Directions
This mock section contains 15 selected-response questions covering the topics most frequently
missed by Korean AP Calculus AB students: related rates, implicit differentiation, the Fundamental
Theorem with variable bounds, particle motion, the Mean Value Theorem, L'Hôpital's Rule, non-uniform
Riemann/trapezoidal sums, optimization, volumes of revolution, differential equations, concavity,
the limit definition of the derivative, piecewise differentiability, average value, and accumulation
functions. A graphing calculator (Desmos) is available for this set. Select an answer choice to see
whether it is correct and read the full explanation immediately. Use the Print button to
generate a clean, ink-saving paper copy.
Question 1Related Rates
Water is pumped into an inverted conical tank at a constant rate of \(2\pi\) cubic feet per minute.
The tank has height \(12\) feet and top radius \(3\) feet. At the instant the water depth is \(4\) feet,
how fast is the water level rising?
(A)\(2\) ft/min
(B)\(8\) ft/min
(C)\(\dfrac{1}{2}\) ft/min
(D)\(\dfrac{32}{\pi}\) ft/min
Since the tank is a cone, similar triangles give \(\dfrac{r}{h}=\dfrac{3}{12}=\dfrac{1}{4}\), so
\(r=\dfrac{h}{4}\). Substituting into \(V=\dfrac{1}{3}\pi r^2 h\) gives
\(V=\dfrac{\pi}{48}h^3\). Differentiating with respect to \(t\):
\(\dfrac{dV}{dt}=\dfrac{\pi}{16}h^2\dfrac{dh}{dt}\). At \(h=4\): \(\dfrac{\pi}{16}(16)\dfrac{dh}{dt}=\pi\dfrac{dh}{dt}=2\pi\),
so \(\dfrac{dh}{dt}=2\) ft/min. The most common Korean-student error is forgetting to square \(h\) after
substituting the similar-triangle ratio (giving choice B), or inverting the ratio \(r/h\) (giving choice C).
Question 2Implicit Differentiation
If \(xy^2+x^2y=6\), find \(\dfrac{dy}{dx}\) at the point \((1,2)\).
(A)\(-\dfrac{8}{5}\)
(B)\(-2\)
(C)\(\dfrac{8}{5}\)
(D)\(-\dfrac{4}{5}\)
Differentiate both sides with respect to \(x\), applying the product rule to each term:
\(y^2+2xy\,y' + 2xy + x^2y' = 0\). Group the \(y'\) terms:
\(y'(2xy+x^2) = -y^2-2xy\), so \(y' = \dfrac{-y^2-2xy}{2xy+x^2}\). At \((1,2)\): numerator
\(=-(4)-2(1)(2)=-8\), denominator \(=2(1)(2)+1=5\), giving \(y'=-\dfrac{8}{5}\). A very common
mistake is applying the product rule to only one of the two terms, or dropping a sign when
moving terms across the equation.
Question 3FTC · Variable Bounds
Let \(h(x)=\displaystyle\int_{x}^{x^2}\sqrt{1+t^2}\,dt\). Find \(h'(1)\).
(A)\(\sqrt{2}\)
(B)\(0\)
(C)\(2\sqrt{2}\)
(D)\(\sqrt{2}-1\)
Both bounds depend on \(x\), so split the integral and apply the Fundamental Theorem with the
chain rule to each bound separately: if \(f(t)=\sqrt{1+t^2}\), then
\(h'(x) = f(x^2)\cdot\dfrac{d}{dx}(x^2) - f(x)\cdot\dfrac{d}{dx}(x) = 2x\sqrt{1+x^4} - \sqrt{1+x^2}\).
At \(x=1\): \(h'(1) = 2\sqrt{2}-\sqrt{2}=\sqrt{2}\). Students frequently forget the chain-rule
factor on the upper bound, or forget the minus sign / derivative on the lower bound entirely.
Question 4Particle Motion
A particle moves along a line with velocity \(v(t)=t^2-4t+3\) for \(0\le t\le 4\), with position
\(x(0)=0\). What is the total distance traveled from \(t=0\) to \(t=4\)?
(A)\(4\)
(B)\(\dfrac{4}{3}\)
(C)\(\dfrac{8}{3}\)
(D)\(0\)
\(v(t)=(t-1)(t-3)=0\) at \(t=1,3\), so the particle changes direction there. The position
function is \(x(t)=\dfrac{t^3}{3}-2t^2+3t\). Computing: \(x(0)=0,\ x(1)=\dfrac{4}{3},\ x(3)=0,\
x(4)=\dfrac{4}{3}\). Total distance \(=\left|\dfrac{4}{3}-0\right|+\left|0-\dfrac{4}{3}\right|+
\left|\dfrac{4}{3}-0\right|=4\). Choice (B), \(\dfrac{4}{3}\), is only the net
displacement \(x(4)-x(0)\) — the classic trap for this question type.
