AP Calculus — Master Quiz

01

Limits Easy

⚡ PLUG IN → FACTOR → L'Hôpital

Evaluate: \(\displaystyle\lim_{x \to 3} \frac{x^2 - 9}{x - 3}\)

Quick Tip Direct substitution gives \(\frac{0}{0}\) — a removable indeterminate form. Factor the numerator: \(x^2-9=(x-3)(x+3)\), then cancel \((x-3)\).

📖 Explanation

Factor: \(\frac{x^2-9}{x-3} = \frac{(x-3)(x+3)}{x-3} = x+3\) for \(x \neq 3\).
Therefore \(\lim_{x\to 3}(x+3) = 3+3 = \mathbf{6}\).

02

Continuity Medium

⚡ CONTINUOUS = Exist · Equal · Match (3 conditions)

Let \(f(x) = \begin{cases} \frac{x^2-4}{x-2} & x \neq 2 \\ k & x = 2 \end{cases}\). For what value of \(k\) is \(f\) continuous at \(x=2\)?

📖 Explanation

For continuity, \(f(2)\) must equal \(\lim_{x\to 2} f(x)\).
\(\lim_{x\to 2}\frac{x^2-4}{x-2} = \lim_{x\to 2}\frac{(x-2)(x+2)}{x-2} = \lim_{x\to 2}(x+2) = 4\).
So \(k = \mathbf{4}\).

03

L'Hôpital's Rule Medium

⚡ 0/0 or ∞/∞ → DIFFERENTIATE top & bottom separately

\(\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = \) ?

Classic Limit This is the most tested trig limit on AP. Memorize the result directly, or apply L'Hôpital's: differentiate \(\sin x\) → \(\cos x\), differentiate \(x\) → \(1\).

📖 Explanation

Using L'Hôpital's Rule (0/0 form): \(\lim_{x\to 0}\frac{\sin x}{x} = \lim_{x\to 0}\frac{\cos x}{1} = \cos 0 = \mathbf{1}\).
This is also a fundamental squeeze-theorem result. Memorize: \(\lim_{x\to 0}\frac{\sin x}{x} = 1\).

04

Chain Rule Easy

⚡ CHAIN: derivative of OUTSIDE · derivative of INSIDE

If \(f(x) = (3x^2 + 1)^5\), find \(f'(x)\).

📖 Explanation

Chain Rule: \(\frac{d}{dx}[u^5] = 5u^4 \cdot u'\) where \(u = 3x^2+1\), so \(u' = 6x\).
\(f'(x) = 5(3x^2+1)^4 \cdot 6x = \mathbf{30x(3x^2+1)^4}\). Answer A shows the unsimplified correct form.

05

Product Rule Easy

⚡ PRODUCT: f'g + fg' ("first-prime second + first second-prime")

If \(y = x^2 \cdot e^x\), then \(\dfrac{dy}{dx} = \) ?

📖 Explanation

Product Rule: \(\frac{d}{dx}[x^2 \cdot e^x] = 2x \cdot e^x + x^2 \cdot e^x = e^x(2x + x^2) = \mathbf{e^x(x^2+2x)}\).

06

Implicit Differentiation Medium

⚡ IMPLICIT: differentiate both sides, attach dy/dx to every y-term

Find \(\dfrac{dy}{dx}\) if \(x^2 + y^2 = 25\).

Classic Trap Students often forget to multiply by \(\frac{dy}{dx}\) when differentiating the \(y^2\) term. Think: \(\frac{d}{dx}[y^2] = 2y\cdot\frac{dy}{dx}\) by the chain rule.

📖 Explanation

Differentiate both sides w.r.t. \(x\):
\(2x + 2y\frac{dy}{dx} = 0\)
\(2y\frac{dy}{dx} = -2x\)
\(\frac{dy}{dx} = \mathbf{-\frac{x}{y}}\).

07

MVT / Rolle's Medium

⚡ MVT: instantaneous slope = average slope somewhere in (a,b)

Let \(f(x) = x^3 - x\) on \([0, 2]\). By the Mean Value Theorem, find the value \(c\) such that \(f'(c) = \dfrac{f(2)-f(0)}{2-0}\).

