Calculus BC — Master Quiz

Question 01 Limits Easy

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KEY: L'Hôpital's Rule — When you see \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), differentiate top & bottom separately. Repeat if needed.

Evaluate the limit:

\[\lim_{x \to 0} \frac{\sin(3x)}{x}\]

Quick Example

\(\lim_{x \to 0} \frac{\sin x}{x} = 1\) → So \(\lim_{x \to 0} \frac{\sin(5x)}{x} = 5\)

✦ Explanation

Use the standard limit: \(\lim_{x \to 0} \frac{\sin(ax)}{x} = a\). Here \(a = 3\), so the answer is 3. Alternatively, L'Hôpital: differentiate top → \(3\cos(3x)\), bottom → \(1\). At \(x=0\): \(3\cos(0)=3\).

Question 02 Continuity Easy

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KEY: 3 Conditions for Continuity
(1) f(a) exists | (2) lim exists | (3) f(a) = lim
All three must hold. One fails → discontinuous.

Which type of discontinuity does \(f(x) = \dfrac{x^2 - 4}{x - 2}\) have at \(x = 2\)?

✦ Explanation

Factor: \(\frac{(x-2)(x+2)}{x-2} = x+2\) for \(x \neq 2\). The limit = 4, but \(f(2)\) is undefined. The "hole" is fixable → removable discontinuity.

Question 03 Chain Rule Easy

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Chain Rule Mantra: "Derivative of outside, keep inside, times derivative of inside."
\(\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)\)

Find \(f'(x)\) if \(f(x) = \sin(x^3)\).

✦ Explanation

Outside: \(\sin(\square)\) → \(\cos(\square)\). Keep inside: \(x^3\). Multiply by inside's derivative \(3x^2\). Result: \(3x^2\cos(x^3)\). ✓

Question 04 Implicit Differentiation Easy

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Implicit Trick: Differentiate both sides w.r.t. \(x\). Every time you differentiate \(y\), attach \(\dfrac{dy}{dx}\). Then solve for \(\dfrac{dy}{dx}\).

Given \(x^2 + y^2 = 25\), find \(\dfrac{dy}{dx}\).

✦ Explanation

Differentiate: \(2x + 2y\frac{dy}{dx} = 0\). Solve: \(\frac{dy}{dx} = -\frac{x}{y}\). This is the slope of the circle at any point \((x,y)\).

Question 05 Related Rates Easy

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Related Rates Recipe: Draw → Label → Find equation → Differentiate (w.r.t. time) → Plug in knowns → Solve.
Units check: rates always have "per time."

The radius of a circle is growing at 2 cm/sec. How fast is the area growing when the radius is 5 cm?
(Area of circle: \(A = \pi r^2\))

✦ Explanation

\(\frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt}\). Plug in \(r=5, \frac{dr}{dt}=2\): \(\frac{dA}{dt} = 2\pi(5)(2) = 20\pi\) cm²/sec.

Question 06 Mean Value Theorem Easy

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MVT in One Line: "Somewhere in the interval, instantaneous slope = average slope."
\(f'(c) = \dfrac{f(b)-f(a)}{b-a}\) for some \(c \in (a,b)\)

\(f(x) = x^2\) on \([1, 3]\). Find the value of \(c\) guaranteed by the Mean Value Theorem.

✦ Explanation

Average slope: \(\frac{f(3)-f(1)}{3-1} = \frac{9-1}{2} = 4\). Set \(f'(c) = 2c = 4\) → \(c = 2\). Check: \(2 \in (1,3)\) ✓

Question 07 Concavity & Inflection Easy

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Concavity Chart: \(f'' > 0\) → concave up (smile 😊) | \(f'' < 0\) → concave down (frown 🙁)
Inflection point: where \(f''\) changes sign (not just = 0!)

\(f(x) = x^4 - 4x^2\). On what interval is \(f\) concave up?

Hint

\(f'(x) = 4x^3 - 8x\). Find \(f''(x)\) first.

✦ Explanation

\(f''(x) = 12x^2 - 8\). Set \(f''=0\): \(x^2 = \frac{2}{3}\) → \(x = \pm\frac{1}{\sqrt{3}}\). Test: \(f'' > 0\) outside these points → concave up on \(\left(-\infty,-\frac{1}{\sqrt{3}}\right) \cup \left(\frac{1}{\sqrt{3}}, \infty\right)\).

