AP Calculus AB — Practice Exam 2

Format: Section I — 20 multiple-choice questions · Section II — 2 free-response questions Suggested time: 40 min (MCQ) + 30 min (FRQ) Coverage: Limits & Continuity · Differentiation (rules, chain, implicit) · Applications of Derivatives · Integration & Accumulation · Differential Equations · Applications of Integration. All arithmetic verified. (Correct answers are spread across A–D by design.)

Section I — Multiple Choice

1. $\displaystyle\lim_{x\to 3}(x^2 - 2x)$ = A) 3 B) 9 C) 6 D) 15

2. $\dfrac{d}{dx}(5x^3 - 2x)$ = A) $15x^2$ B) $5x^2 - 2$ C) $15x^3 - 2$ D) $15x^2 - 2$

3. $\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}$ = A) 0 B) $\infty$ C) 1 D) undefined

4. If $f(x) = e^{3x}$, then $f'(x)$ = A) $3e^{3x}$ B) $e^{3x}$ C) $3x e^{3x}$ D) $e^{3}$

5. $\dfrac{d}{dx}\big(\sin(x^2)\big)$ = A) $\cos(x^2)$ B) $-2x\cos(x^2)$ C) $2x\sin(x^2)$ D) $2x\cos(x^2)$

6. $\displaystyle\int (6x^2 + 4),dx$ = A) $2x^3 + 4x + C$ B) $12x + C$ C) $6x^3 + 4x + C$ D) $2x^3 + C$

7. The function $f(x) = \dfrac{1}{x-2}$ is discontinuous at $x =$ A) 0 B) 1 C) 2 D) $-2$

8. If $g(x) = x^3 - 3x$, the critical points occur at $x =$ A) 0 only B) $\pm 1$ C) $\pm 3$ D) $\pm\sqrt{3}$

9. $\displaystyle\int_0^2 3x^2,dx$ = A) 4 B) 6 C) 8 D) 12

10. Using the product rule, $\dfrac{d}{dx}(x^2 e^x)$ = A) $2x e^x$ B) $x^2 e^x$ C) $2xe^x + x^2 e^x$ D) $2x + e^x$

11. If $\dfrac{dy}{dx} = ky$ and $y(0) = 100$, the solution is: A) $y = 100 + kt$ B) $y = kt^2$ C) $y = 100k^t$ D) $y = 100 e^{kt}$

12. The slope of the tangent line to $y = x^2$ at $x = 5$ is: A) 5 B) 10 C) 25 D) 2

13. $\displaystyle\lim_{x\to\infty}\dfrac{4x^2 + 1}{2x^2 - x}$ = A) 0 B) 4 C) $\infty$ D) 2

14. For implicit differentiation of $x^2 + y^2 = 25$, $\dfrac{dy}{dx}$ = A) $\dfrac{x}{y}$ B) $-\dfrac{x}{y}$ C) $-\dfrac{y}{x}$ D) $\dfrac{y}{x}$

15. The average value of $f(x) = 2x$ on $[0, 4]$ is: A) 2 B) 4 C) 8 D) 16

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

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

18. A particle's velocity is $v(t) = 3t^2 - 6t$. Its acceleration at $t = 2$ is: A) 0 B) 12 C) 6 D) $-6$

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

20. By the Fundamental Theorem of Calculus, $\dfrac{d}{dx}\displaystyle\int_1^{x} t^3,dt$ = A) $\dfrac{x^4}{4}$ B) $3x^2$ C) $t^3$ D) $x^3$

Section II — Free Response

FRQ 1 (Applications of Derivatives — 5 points). A particle moves along a line with position $s(t) = t^3 - 6t^2 + 9t$ for $t \ge 0$ (in meters, seconds). (a) Find the velocity $v(t)$ and acceleration $a(t)$. (2 pts) (b) Find all times when the particle is at rest. (1 pt) (c) Determine the interval(s) on which the particle moves in the positive direction. (1 pt) (d) Find the total distance traveled on $[0, 2]$. (1 pt)

