This is a \(\tfrac{0}{0}\) form, so L'Hôpital's Rule applies. Differentiating numerator and denominator: \[ \lim_{x\to 0}\frac{2e^{2x}-2}{2x} \] which is still \(\tfrac{0}{0}\), so apply L'Hôpital again: \[ \lim_{x\to 0}\frac{4e^{2x}}{2}=\frac{4}{2}=2 \] Check with the Maclaurin series: \(e^{2x}=1+2x+2x^2+\tfrac{4}{3}x^3+\cdots\), so the numerator is \(2x^2+\tfrac{4}{3}x^3+\cdots\), and dividing by \(x^2\) gives \(2\) as \(x\to 0\). Both methods agree: the answer is 2.

Why students miss this: Choice B (1) comes from forgetting the chain-rule factor of 2 when differentiating \(e^{2x}\) — treating it like \(e^{x}\). Choice D (4) comes from double-counting that same factor of 2 (applying it twice instead of once per derivative). Choice A (0) results from over-cancelling and assuming the numerator vanishes faster than it actually does.

Let \(x\) = distance from wall to base, \(y\) = height of top of ladder. Then \(x^2+y^2=13^2=169\). Differentiating with respect to \(t\): \[ 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0 \] At \(x=5\): \(y=\sqrt{169-25}=\sqrt{144}=12\). Substituting \(\dfrac{dx}{dt}=2\): \[ 2(5)(2)+2(12)\frac{dy}{dt}=0 \;\Rightarrow\; \frac{dy}{dt}=-\frac{20}{24}=-\frac{5}{6}\text{ ft/s} \] The negative sign means the top is descending — the answer is \(-\tfrac{5}{6}\) ft/s.

Why students miss this: Choice B drops the negative sign (forgetting the top is moving down as the base moves away). Choice C swaps \(x\) and \(y\) in the final ratio. Choice D uses \(y=13\) instead of correctly solving \(y=12\) from the Pythagorean relation.

Arc length: \(\displaystyle L=\int_0^2 \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt\), with \(\dfrac{dx}{dt}=3t^2\) and \(\dfrac{dy}{dt}=2t\). \[ L=\int_0^2 \sqrt{9t^4+4t^2}\,dt=\int_0^2 t\sqrt{9t^2+4}\,dt \] Using \(u=9t^2+4\), \(du=18t\,dt\): \[ L=\frac{1}{27}(9t^2+4)^{3/2}\Big|_0^2=\frac{1}{27}\left(40^{3/2}-4^{3/2}\right)=\frac{80\sqrt{10}-8}{27}\approx 9.073 \] Why students miss this: Choice A (9.371) is what you get if you forget to subtract the lower-limit contribution \(-\tfrac{8}{27}\) (treating the antiderivative at \(t=0\) as zero, when it's actually \(\tfrac{1}{27}(4)^{3/2}=\tfrac{8}{27}\)). Choice C comes from adding \(\dfrac{dx}{dt}+\dfrac{dy}{dt}\) instead of squaring each term first. Choice D is a rough arithmetic guess.

Polar area formula: \(\displaystyle A=\frac{1}{2}\int_0^{2\pi} r^2\,d\theta\). \[ A=\frac{1}{2}\int_0^{2\pi}(2+2\cos\theta)^2\,d\theta=\frac{1}{2}\int_0^{2\pi}(4+8\cos\theta+4\cos^2\theta)\,d\theta \] Using \(\cos^2\theta=\tfrac{1+\cos2\theta}{2}\), the integral evaluates to \(\tfrac12(12\pi)=6\pi\).

Why students miss this: Choice D (\(12\pi\)) comes from forgetting the \(\tfrac12\) factor in the polar area formula entirely. Choice A (\(3\pi\)) comes from mistakenly halving again or integrating over only \([0,\pi]\) without doubling back for symmetry. Choice C is a common arithmetic slip when expanding \((2+2\cos\theta)^2\).

Lagrange Error Bound: \(\displaystyle |R_n(x)|\le \frac{M}{(n+1)!}|x-a|^{n+1}\), where \(n=3\) (degree of polynomial), so \(n+1=4\). \[ |R_3(0.5)|\le \frac{1}{4!}(0.5)^4=\frac{1}{24}\cdot\frac{1}{16}=\frac{1}{384} \] Why students miss this: Choice A (\(\tfrac{1}{48}\)) uses \(3!\) instead of \(4!\) — forgetting that the error term uses \((n+1)!\), not \(n!\). Choice B (\(\tfrac{1}{192}\)) uses \(x^3\) instead of \(x^4\) — using the degree of the polynomial instead of the degree of the next term. Choice D over-multiplies by an extra factor of \(0.5\).

