Find the value of the limit:
\[ \lim_{x \to 0} \frac{1-\cos(3x)}{x^2} \]
This is a \(0/0\) form, so L'Hôpital's Rule applies. Differentiating numerator and denominator:
\[ \lim_{x\to 0}\frac{3\sin(3x)}{2x} = \frac{3}{2}\lim_{x\to 0}\frac{\sin(3x)}{x} = \frac{3}{2}\cdot 3 = \frac{9}{2} \]
The most common Korean-student error here is forgetting the chain-rule factor of 3 when differentiating \(\cos(3x)\), which gives the wrong answer \(3/2\) (choice C). Always multiply by the derivative of the inner function.
If \(x^2y + y^3 = 10\), what is \(\dfrac{dy}{dx}\) at the point \((1,2)\)?
Differentiate both sides with respect to \(x\), remembering the product rule on \(x^2y\):
\[ 2xy + x^2\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0 \]
\[ \frac{dy}{dx} = \frac{-2xy}{x^2+3y^2} \]
At \((1,2)\): numerator \(=-2(1)(2)=-4\), denominator \(=1+3(4)=13\), so \(\dfrac{dy}{dx}=-\dfrac{4}{13}\). The usual trap is forgetting the \(x^2\) term from the product rule and only writing \(2xy\cdot y' \), which leads to the wrong sign or wrong denominator.
Water drains from a conical tank, vertex pointing down, whose top radius is 4 ft and whose height is 12 ft. When the depth of water is \(h=6\) ft, the volume is decreasing at a rate of \(2\ \text{ft}^3/\text{min}\). At that instant, how fast is the depth \(h\) decreasing?
By similar triangles, \(r = h/3\), so \(V = \dfrac{1}{3}\pi r^2 h = \dfrac{\pi h^3}{27}\). Differentiate:
\[ \frac{dV}{dt} = \frac{\pi h^2}{9}\frac{dh}{dt} \]
At \(h=6\): \(\dfrac{\pi(36)}{9}=4\pi\), so \(-2 = 4\pi \dfrac{dh}{dt}\), giving \(\dfrac{dh}{dt} = -\dfrac{1}{2\pi}\) ft/min. The classic trap is using the wrong similar-triangle ratio (mixing up radius 4 and height 12) or forgetting to cube \(h\) before differentiating.
An open-top box with a square base is to have volume \(32\ \text{ft}^3\). Let \(x\) be the side length of the base. What value of \(x\) minimizes the total surface area of the box?
Height \(h = 32/x^2\). Surface area (no top): \(S = x^2 + 4xh = x^2 + \dfrac{128}{x}\).
\[ S'(x) = 2x - \frac{128}{x^2} = 0 \implies x^3 = 64 \implies x = 4 \]
Since \(S''(x) = 2 + 256/x^3 > 0\), this is a minimum. A common mistake is writing surface area as if the box were closed (adding an extra \(x^2\) for a top that doesn't exist), which shifts the critical value.
A particle moves along a line with velocity \(v(t) = t^2 - 4t + 3\) for \(0 \le t \le 4\). What is the total distance traveled by the particle over this interval?
\(v(t)=(t-1)(t-3)\) changes sign at \(t=1\) and \(t=3\). With \(s(t)=\frac{t^3}{3}-2t^2+3t\): \(s(0)=0,\ s(1)=\frac{4}{3},\ s(3)=0,\ s(4)=\frac{4}{3}\).
Total distance \(= |s(1)-s(0)| + |s(3)-s(1)| + |s(4)-s(3)| = \frac{4}{3}+\frac{4}{3}+\frac{4}{3} = 4\).
This question is a classic trap: the net displacement \(s(4)-s(0) = 4/3\) (choice A) is not the total distance. Whenever velocity changes sign on the interval, you must split the integral at the zeros and add absolute values.
For \(f(x) = x^3 - x\) on \([0,2]\), find the value(s) of \(c\) guaranteed by the Mean Value Theorem.
Average rate of change: \(\dfrac{f(2)-f(0)}{2-0} = \dfrac{6-0}{2}=3\). Set \(f'(c)=3c^2-1=3\), so \(c^2=\dfrac43\), \(c=\pm\dfrac{2}{\sqrt3}\). Only the positive root \(c=\dfrac{2\sqrt3}{3}\approx 1.155\) lies in \((0,2)\), so it is the only valid answer.
The trap is forgetting to discard the negative root, or forgetting to rationalize \(2/\sqrt3\) into \(2\sqrt3/3\) and therefore not recognizing the matching choice.
Let \(g(x) = \displaystyle\int_{x^2}^{x^3} \ln(t)\, dt\) for \(x>0\). Find \(g'(2)\).
By the extended Fundamental Theorem (Leibniz rule):
\[ g'(x) = \ln(x^3)\cdot 3x^2 - \ln(x^2)\cdot 2x = 9x^2\ln x - 4x\ln x \]
At \(x=2\): \(g'(2) = 2\ln 2 (18-4) = 28\ln 2\). The most common error is forgetting to multiply each term by the derivative of its bound (\(3x^2\) and \(2x\)), which incorrectly leaves just \(\ln(x^3)-\ln(x^2)=\ln x\) (choice D).
The table gives values of \(f\) and \(f'\):
\(x=1:\ f(1)=2,\ f'(1)=3 \quad\) \(x=2:\ f(2)=5,\ f'(2)=-1 \quad\) \(x=3:\ f(3)=4,\ f'(3)=2\)
If \(g(x) = f(f(x))\), what is \(g'(1)\)?