Question 5Mean Value Theorem
For \(f(x)=x^3-3x\) on \([0,2]\), find the value of \(c\) guaranteed by the Mean Value Theorem.
(A)\(\dfrac{2\sqrt{3}}{3}\)
(B)\(1\)
(C)\(\sqrt{3}\)
(D)\(\dfrac{2}{3}\)
\(f\) is a polynomial, so it is continuous and differentiable everywhere, and MVT applies.
Average rate of change: \(\dfrac{f(2)-f(0)}{2-0}=\dfrac{2-0}{2}=1\). Since
\(f'(x)=3x^2-3\), set \(3x^2-3=1 \Rightarrow x^2=\dfrac{4}{3}\Rightarrow x=\pm\dfrac{2}{\sqrt3}\).
Only \(x=\dfrac{2\sqrt3}{3}\approx1.155\) lies in the open interval \((0,2)\), so this is \(c\).
Forgetting to discard the negative root, or forgetting to rationalize, are the two common slip-ups here.
Question 6L'Hôpital's Rule
Evaluate \(\displaystyle\lim_{x\to 0}\dfrac{e^{2x}-1-2x}{x^2}\).
(A)\(2\)
(B)\(0\)
(C)\(1\)
(D)\(4\)
Direct substitution gives \(\dfrac{0}{0}\), so L'Hôpital's Rule applies:
\(\displaystyle\lim_{x\to0}\dfrac{2e^{2x}-2}{2x}\). This is again \(\dfrac{0}{0}\) at \(x=0\), so
apply L'Hôpital a
second time: \(\displaystyle\lim_{x\to0}\dfrac{4e^{2x}}{2}=\dfrac{4}{2}=2\).
Many students stop after the first application (still indeterminate) or plug in \(x=0\) too early
and conclude the limit is \(0\).
Question 7Trapezoidal Sum (Non-Uniform)
Selected values of a continuous function \(f\) are given below.
| \(x\) |
0 |
1 |
3 |
6 |
| \(f(x)\) |
5 |
7 |
11 |
17 |
Use a trapezoidal sum with these subintervals to approximate \(\displaystyle\int_0^6 f(x)\,dx\).
(A)\(66\)
(B)\(60\)
(C)\(72\)
(D)\(33\)
The subintervals are
not equal width, so each trapezoid must be computed with its own
width: \([0,1]\): width \(1\), area \(=1\cdot\dfrac{5+7}{2}=6\); \([1,3]\): width \(2\), area
\(=2\cdot\dfrac{7+11}{2}=18\); \([3,6]\): width \(3\), area \(=3\cdot\dfrac{11+17}{2}=42\).
Total \(=6+18+42=66\). The most common Korean-student error is treating the table as evenly
spaced (using width \(2\) for every piece), which produces an incorrect total.
Question 8Optimization
An open-top box with a square base is to have a volume of \(32\) cubic units. Find the minimum
possible surface area of the box.
(A)\(48\)
(B)\(64\)
(C)\(32\)
(D)\(96\)
Let the base side be \(x\) and height \(h\). Volume: \(x^2h=32\Rightarrow h=\dfrac{32}{x^2}\).
Since the top is open, surface area \(S=x^2+4xh=x^2+\dfrac{128}{x}\). Minimize:
\(S'(x)=2x-\dfrac{128}{x^2}=0\Rightarrow x^3=64\Rightarrow x=4\), so \(h=2\). Then
\(S(4)=16+32=48\). A second derivative check confirms this is a minimum. Common errors include
forgetting the box is open-top (using \(6x^2\)-type formulas for a closed box) or an algebra slip
solving \(x^3=64\).
Question 9Volume of Revolution
The region bounded by \(y=2x\) and \(y=x^2\) is revolved about the \(x\)-axis. Find the volume of
the resulting solid.
(A)\(\dfrac{64\pi}{15}\)
(B)\(\dfrac{32\pi}{5}\)
(C)\(\dfrac{16\pi}{3}\)
(D)\(\dfrac{8\pi}{3}\)
The curves meet where \(2x=x^2\), i.e. \(x=0,2\). On \([0,2]\), \(y=2x\) is the outer radius and
\(y=x^2\) is the inner radius, so by the washer method:
\(V=\pi\displaystyle\int_0^2\left[(2x)^2-(x^2)^2\right]dx=\pi\int_0^2(4x^2-x^4)\,dx
=\pi\left[\dfrac{4x^3}{3}-\dfrac{x^5}{5}\right]_0^2=\pi\left(\dfrac{32}{3}-\dfrac{32}{5}\right)
=\dfrac{64\pi}{15}\). The most common error is reversing which curve is the outer vs. inner
radius, which changes the sign and the final answer.
Question 10Differential Equations · Growth
A population grows at a rate proportional to its size. If \(P(0)=500\) and \(P(3)=1500\), find \(P(6)\).