Question 08 U-Substitution Easy

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U-Sub Trigger: See a function and its derivative together → let \(u\) = inner function.
After substitution, no \(x\) should remain — only \(u\) and \(du\).

\[\int 2x \cdot e^{x^2}\, dx\]

✦ Explanation

Let \(u = x^2\), so \(du = 2x\, dx\). The integral becomes \(\int e^u\, du = e^u + C = e^{x^2} + C\).

Question 09 Integration by Parts Easy

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LIATE Order (choose \(u\) from left): Logarithm · Inverse trig · Algebraic · Trig · Exponential
Formula: \(\int u\, dv = uv - \int v\, du\)

\[\int x e^x\, dx\]

✦ Explanation

LIATE: \(u = x\) (Algebraic), \(dv = e^x dx\). Then \(du = dx\), \(v = e^x\).
\(\int xe^x dx = xe^x - \int e^x dx = xe^x - e^x + C = e^x(x-1) + C\).

Question 10 FTC Part 2 Easy

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FTC Part 2: \(\dfrac{d}{dx}\!\int_a^{g(x)}\! f(t)\,dt = f(g(x)) \cdot g'(x)\)
Plug upper limit into integrand, multiply by derivative of upper limit.

Find \(\dfrac{d}{dx}\!\displaystyle\int_1^{x^2} \cos(t)\, dt\).

✦ Explanation

Upper limit \(g(x) = x^2\), \(g'(x) = 2x\). Integrand \(f(t) = \cos t\). By FTC2: \(\cos(x^2) \cdot 2x = 2x\cos(x^2)\).

Question 11 Ratio Test Easy

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Ratio Test: Compute \(L = \lim_{n\to\infty}\left|\dfrac{a_{n+1}}{a_n}\right|\)
\(L < 1\) → converges | \(L > 1\) → diverges | \(L = 1\) → inconclusive

Does the series \(\displaystyle\sum_{n=1}^{\infty} \dfrac{n!}{3^n}\) converge or diverge?

✦ Explanation

\(\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)!}{3^{n+1}} \cdot \frac{3^n}{n!} = \frac{n+1}{3} \to \infty\). Since \(L = \infty > 1\), the series diverges.

Question 12 Taylor Series Easy

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Must-Memorize Maclaurin Series:
\(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\)
\(\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\)
\(\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots\)

The Maclaurin series for \(e^x\) is \(1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots\). What is the Maclaurin series for \(e^{-x^2}\)?

✦ Explanation

Substitute \(x \to -x^2\) into the \(e^x\) series: \(e^{-x^2} = \sum_{n=0}^{\infty}\frac{(-x^2)^n}{n!} = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{n!} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!} + \cdots\)

Question 13 Geometric Series Easy

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Geometric Series Sum: \(\displaystyle\sum_{n=0}^{\infty} ar^n = \dfrac{a}{1-r}\) when \(|r| < 1\)
Spot the pattern: constant ratio between consecutive terms.

\[\sum_{n=0}^{\infty} \frac{2}{3^n} = \;?\]

✦ Explanation

\(a = 2\), \(r = \frac{1}{3}\). Since \(|r| = \frac{1}{3} < 1\), sum \(= \frac{a}{1-r} = \frac{2}{1 - \frac{1}{3}} = \frac{2}{\frac{2}{3}} = 3\).

Question 14 Parametric Slope Easy

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Parametric Slope: \(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}\)
Divide, don't forget the chain rule — never differentiate \(y\) directly with \(x\).

\(x = t^2,\; y = t^3 - t\). Find \(\dfrac{dy}{dx}\) at \(t = 1\).

✦ Explanation

\(\frac{dx}{dt} = 2t,\; \frac{dy}{dt} = 3t^2 - 1\). At \(t=1\): \(\frac{dy}{dx} = \frac{3(1)^2 - 1}{2(1)} = \frac{2}{2} = 1\).

