Approximate H(2.5):

H(2.5) ≈ 36 + 2(2.5 − 2) = 36 + 2(0.5) = 37 °C.

(c) Over [5, 9], the average rate of change of H is

(H(9) − H(5)) / (9 − 5) = (45 − 40) / 4 = 5/4 = 1.25 °C/hour.

H is twice differentiable, so it is continuous on [5, 9] and differentiable on (5, 9). By the Mean Value Theorem, there exists a value c in (5, 9) such that

H′(c) = (H(9) − H(5)) / (9 − 5) = 1.25 °C/hour.

Yes — there must be such a time c, guaranteed by the Mean Value Theorem.

(d) The trapezoidal sum over the subintervals [0,2], [2,5], [5,9], [9,12] is

∫₀¹² H(t) dt ≈ (2)·(30+36)/2 + (3)·(36+40)/2 + (4)·(40+45)/2 + (3)·(45+44)/2
            = 66 + 114 + 170 + 133.5
            = 483.5 (°C·hours).

The average temperature is then

(1/12) ∫₀¹² H(t) dt ≈ 483.5/12 ≈ 40.292 °C.

The average temperature over the 12 hours is approximately 40.292 °C (≈ 40.29 °C).

FRQ 6 — Rubric (9 points)

| Part | Pts | Earned for |

|---|---|---|

| (a) | 2 | H′(4) ≈ (40 − 36)/(5 − 2) = 4/3 with computation shown (1); interpretation: temperature increasing at ≈ 4/3 °C/hour at t = 4, with units (1) |

| (b) | 2 | Tangent line y = 36 + 2(t − 2) (1); H(2.5) ≈ 37 °C (1) |

| (c) | 2 | Computes avg rate (45 − 40)/(9 − 5) = 1.25 (1); cites MVT with hypotheses (continuous/differentiable) to conclude "yes" (1) |

| (d) | 3 | Correct trapezoidal setup with the four subintervals (1); evaluates ∫₀¹² H ≈ 483.5 (1); divides by 12 for average ≈ 40.292 °C with units (1) |

Where students lose points. In (a), omitting units or interpreting H′(4) as a temperature rather than a rate. In (b), using a table secant slope instead of the given H′(2) = 2. In (c), citing the IVT (wrong theorem — IVT is about function values, MVT is about the derivative) or failing to state the continuity/differentiability hypotheses. In (d), using equal-width subintervals (the data are unevenly spaced — widths are 2, 3, 4, 3), or forgetting the final 1/12 factor (that gives the integral, 483.5, not the average).

SCORE-CONVERSION GUIDE (Whole Exam)

This conversion is approximate and for practice only. The College Board sets cut scores anew each year; use this band map to gauge readiness, not to predict an exact score.

The AP Calculus AB exam weights the two sections equally: Section I (Multiple Choice) is 50% and Section II (Free Response) is 50%.

Step 1 — Section I (Multiple Choice), 45 questions

MC weighted score = (number correct out of 45) × (60 / 45) = (number correct) × 1.3333

This rescales the 45 MC questions to a maximum of 60 points. (There is no penalty for wrong or blank answers.)

Step 2 — Section II (Free Response), 6 FRQs × 9 points = 54 raw points

FRQ weighted score = (raw FRQ points out of 54) × (60 / 54) = (raw points) × 1.1111

This rescales the 54 FRQ points to a maximum of 60 points.

Step 3 — Composite score (out of 120)

Composite = MC weighted score + FRQ weighted score      (maximum 120)

Step 4 — Approximate AP score band

| Composite (out of 120) | Approx. AP Score | Interpretation |

|---|---|---|

| ≈ 80 – 120 | 5 | Extremely well qualified |

| ≈ 62 – 79 | 4 | Well qualified |

| ≈ 43 – 61 | 3 | Qualified (typical "passing" threshold for credit) |

| ≈ 28 – 42 | 2 | Possibly qualified |

| ≈ 0 – 27 | 1 | No recommendation |

Worked example. Suppose a student answers 32 of 45 MC correct and earns 38 of 54 FRQ points.

MC weighted  = 32 × 1.3333 ≈ 42.67
FRQ weighted = 38 × 1.1111 ≈ 42.22
Composite    ≈ 42.67 + 42.22 ≈ 84.9   →   AP score ≈ 5 (near the 4/5 border)

How to read your result. The single most actionable number is your raw FRQ total out of 54: the free-response section is where precise justification language and correct units earn (or lose) the points that separate a 3 from a 5. If your composite lands in the 3 band, target the FRQ justification and interpretation points first — they are the cheapest points to recover.

CalcIQ · Mock Exam 2 · Section II (Free Response) · Full-length simulation covering Units 1–8

This practice exam is independent study material for AP Calculus AB preparation and is not endorsed by or affiliated with the College Board, which produces the AP Calculus AB exam. "AP" and "Advanced Placement" are registered trademarks of the College Board. The score-conversion band map above is an unofficial approximation for self-assessment only.

Accuracy review: Every integral, derivative, area, volume, average value, and particular solution in all six free-response questions was independently recomputed and symbolically verified (sympy), and every antiderivative was differentiated back to confirm it. Calculator-active values (FRQ 1: total in ≈ 44.323, W′(3) ≈ 3.110, W(8) ≈ 61.182, max at t ≈ 5.935 with W ≈ 64.228; FRQ 2: a(2) ≈ −0.311, displacement ≈ −3.404, x(6) ≈ −1.404, total distance ≈ 9.414, average velocity ≈ −0.567) were confirmed by numerical integration and differentiation. Non-calculator values (FRQ 3: f(4)=13, f(8)=5, max at x=4, inflections at x=2,6, f(2)=9, f(2.1)≈9.4; FRQ 4: area 9/2, square cross-sections 81/10, washer 108π/5, average gap 3/2; FRQ 5: y=3e^{x²}, f(1)=3e, f″(0)=6>0; FRQ 6: H′(4)≈4/3, tangent 36+2(t−2), MVT value 1.25, trapezoid 483.5 with average ≈40.292) were verified exactly. No BC-only topics appear: no Euler's method, no logistic growth, no shell method. Content pending mathematical-accuracy review (Isaac, retired actuary).

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