(A)\(4500\)
(B)\(3000\)
(C)\(6000\)
(D)\(2250\)
"Proportional to its size" means \(\dfrac{dP}{dt}=kP\), so \(P(t)=500e^{kt}\) — this is
exponential, not linear, growth. From \(P(3)=1500\): \(e^{3k}=3\). Then
\(e^{6k}=(e^{3k})^2=9\), so \(P(6)=500\cdot9=4500\). Choice (B), \(3000\), comes from incorrectly
assuming the population increases by the same amount (\(1000\)) every \(3\) years — a linear
model, which does not apply here.
Question 11Concavity · Inflection Points
If \(f'(x)=x^3-3x^2-9x+5\), at what \(x\)-value(s) does the graph of \(f\) have inflection points?
(A)\(x=-1\) and \(x=3\)
(B)\(x=-1\) only
(C)\(x=3\) only
(D)\(x=0\) and \(x=3\)
Inflection points of \(f\) occur where \(f''\) changes sign — that is, where \(f'\) has a local
max or min. Since we are given \(f'(x)\) directly, differentiate once more:
\(f''(x)=3x^2-6x-9=3(x-3)(x+1)\). This is zero at \(x=-1\) and \(x=3\), and testing intervals
confirms the sign of \(f''\) actually changes at both points, so both are genuine inflection
points. A common mistake here is confusing "given \(f'\)" with "given \(f\)" and differentiating
one too few or too many times.
Question 12Limit Definition of the Derivative
Evaluate \(\displaystyle\lim_{h\to 0}\dfrac{\cos\left(\frac{\pi}{3}+h\right)-\cos\left(\frac{\pi}{3}\right)}{h}\).
(A)\(-\dfrac{\sqrt3}{2}\)
(B)\(\dfrac{1}{2}\)
(C)\(\dfrac{\sqrt3}{2}\)
(D)\(-\dfrac{1}{2}\)
This expression is exactly the limit definition of the derivative,
\(\displaystyle\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}=f'(a)\), with \(f(x)=\cos x\) and
\(a=\dfrac{\pi}{3}\). Since \(f'(x)=-\sin x\), the limit equals
\(-\sin\!\left(\dfrac{\pi}{3}\right)=-\dfrac{\sqrt3}{2}\). No actual limit computation with
trig identities is needed — recognizing the pattern is the key skill this question tests. A
common error is confusing this with the value of \(\cos(\pi/3)\) itself or forgetting the
negative sign from differentiating cosine.
Question 13Piecewise Differentiability
For what values of \(a\) and \(b\) is
\(f(x)=\begin{cases}x^2+1, & x\le 1\\ ax+b, & x>1\end{cases}\)
differentiable at \(x=1\)?
(A)\(a=2,\ b=0\)
(B)\(a=2,\ b=2\)
(C)\(a=1,\ b=1\)
(D)\(a=0,\ b=2\)
Differentiability requires
both continuity and matching slopes at \(x=1\).
Continuity: \(1^2+1=a(1)+b \Rightarrow a+b=2\). Slope matching: the left-hand derivative of
\(x^2+1\) is \(2x\), which equals \(2\) at \(x=1\); this must equal the slope of the line, \(a\).
So \(a=2\), and then \(b=2-2=0\). Many students only enforce continuity and forget that
differentiability additionally requires the derivatives from each side to agree.
Question 14Average Value of a Function
Find the average value of \(f(x)=x^2\) on the interval \([1,4]\).
(A)\(7\)
(B)\(21\)
(C)\(6.25\)
(D)\(8.5\)
Average value \(=\dfrac{1}{b-a}\displaystyle\int_a^b f(x)\,dx=\dfrac{1}{4-1}\int_1^4 x^2\,dx
=\dfrac{1}{3}\left[\dfrac{x^3}{3}\right]_1^4=\dfrac{1}{3}\cdot\dfrac{63}{3}=7\). Choice (B),
\(21\), is the value of the integral
without dividing by the interval length — the single
most common mistake on average-value questions. Choice (D) mistakenly averages the two
endpoint function values instead of integrating.
Question 15Accumulation Functions
Let \(g(x)=\displaystyle\int_0^x (t^2-4)\,dt\) for \(x\ge 0\). At what value of \(x\) does \(g\)
have a local minimum?
(A)\(x=2\)
(B)\(x=-2\)
(C)\(x=0\)
(D)\(x=4\)
By the Fundamental Theorem of Calculus, \(g'(x)=x^2-4\), which is zero at \(x=\pm2\). But the
domain here is restricted to \(x\ge0\), so \(x=-2\) is not in the domain and choice (B) can be
eliminated immediately — a detail many students miss. At \(x=2\): \(g''(x)=2x=4>0\), confirming
a local minimum there.
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