Question 15 Polar Area Easy

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Polar Area Formula: \(A = \dfrac{1}{2}\displaystyle\int_\alpha^\beta [r(\theta)]^2\, d\theta\)
The \(\frac{1}{2}\) is easy to forget — think of tiny triangle sectors!

Find the area enclosed by the polar curve \(r = 2\cos\theta\) for \(0 \leq \theta \leq \pi\).

✦ Explanation

\(A = \frac{1}{2}\int_0^{\pi}(2\cos\theta)^2 d\theta = \frac{1}{2}\int_0^{\pi}4\cos^2\theta\,d\theta = 2\int_0^\pi\frac{1+\cos 2\theta}{2}d\theta = \int_0^\pi(1+\cos2\theta)d\theta = [\theta + \frac{\sin2\theta}{2}]_0^\pi = \pi\). (This is a circle of radius 1 → area \(=\pi\) ✓)

Question 16 Separable ODE Easy

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Separable ODE Steps: (1) Move all \(y\)'s left, all \(x\)'s right. (2) Integrate both sides. (3) Solve for \(y\). (4) Apply IC (initial condition).

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

✦ Explanation

Separate: \(\frac{dy}{y} = 3dx\). Integrate: \(\ln|y| = 3x + C\). Exponentiate: \(y = Ae^{3x}\). Apply IC \(y(0)=2\): \(A=2\). So \(y = 2e^{3x}\).

Question 17 Logistic Growth Easy

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Logistic Equation: \(\dfrac{dP}{dt} = kP\!\left(1 - \dfrac{P}{M}\right)\)
Growth is fastest when \(P = \dfrac{M}{2}\). Levels off at carrying capacity \(M\).

A population follows \(\dfrac{dP}{dt} = 0.4P\!\left(1-\dfrac{P}{500}\right)\). At what population is the growth rate the fastest?

✦ Explanation

Carrying capacity \(M = 500\). Growth rate is maximized at \(P = \frac{M}{2} = 250\). At this point the logistic curve has its steepest slope (inflection point of the \(P\) vs \(t\) graph).

Question 18 Arc Length Easy

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Arc Length Formulas:
Cartesian: \(L = \int_a^b \sqrt{1 + [f'(x)]^2}\, dx\)
Parametric: \(L = \int_\alpha^\beta \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\, dt\)

Which integral gives the arc length of \(y = x^{3/2}\) from \(x = 0\) to \(x = 4\)?

✦ Explanation

\(f'(x) = \frac{3}{2}x^{1/2}\). So \([f'(x)]^2 = \frac{9}{4}x\). Arc length \(= \int_0^4 \sqrt{1 + \frac{9x}{4}}\, dx\).

Question 19 Alternating Series Error Easy

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Alternating Series Remainder: The error is less than or equal to the absolute value of the first omitted term.
\(|S - S_n| \leq |a_{n+1}|\)

The alternating series \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}\) is approximated by its first 3 terms. What is the maximum error?

First few terms

\(S_3 = 1 - \dfrac{1}{4} + \dfrac{1}{9}\). The next term would be \(\dfrac{1}{16}\).

✦ Explanation

By the Alternating Series Estimation Theorem, the error \(\leq |a_4| = \frac{1}{4^2} = \frac{1}{16}\). The 4th term (first omitted) bounds the error.

Question 20 Volume: Washer Method Easy

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Washer Method: \(V = \pi\displaystyle\int_a^b\!\left([R(x)]^2 - [r(x)]^2\right)dx\)
\(R\) = outer radius, \(r\) = inner radius. Always subtract inner from outer — don't forget the \(\pi\)!

The region between \(y = \sqrt{x}\) and \(y = x\) (for \(0 \leq x \leq 1\)) is rotated about the \(x\)-axis. Which integral gives the volume?

✦ Explanation

Outer radius \(R = \sqrt{x}\), inner radius \(r = x\). Washer volume: \(\pi\int_0^1\left[(\sqrt{x})^2 - x^2\right]dx = \pi\int_0^1(x - x^2)dx\). Evaluate: \(\pi\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 = \pi\left(\frac{1}{2}-\frac{1}{3}\right) = \frac{\pi}{6}\).

Calculus BCMaster Quiz

Calculus BC
Master Quiz