Unit 1 · Limits & Continuity
Q 01 / 20 AB ★★☆ Medium
Evaluate: \(\displaystyle\lim_{x \to 0} \frac{\sin(3x)}{5x}\)
Standard limit identity: \(\displaystyle\lim_{u\to 0}\frac{\sin u}{u}=1\)
Rewrite by introducing the factor \(\frac{3}{3}\): \(\dfrac{\sin(3x)}{5x} = \dfrac{3}{5}\cdot\dfrac{\sin(3x)}{3x}\)
As \(x\to 0\), we have \(3x\to 0\), so \(\dfrac{\sin(3x)}{3x}\to 1\).
Therefore the limit \(=\dfrac{3}{5}\cdot 1 = \dfrac{3}{5}\).   ✅ Answer: C
Q 02 / 20 AB ★★☆ Medium
The function \(f(x)=\dfrac{x^2-4}{x-2}\) has a removable discontinuity at \(x=2\). What value must \(f(2)\) be assigned to make \(f\) continuous there?
Factor the numerator: \(x^2-4=(x-2)(x+2)\).
Cancel \((x-2)\): \(f(x)=x+2\) for \(x\ne 2\).
\(\displaystyle\lim_{x\to 2}f(x)=\lim_{x\to 2}(x+2)=4\).
For continuity, define \(f(2)=4\).   ✅ Answer: C
Q 03 / 20 AB ★★★ Hard
Using L'Hôpital's Rule, evaluate \(\displaystyle\lim_{x\to 0}\frac{e^x - 1 - x}{x^2}\).
At \(x=0\): numerator \(=1-1-0=0\), denominator \(=0\). Form \(\tfrac{0}{0}\) → apply L'Hôpital.
1st application: \(\dfrac{e^x-1}{2x}\). At \(x=0\): still \(\tfrac{0}{0}\) → apply again.
2nd application: \(\dfrac{e^x}{2}\). At \(x=0\): \(\dfrac{e^0}{2}=\dfrac{1}{2}\).
✅ Answer: C
Unit 2 · Differentiation Rules
Q 04 / 20 AB ★☆☆ Easy
If \(f(x)=x^3\sin x\), find \(f'(x)\).
Product Rule: \((uv)'=u'v+uv'\) with \(u=x^3,\;v=\sin x\).
\(u'=3x^2,\quad v'=\cos x\).
\(f'(x)=3x^2\sin x + x^3\cos x\).   ✅ Answer: B
Q 05 / 20 AB ★★☆ Medium
Find \(\dfrac{dy}{dx}\) if \(y=\ln(\cos x)\).
Chain Rule: \(\dfrac{d}{dx}[\ln u]=\dfrac{u'}{u}\) where \(u=\cos x,\;u'=-\sin x\).
\(\dfrac{dy}{dx}=\dfrac{-\sin x}{\cos x}=-\tan x\).   ✅ Answer: C
Q 06 / 20 AB ★★☆ Medium
Use implicit differentiation to find \(\dfrac{dy}{dx}\) given \(x^2+y^2=25\).
Differentiate both sides w.r.t. \(x\): \(2x+2y\dfrac{dy}{dx}=0\).
Isolate: \(2y\dfrac{dy}{dx}=-2x \;\Rightarrow\; \dfrac{dy}{dx}=-\dfrac{x}{y}\).   ✅ Answer: B
Q 07 / 20 AB ★★★ Hard
Find \(f'(x)\) if \(f(x)=e^{x^2}\tan x\).
Product Rule: \(f'=(e^{x^2})'\tan x + e^{x^2}(\tan x)'\).
By Chain Rule: \((e^{x^2})'=2x\,e^{x^2}\). Also \((\tan x)'=\sec^2 x\).
\(f'(x)=2x\,e^{x^2}\tan x+e^{x^2}\sec^2 x=e^{x^2}(2x\tan x+\sec^2 x)\).   ✅ Answer: C
Unit 3 · Applications of Derivatives
Q 08 / 20 AB ★★☆ Medium
The function \(f(x)=2x^3-9x^2+12x-4\) has a local minimum at \(x=\)
\(f'(x)=6x^2-18x+12=6(x^2-3x+2)=6(x-1)(x-2)\).