FRQ 2 (Integration & Accumulation — 4 points). Water flows into a tank at a rate $R(t) = 4t + 2$ liters per minute for $0 \le t \le 5$. (a) Write an expression for the total water added in the first 5 minutes and evaluate it. (2 pts) (b) If the tank initially holds 10 liters, how much water is in the tank at $t = 5$? (1 pt) (c) Find the average rate of inflow over $[0, 5]$. (1 pt)

Answer key (Section I)

Key distribution: A×5, B×4, C×5, D×6.

Worked solutions (Section I)

1. (A) $3^2 - 2(3) = 9 - 6 = 3$. 2. (D) Power rule: $15x^2 - 2$. 3. (C) Standard limit $= 1$. 4. (A) Chain rule: $e^{3x}\cdot 3 = 3e^{3x}$. 5. (D) Chain rule: $\cos(x^2)\cdot 2x = 2x\cos(x^2)$. 6. (A) $\int 6x^2,dx = 2x^3$; $\int 4,dx = 4x$ → $2x^3 + 4x + C$. 7. (C) Denominator zero at $x = 2$ → discontinuity. 8. (B) $g'(x) = 3x^2 - 3 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$. 9. (C) $[x^3]_0^2 = 8$. 10. (C) Product rule: $2x e^x + x^2 e^x$. 11. (D) Exponential growth solution $y = 100e^{kt}$. 12. (B) $y' = 2x = 2(5) = 10$. 13. (D) Ratio of leading coefficients $4/2 = 2$. 14. (B) $2x + 2y,y' = 0 \Rightarrow y' = -x/y$. 15. (B) $\frac{1}{4-0}\int_0^4 2x,dx = \frac{1}{4}[x^2]_0^4 = \frac{16}{4} = 4$. 16. (D) $f' > 0$ increasing; $f'' < 0$ concave down. 17. (A) $\int \frac{1}{x},dx = \ln|x| + C$. 18. (C) $a(t) = v'(t) = 6t - 6$; at $t=2$, $a = 6$. 19. (A) $\int_0^1 (x - x^2),dx = [\tfrac{x^2}{2} - \tfrac{x^3}{3}]_0^1 = \tfrac{1}{2} - \tfrac{1}{3} = \tfrac{1}{6}$. 20. (D) FTC Part 1: derivative of the accumulation function is the integrand at $x$ → $x^3$.

FRQ solutions

FRQ 1 (5 pts). (a) (2) $v(t) = 3t^2 - 12t + 9$; $a(t) = 6t - 12$. (b) (1) $v(t) = 0 \Rightarrow 3(t^2 - 4t + 3) = 0 \Rightarrow (t-1)(t-3) = 0 \Rightarrow t = 1, 3$. (c) (1) $v(t) > 0$ where $(t-1)(t-3) > 0$: on $[0,1)$ and $(3,\infty)$. (d) (1) On $[0,2]$ the particle moves forward then back. $s(0)=0,\ s(1)=4,\ s(2)=2$. Distance $= |4-0| + |2-4| = 4 + 2 = 6$ meters.

FRQ 2 (4 pts). (a) (2) $\int_0^5 (4t+2),dt = [2t^2 + 2t]_0^5 = 2(25) + 10 = 60$ liters. (b) (1) $10 + 60 = 70$ liters. (c) (1) Average rate $= \frac{1}{5}\int_0^5 (4t+2),dt = \frac{60}{5} = 12$ liters/min.

Scoring guidance

• Section I: 20 points (1 each).
• Section II: FRQ 1 = 5 pts, FRQ 2 = 4 pts.

Pedagogy: the FRQs reward process — earn partial credit by writing $v(t)$, setting it to zero, and testing sign intervals even if the final distance is off. Show every antiderivative and evaluation step.