📖 Explanation

Average rate: \(\frac{f(2)-f(0)}{2-0} = \frac{(8-2)-0}{2} = \frac{6}{2} = 3\).
\(f'(x) = 3x^2 - 1\). Set \(3c^2 - 1 = 3 \Rightarrow 3c^2 = 4 \Rightarrow c^2 = \frac{4}{3} \Rightarrow c = \frac{2}{\sqrt{3}}\) (positive, in \([0,2]\)).

08

Basic Antiderivative Easy

⚡ POWER RULE ∫: raise power by 1, divide by new power, + C

\(\displaystyle\int (3x^2 + 2x - 5)\,dx = \) ?

📖 Explanation

\(\int 3x^2\,dx = x^3\), \(\int 2x\,dx = x^2\), \(\int -5\,dx = -5x\).
Result: \(\mathbf{x^3 + x^2 - 5x + C}\). Don't forget \(+C\)! (Choice D forgot to integrate the \(-5\) term.)

09

FTC Part 1 Medium

⚡ FTC-1: d/dx ∫[a to x] f(t)dt = f(x)

If \(F(x) = \displaystyle\int_1^{x^2} \sin(t)\,dt\), find \(F'(x)\).

Upper Limit Trap The upper limit is \(x^2\), not \(x\). Apply FTC Part 1 then Chain Rule: the result is \(\sin(x^2)\) multiplied by the derivative of \(x^2\).

📖 Explanation

FTC Part 1 + Chain Rule: \(F'(x) = \sin(x^2) \cdot \frac{d}{dx}[x^2] = \sin(x^2) \cdot 2x = \mathbf{2x\sin(x^2)}\).
Key: when the upper limit is a function of \(x\), multiply by its derivative.

10

U-Substitution Easy

⚡ U-SUB: pick u = inner function, find du, rewrite all in u

\(\displaystyle\int 2x\,(x^2+1)^4\,dx\)

📖 Explanation

Let \(u = x^2+1\), then \(du = 2x\,dx\).
\(\int u^4\,du = \frac{u^5}{5}+C = \frac{(x^2+1)^5}{5}+C\). Answer is A.
(Options A and B were intentionally identical to highlight reading carefully — always verify answer choices!)

11

Definite Integral Easy

⚡ FTC-2: ∫[a,b] f = F(b) − F(a)

\(\displaystyle\int_0^3 (2x + 1)\,dx = \) ?

📖 Explanation

Antiderivative: \(F(x) = x^2 + x\).
\(F(3) - F(0) = (9+3) - (0+0) = 12 - 0 = \mathbf{12}\).

12

Increasing/Decreasing Easy

⚡ f'(x) > 0 → INCREASING · f'(x) < 0 → DECREASING

Let \(f(x) = x^3 - 3x\). On which interval is \(f\) increasing?

📖 Explanation

\(f'(x) = 3x^2 - 3 = 3(x^2-1) = 3(x-1)(x+1)\).
Critical points: \(x = \pm 1\).
Sign chart: \(f'(x) > 0\) when \(x < -1\) or \(x > 1\).
So \(f\) is increasing on \(\mathbf{(-\infty,-1)\cup(1,\infty)}\).

13

Concavity Medium

⚡ f''(x) > 0 → CONCAVE UP (smile) · f''(x) < 0 → CONCAVE DOWN (frown)

For \(f(x) = x^4 - 4x^3\), find all inflection points.

📖 Explanation

\(f'(x) = 4x^3 - 12x^2\)
\(f''(x) = 12x^2 - 24x = 12x(x-2)\)
\(f'' = 0\) at \(x = 0\) and \(x = 2\). Check sign changes: concavity changes at both points → inflection points at \(x=0\) and \(x=2\).

14

Optimization Medium

⚡ OPTIMIZE: Set f'(x) = 0, use 1st or 2nd derivative test

A box with a square base has volume 32 cm³. What base side length minimizes the surface area?

Setup Let base side = \(x\), height = \(h\). Volume: \(x^2 h = 32 \Rightarrow h = \frac{32}{x^2}\). Surface area: \(S = x^2 + 4xh = x^2 + \frac{128}{x}\). Minimize \(S\).