Critical points: \(x=1\) and \(x=2\).
Sign chart of \(f'\): positive on \((-\infty,1)\), negative on \((1,2)\), positive on \((2,\infty)\).
\(f'\) changes − → + at \(x=2\) → local minimum.   ✅ Answer: C
Q 09 / 20 AB ★★★ Hard
A particle moves along the \(x\)-axis with position \(x(t)=t^3-6t^2+9t\) for \(t\ge 0\). At what value(s) of \(t\) does the particle change direction?
Velocity: \(v(t)=x'(t)=3t^2-12t+9=3(t-1)(t-3)\).
Zeros at \(t=1\) and \(t=3\). Check sign changes:
\(v>0\) on \((0,1)\), \(v<0\) on \((1,3)\), \(v>0\) on \((3,\infty)\). Sign changes at both zeros.
The particle reverses direction at \(t=1\) and \(t=3\).   ✅ Answer: C
Q 10 / 20 AB ★★★ Hard
The Mean Value Theorem guarantees a \(c\in(1,4)\) with \(f'(c)=\dfrac{f(4)-f(1)}{3}\) for \(f(x)=\sqrt{x}\). Find the exact value of \(c\).
Average rate: \(\dfrac{f(4)-f(1)}{4-1}=\dfrac{2-1}{3}=\dfrac{1}{3}\).
\(f'(x)=\dfrac{1}{2\sqrt{x}}\). Set \(f'(c)=\dfrac{1}{3}\): \(\dfrac{1}{2\sqrt{c}}=\dfrac{1}{3}\).
\(2\sqrt{c}=3\;\Rightarrow\;\sqrt{c}=\dfrac{3}{2}\;\Rightarrow\;c=\dfrac{9}{4}\in(1,4)\;\checkmark\).   ✅ Answer: B
Q 11 / 20 AB ★★☆ Medium
If \(f''(x)>0\) on \((a,b)\), which statement about \(f\) on \((a,b)\) is necessarily true?
\(f''>0\) means \(f'\) is increasing — the slope of the tangent is growing.
By definition this is concave up (graph curves like a bowl opening upward \(\cup\)).
Note: \(f''>0\) tells us nothing about the sign of \(f'\) itself (f could be increasing or decreasing but curving upward).   ✅ Answer: C
Unit 4 · Integration Techniques
Q 12 / 20 AB ★☆☆ Easy
Evaluate \(\displaystyle\int_0^{\pi/2}\cos x\,dx\).
Antiderivative of \(\cos x\) is \(\sin x\).
\(\Big[\sin x\Big]_0^{\pi/2}=\sin\!\left(\tfrac{\pi}{2}\right)-\sin(0)=1-0=1\).   ✅ Answer: C
Q 13 / 20 AB ★★☆ Medium
If \(\displaystyle F(x)=\int_1^x\sqrt{t^2+1}\,dt\), then \(F'(x)=\)
Fundamental Theorem of Calculus, Part 1: \(\dfrac{d}{dx}\!\int_a^x f(t)\,dt = f(x)\).
Here \(f(t)=\sqrt{t^2+1}\). Replace \(t\) with \(x\): \(F'(x)=\sqrt{x^2+1}\).   ✅ Answer: B
Q 14 / 20 AB ★★☆ Medium
Evaluate \(\displaystyle\int x\,e^{x^2}\,dx\).
U-substitution: let \(u=x^2\;\Rightarrow\;du=2x\,dx\;\Rightarrow\;x\,dx=\tfrac{du}{2}\).
\(\displaystyle\int x\,e^{x^2}dx=\int e^u\cdot\frac{du}{2}=\frac{1}{2}e^u+C=\frac{1}{2}e^{x^2}+C\).
Verify: \(\dfrac{d}{dx}\!\left[\tfrac{1}{2}e^{x^2}\right]=\tfrac{1}{2}\cdot 2x\,e^{x^2}=x\,e^{x^2}\;\checkmark\).   ✅ Answer: B
Q 15 / 20 AB ★★★ Hard
Evaluate \(\displaystyle\int x\ln x\,dx\) using integration by parts.