Chain rule: \(g'(x) = f'(f(x))\cdot f'(x)\). Since \(f(1)=2\), \(g'(1) = f'(2)\cdot f'(1) = (-1)(3) = -3\).
The most common mistake is plugging \(x=1\) into \(f'\) twice (giving \(f'(1)\cdot f'(1)=9\)) instead of first evaluating \(f(1)=2\) and using \(f'(2)\) for the outer derivative.
Let \(f(x) = \begin{cases} ax^2+1, & x \le 2 \\ 4x-3, & x>2 \end{cases}\). Find the value of \(a\) for which \(f\) is differentiable at \(x=2\).
First, continuity requires \(4a+1 = 5\), so \(a=1\). Checking differentiability with \(a=1\): the left derivative is \(2ax=2(1)(2)=4\) and the right derivative (slope of \(4x-3\)) is \(4\). They match, so \(f\) is differentiable at \(x=2\) exactly when \(a=1\).
Students often stop after checking continuity and forget to verify that the derivatives also match — in this problem they happen to match automatically, but it is essential to check both conditions rather than assume.
Evaluate: \(\displaystyle\lim_{x\to 0} \frac{e^x - 1 - x}{x^2}\)
This is \(0/0\). Applying L'Hôpital once gives \(\dfrac{e^x-1}{2x}\), which is still \(0/0\) at \(x=0\), so apply it a second time: \(\dfrac{e^x}{2} \to \dfrac12\).
The common error is stopping after one application and evaluating \(\dfrac{e^0-1}{2(0)} = \dfrac{0}{0}\) as if it were a final numeric answer, or incorrectly treating it as 0 without recognizing the indeterminate form persists.
Evaluate: \(\displaystyle\int_0^{\pi/2} \sin^3(x)\cos(x)\, dx\)
Let \(u=\sin x\), \(du=\cos x\, dx\). When \(x=0\), \(u=0\); when \(x=\pi/2\), \(u=1\).
\[ \int_0^1 u^3\, du = \left[\frac{u^4}{4}\right]_0^1 = \frac14 \]
A frequent error is forgetting to change the bounds of integration to match the new variable \(u\), and instead re-substituting \(\sin x\) back in and plugging the original \(x\)-bounds directly into the antiderivative in \(u\).
How many inflection points does \(f(x) = x^4 - 4x^3\) have?
\(f''(x) = 12x^2 - 24x = 12x(x-2)\), which is zero at \(x=0\) and \(x=2\). Testing signs: \(f''>0\) on \((-\infty,0)\), \(f''<0\) on \((0,2)\), \(f''>0\) on \((2,\infty)\). Concavity changes at both \(x=0\) and \(x=2\), so there are 2 inflection points.
The trap is assuming every zero of \(f''\) is automatically an inflection point without checking that concavity actually changes sign there — in this problem both do change sign, but that check must never be skipped.
Find the average value of \(f(x)=x^2\) on the interval \([1,4]\).
Average value \(= \dfrac{1}{4-1}\displaystyle\int_1^4 x^2\,dx = \dfrac13\left[\dfrac{x^3}{3}\right]_1^4 = \dfrac13\left(\dfrac{63}{3}\right)=\dfrac13(21)=7\).
The most common mistake is computing the definite integral (21) but forgetting to divide by the length of the interval \((b-a)=3\), which gives the wrong answer of 21 (choice B) — that is the net area, not the average value.
The region bounded by \(y=\sqrt{x}\), \(y=0\), and \(x=4\) is revolved about the x-axis. Find the volume of the resulting solid.
\[ V = \pi\int_0^4 (\sqrt{x})^2\, dx = \pi\int_0^4 x\, dx = \pi\left[\frac{x^2}{2}\right]_0^4 = \pi(8) = 8\pi \]
A common error is squaring \(\sqrt x\) incorrectly (writing \(x^{1/2}\) still instead of \(x\)) or integrating \(x\) as \(x^2/3\) by confusing the disk-method exponent rules.
Let \(f(x) = x^3+2x+1\). Given \(f(1)=4\), find \((f^{-1})'(4)\).
\((f^{-1})'(4) = \dfrac{1}{f'(1)}\) since \(f(1)=4\). Here \(f'(x)=3x^2+2\), so \(f'(1)=5\), giving \((f^{-1})'(4)=\dfrac15\).
The classic error is evaluating \(f'\) at \(x=4\) instead of at \(x=1\) — remember the inverse-function derivative rule uses \(f'\) at the input that maps to the given output, not the output value itself.
If \(y = x^{\sin x}\) for \(x>0\), find \(\dfrac{dy}{dx}\) at \(x=\dfrac{\pi}{2}\).
Take logs: \(\ln y = \sin x \ln x\). Differentiate implicitly:
\[ \frac{y'}{y} = \cos x \ln x + \frac{\sin x}{x} \]
At \(x=\pi/2\): \(\cos(\pi/2)=0\) kills the first term, and \(\sin(\pi/2)/x = 1/(\pi/2)=2/\pi\). Also \(y = (\pi/2)^{\sin(\pi/2)} = (\pi/2)^1 = \pi/2\). So \(y' = y\cdot\dfrac{2}{\pi} = \dfrac{\pi}{2}\cdot\dfrac{2}{\pi} = 1\).
The usual mistake is treating \(x^{\sin x}\) as a normal power function and using the power rule directly (\(\sin x \cdot x^{\sin x - 1}\)), which is invalid whenever the exponent is itself a function of \(x\) — logarithmic differentiation is required.