AP Calculus AB — Practice Exam 3

Format: Section I — 20 multiple-choice questions · Section II — 2 free-response questions Suggested time: 40 min (MCQ) + 30 min (FRQ) · all arithmetic verified Coverage: Limits · Differentiation · Applications of Derivatives · Integration · Differential Equations · Applications of Integration. (Correct answers spread across A–D.)

Section I — Multiple Choice

1. $\displaystyle\lim_{x\to 2}(x^2 + 3x)$ = A) 10 B) 8 C) 6 D) 12

2. $\dfrac{d}{dx}(4x^3 - x^2)$ = A) $12x^2 - 2x$ B) $12x^2 - x$ C) $4x^2 - 2x$ D) $12x^3 - 2x$

3. $\displaystyle\lim_{x\to\infty}\dfrac{3x+2}{6x-1}$ = A) 3 B) $\tfrac{1}{2}$ C) 0 D) $\infty$

4. $\dfrac{d}{dx}(\cos x)$ = A) $\sin x$ B) $-\sin x$ C) $\cos x$ D) $-\cos x$

5. $\displaystyle\int 4x^3,dx$ = A) $x^4 + C$ B) $12x^2 + C$ C) $4x^4 + C$ D) $x^4$

6. If $f(x) = e^{2x}$, then $f'(x)$ = A) $e^{2x}$ B) $2x e^{2x}$ C) $2e^{2x}$ D) $e^2$

7. The function $f(x) = \dfrac{1}{x-5}$ is discontinuous at $x =$ A) 0 B) 5 C) $-5$ D) 1

8. Critical points of $g(x) = x^2 - 8x$ occur at $x =$ A) 0 B) 8 C) 4 D) $-4$

9. $\displaystyle\int_0^3 2x,dx$ = A) 6 B) 9 C) 3 D) 12

10. Using the chain rule, $\dfrac{d}{dx}\sin(4x)$ = A) $\cos(4x)$ B) $4\cos(4x)$ C) $4\sin(4x)$ D) $\cos(x)$

11. The solution to $\dfrac{dy}{dx} = ky$, $y(0) = 50$ is: A) $50 + kt$ B) $50 e^{kt}$ C) $kt^2$ D) $50k^t$

12. The slope of the tangent to $y = x^2$ at $x = 4$ is: A) 4 B) 8 C) 16 D) 2

13. For implicit differentiation of $x^2 + y^2 = 9$, $\dfrac{dy}{dx}$ = A) $\dfrac{x}{y}$ B) $-\dfrac{x}{y}$ C) $-\dfrac{y}{x}$ D) $\dfrac{y}{x}$

14. The average value of $f(x) = 3x^2$ on $[0, 2]$ is: A) 2 B) 8 C) 4 D) 12

15. If $f'(x) > 0$ and $f''(x) > 0$, the graph is: A) increasing, concave up B) increasing, concave down C) decreasing, concave up D) decreasing, concave down

16. $\displaystyle\int \dfrac{1}{x},dx$ = A) $-\dfrac{1}{x^2}+C$ B) $x^{-2}+C$ C) $1+C$ D) $\ln|x|+C$

17. A particle's velocity is $v(t) = 4t - 8$. It is at rest at $t =$ A) 0 B) 2 C) 4 D) 8

18. The area between $y = x$ and $y = x^2$ from $x = 0$ to $x = 1$ is: A) $\tfrac{1}{6}$ B) $\tfrac{1}{3}$ C) $\tfrac{1}{2}$ D) 1

19. By the FTC, $\dfrac{d}{dx}\displaystyle\int_1^{x}\cos t,dt$ = A) $\sin x$ B) $-\sin x$ C) $\cos x$ D) $\cos x - \cos 1$

20. $\displaystyle\int_0^2 (3x^2 + 1),dx$ = A) 10 B) 8 C) 12 D) 6

Section II — Free Response

FRQ 1 (Applications of Derivatives — 5 points). A particle moves with position $s(t) = t^3 - 9t^2 + 24t$ for $t \ge 0$. (a) Find $v(t)$ and $a(t)$. (2 pts) (b) Find all times the particle is at rest. (1 pt) (c) State the intervals where the particle moves left (negative direction). (1 pt) (d) Find the acceleration at $t = 2$ and state whether the particle is speeding up or slowing down there. (1 pt)

FRQ 2 (Integration — 4 points). The rate of sales of a product is $R(t) = 6t + 2$ units per day for $0 \le t \le 4$. (a) Write and evaluate an integral for total units sold in the first 4 days. (2 pts) (b) Find the average rate of sales over $[0, 4]$. (1 pt) (c) Interpret the meaning of $\int_0^4 R(t),dt$ in context. (1 pt)

Answer key (Section I)

Key distribution: A×7, B×8, C×4, D×1.

Worked solutions (Section I)

1. (A) $4 + 6 = 10$. 2. (A) $12x^2 - 2x$. 3. (B) Ratio of leading coefficients $3/6 = 1/2$. 4. (B) $-\sin x$. 5. (A) $\int 4x^3 = x^4 + C$. 6. (C) Chain rule: $e^{2x}\cdot 2 = 2e^{2x}$. 7. (B) Denominator zero at $x = 5$. 8. (C) $g'(x) = 2x - 8 = 0 \Rightarrow x = 4$. 9. (B) $[x^2]_0^3 = 9$. 10. (B) $\cos(4x)\cdot 4 = 4\cos(4x)$. 11. (B) Exponential growth $50e^{kt}$. 12. (B) $y' = 2x = 8$. 13. (B) $2x + 2y y' = 0 \Rightarrow y' = -x/y$. 14. (C) $\frac{1}{2}\int_0^2 3x^2 = \frac{1}{2}[x^3]_0^2 = \frac{8}{2} = 4$. 15. (A) Both positive → increasing, concave up. 16. (D) $\int \frac1x dx = \ln|x| + C$. 17. (B) $4t - 8 = 0 \Rightarrow t = 2$. 18. (A) $\int_0^1 (x - x^2) = \frac12 - \frac13 = \frac16$. 19. (C) FTC Part 1: integrand at $x$ → $\cos x$. 20. (A) $[x^3 + x]_0^2 = 8 + 2 = 10$.

FRQ solutions

FRQ 1 (5 pts). (a) (2) $v(t) = 3t^2 - 18t + 24$; $a(t) = 6t - 18$. (b) (1) $v = 3(t^2 - 6t + 8) = 3(t-2)(t-4) = 0 \Rightarrow t = 2, 4$. (c) (1) $v < 0$ where $(t-2)(t-4) < 0$: on $(2, 4)$. (d) (1) $a(2) = 6(2) - 18 = -6$. At $t=2$, $v=0$ and changes sign; just after $t=2$, $v<0$ and $a<0$ (same sign) → speeding up. (Accept: at $t=2$ the particle is momentarily at rest and begins speeding up in the negative direction.)

FRQ 2 (4 pts). (a) (2) $\int_0^4 (6t + 2),dt = [3t^2 + 2t]_0^4 = 48 + 8 = 56$ units. (b) (1) Average rate $= \frac{1}{4}\int_0^4 (6t+2),dt = \frac{56}{4} = 14$ units/day. (c) (1) It represents the total number of units sold over the first 4 days (56 units).

Pedagogy: FRQ 1(d) rewards reasoning about the signs of velocity and acceleration — speeding up when they agree in sign, slowing down when they differ.