IBP: \(\int u\,dv=uv-\int v\,du\). Choose \(u=\ln x\Rightarrow du=\frac{1}{x}dx\) and \(dv=x\,dx\Rightarrow v=\frac{x^2}{2}\).
\(\displaystyle\int x\ln x\,dx=\frac{x^2}{2}\ln x - \int\frac{x^2}{2}\cdot\frac{1}{x}dx=\frac{x^2}{2}\ln x-\int\frac{x}{2}dx\).
\(=\dfrac{x^2}{2}\ln x-\dfrac{x^2}{4}+C\).
Verify: differentiate and recover \(x\ln x\;\checkmark\).   ✅ Answer: B
Unit 5 · Applications of Integration
Q 16 / 20 AB ★★☆ Medium
Find the area of the region enclosed by \(y=x^2\) and \(y=x+2\).
Intersections: \(x^2=x+2\Rightarrow x^2-x-2=0\Rightarrow(x-2)(x+1)=0\), so \(x=-1,\,2\).
On \([-1,2]\): upper curve is \(y=x+2\), lower is \(y=x^2\).
\(A=\displaystyle\int_{-1}^{2}(x+2-x^2)\,dx=\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]_{-1}^{2}\).
At \(x=2\): \(2+4-\tfrac{8}{3}=\tfrac{10}{3}\).   At \(x=-1\): \(\tfrac{1}{2}-2+\tfrac{1}{3}=-\tfrac{7}{6}\).
\(A=\dfrac{10}{3}-\!\left(-\dfrac{7}{6}\right)=\dfrac{20}{6}+\dfrac{7}{6}=\dfrac{27}{6}=\dfrac{9}{2}\).   ✅ Answer: B
Q 17 / 20 AB ★★★ Hard
Find the volume of the solid formed by rotating the region bounded by \(y=\sqrt{x}\), \(x=4\), and \(y=0\) around the \(x\)-axis.
Disk Method: \(V=\pi\displaystyle\int_a^b[f(x)]^2\,dx\).
\(V=\pi\displaystyle\int_0^4(\sqrt{x})^2\,dx=\pi\int_0^4 x\,dx=\pi\left[\frac{x^2}{2}\right]_0^4=\pi\cdot\frac{16}{2}=8\pi\).   ✅ Answer: B
Q 18 / 20 AB ★★★ Hard
A particle starts from rest (\(v(0)=0\)) and has acceleration \(a(t)=6t-2\). What is the displacement from \(t=0\) to \(t=3\)?
Integrate \(a(t)\): \(v(t)=\int(6t-2)\,dt=3t^2-2t+C\). Since \(v(0)=0\Rightarrow C=0\).
Displacement \(=\displaystyle\int_0^3 v(t)\,dt=\int_0^3(3t^2-2t)\,dt=\Big[t^3-t^2\Big]_0^3\).
\(=(27-9)-(0)=18\).   ✅ Answer: B
Unit 6 · BC Only — Infinite Series & Convergence
Q 19 / 20 BC ★★★ Hard
Using the Maclaurin series \(e^x=\displaystyle\sum_{n=0}^{\infty}\dfrac{x^n}{n!}\), write the first three non-zero terms of the Maclaurin series for \(g(x)=e^{-x^2}\).
Substitute \(-x^2\) for \(x\) in \(e^x=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots\).
\(e^{-x^2}=1+(-x^2)+\dfrac{(-x^2)^2}{2!}+\cdots=1-x^2+\dfrac{x^4}{2}-\dfrac{x^6}{6}+\cdots\).
First three non-zero terms: \(1-x^2+\dfrac{x^4}{2!}\).   ✅ Answer: A
Q 20 / 20 BC ★★★ Hard
Determine the convergence behavior of \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\).
Alternating Series Test: \(b_n=\tfrac{1}{n}\) is positive, decreasing, and \(\lim_{n\to\infty}\tfrac{1}{n}=0\). → Series converges.
Absolute convergence test: \(\sum\tfrac{1}{n}\) is the harmonic series → diverges.
Converges but NOT absolutely → conditionally convergent.   ✅ Answer: C
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