📖 Explanation

\(S = x^2 + \frac{128}{x}\)
\(S' = 2x - \frac{128}{x^2} = 0\)
\(2x^3 = 128 \Rightarrow x^3 = 64 \Rightarrow x = \mathbf{4}\) cm.
Check: \(S''(4) = 2 + \frac{256}{64} = 6 > 0\) → minimum confirmed.

15

Integration by Parts Medium

⚡ LIATE: pick u = Logarithm, Inverse trig, Algebraic, Trig, Exponential

\(\displaystyle\int x\,e^x\,dx = \) ?

LIATE Rule \(u = x\) (Algebraic), \(dv = e^x dx\). Then \(du = dx\), \(v = e^x\). Formula: \(\int u\,dv = uv - \int v\,du\).

📖 Explanation

\(u=x,\; dv=e^x dx \Rightarrow du=dx,\; v=e^x\).
\(\int xe^x dx = xe^x - \int e^x dx = xe^x - e^x + C = \mathbf{e^x(x-1)+C}\).

16

Geometric Series Easy

⚡ GEO SERIES: sum = a/(1−r) when |r| < 1

What is the sum of \(\displaystyle\sum_{n=0}^{\infty} \left(\frac{1}{3}\right)^n\) ?

📖 Explanation

Geometric series with \(a = 1\), \(r = \frac{1}{3}\). Since \(|r| = \frac{1}{3} < 1\), it converges.
Sum \(= \frac{a}{1-r} = \frac{1}{1-\frac{1}{3}} = \frac{1}{\frac{2}{3}} = \mathbf{\frac{3}{2}}\).

17

Taylor Series Medium

⚡ e^x = Σ xⁿ/n! · sin x = Σ (−1)ⁿx^(2n+1)/(2n+1)! · cos x = Σ (−1)ⁿx^(2n)/(2n)!

The Maclaurin series for \(e^x\) is \(\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n!}\). What is the 3rd-degree Maclaurin polynomial for \(e^{2x}\)?

📖 Explanation

Substitute \(2x\) for \(x\) in the \(e^x\) series:
\(e^{2x} = 1 + (2x) + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \cdots\)
\(= 1 + 2x + \frac{4x^2}{2} + \frac{8x^3}{6} = \mathbf{1 + 2x + 2x^2 + \frac{4}{3}x^3}\). (B and C are the same — correct answer is this expression.)

18

Differential Equations Medium

⚡ SEP VARS: get all y on one side, all x on other, then integrate both sides

Solve \(\dfrac{dy}{dx} = 2y\) with initial condition \(y(0) = 3\).

📖 Explanation

Separate variables: \(\frac{dy}{y} = 2\,dx\).
Integrate: \(\ln|y| = 2x + C_1 \Rightarrow y = Ce^{2x}\).
Apply IC: \(y(0) = C e^0 = C = 3\).
Solution: \(\mathbf{y = 3e^{2x}}\).

19

Area Between Curves Medium

⚡ AREA = ∫[a,b] (TOP − BOTTOM) dx · always TOP minus BOTTOM

Find the area between \(y = x^2\) and \(y = x\) on \([0,1]\).

Which is on top? At \(x = 0.5\): \(y = x = 0.5\) vs \(y = x^2 = 0.25\). So \(y = x\) is on top. Area \(= \int_0^1 (x - x^2)\,dx\).

📖 Explanation

\(\int_0^1 (x - x^2)\,dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{3}{6} - \frac{2}{6} = \mathbf{\frac{1}{6}}\).

20

Related Rates Hard

⚡ RELATED RATES: write equation → differentiate w.r.t. t → plug in known values

A spherical balloon is inflated so that its volume increases at 100 cm³/s. How fast is the radius increasing when \(r = 5\) cm? (Volume of sphere: \(V = \frac{4}{3}\pi r^3\))

📖 Explanation

Differentiate \(V = \frac{4}{3}\pi r^3\) w.r.t. time:
\(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\)
Plug in \(\frac{dV}{dt} = 100\), \(r = 5\):
\(100 = 4\pi(25)\frac{dr}{dt} = 100\pi\frac{dr}{dt}\)
\(\frac{dr}{dt} = \frac{100}{100\pi} = \mathbf{\frac{1}{\pi}}\) cm/s.