By the Ratio Test, the series converges when \(\left|\dfrac{x-3}{4}\right|<1\), i.e. \(|x-3|<4\), giving radius of convergence 4 and open interval \((-1,7)\). Endpoints must be checked separately:

• At \(x=-1\): series becomes \(\displaystyle\sum \frac{(-1)^n}{n}\), the alternating harmonic series — converges (conditionally).
• At \(x=7\): series becomes \(\displaystyle\sum \frac{1}{n}\), the harmonic series — diverges.

So the true interval of convergence is \([-1,7)\).

Why students miss this: The ratio test only tells you about the open interval — it never determines endpoint behavior. Skipping the endpoint check (Choice A) or assuming both endpoints behave the same way (Choices B and D) are the most common errors on this exact problem type.

On \([0,2]\), \(2x \ge x^2\), so \(2x\) is the outer radius and \(x^2\) is the inner radius. \[ V=\pi\int_0^2 \left[(2x)^2-(x^2)^2\right]dx=\pi\int_0^2(4x^2-x^4)\,dx=\pi\left[\frac{4x^3}{3}-\frac{x^5}{5}\right]_0^2 \] \[ =\pi\left(\frac{32}{3}-\frac{32}{5}\right)=\pi\cdot\frac{160-96}{15}=\frac{64\pi}{15} \] Why students miss this: Choice B (\(\tfrac{8\pi}{3}\)) is actually the shell-method answer for rotation about the y-axis — a common mix-up when a student applies the wrong axis formula. Choice C forgets to double a term when expanding, and Choice D accidentally doubles the entire integral.

Shell method: \(\displaystyle V=2\pi\int_0^2 x\big[(2x)-(x^2)\big]dx=2\pi\int_0^2(2x^2-x^3)\,dx\) \[ =2\pi\left[\frac{2x^3}{3}-\frac{x^4}{4}\right]_0^2=2\pi\left(\frac{16}{3}-4\right)=2\pi\cdot\frac{4}{3}=\frac{8\pi}{3} \] Why students miss this: Choice A is the washer-method result for the x-axis rotation — mixing up which method belongs to which axis is the single most common BC volume error. Choice C forgets the factor of \(2\pi\) is applied to the whole bracket, not just one term.

Let \(u=\ln x\), \(du=\tfrac{1}{x}dx\). When \(x=e\), \(u=1\); as \(x\to\infty\), \(u\to\infty\). \[ \int_{1}^{\infty}\frac{1}{u^2}\,du=\left[-\frac{1}{u}\right]_1^{\infty}=0-(-1)=1 \] The integral converges to 1.

Why students miss this: Many students assume any integral involving \(1/x\)-type behavior automatically diverges (like the harmonic series) — but the extra factor of \((\ln x)^2\) in the denominator makes this a convergent \(p\)-integral in disguise (\(p=2>1\) after substitution). Choice C is an arithmetic slip forgetting to negate \(-1/u\) correctly at the boundary.

Separate variables: \(\displaystyle \frac{dy}{y^2}=x\,dx \Rightarrow -\frac{1}{y}=\frac{x^2}{2}+C\) Using \(y(0)=-1\): \(-\dfrac{1}{-1}=0+C \Rightarrow C=1\). So \(-\dfrac{1}{y}=\dfrac{x^2}{2}+1\), giving \[ y=\frac{-1}{\frac{x^2}{2}+1}=\frac{-2}{x^2+2} \] At \(x=1\): \(y(1)=\dfrac{-2}{3}\).

Why students miss this: Choice A drops the negative sign carried from the initial condition \(y(0)=-1\). Choice B mishandles the constant of integration (a very common separable-DE slip). Always verify by plugging your final solution back into the original initial condition.

Carrying capacity \(L=500\); fastest growth occurs at \(P=L/2=250\) (the inflection point of the logistic curve). The closed-form solution is \[ P(t)=\frac{L}{1+Ae^{-kt}},\qquad A=\frac{L-P_0}{P_0}=\frac{450}{50}=9,\quad k=0.3 \] Setting \(P(t)=250\): \(1+9e^{-0.3t}=2 \Rightarrow e^{-0.3t}=\tfrac19 \Rightarrow t=\dfrac{\ln 9}{0.3}\approx 7.324\).

Why students miss this: Choice A comes from a decimal-place slip (computing \(\ln 9\) but forgetting to divide by the small rate \(0.3\) correctly, off by a factor of ten). Recognizing that maximum growth always occurs at exactly half the carrying capacity — without needing calculus at that specific step — is the key shortcut BC students should memorize.

Velocity: \(v(t)=3t^2-12t+9=3(t-1)(t-3)\), zero at \(t=1\) and \(t=3\) — both inside \([0,5]\), so the particle changes direction twice. Evaluate position at each critical time: \[ x(0)=0,\quad x(1)=4,\quad x(3)=0,\quad x(5)=20 \] Total distance \(=|4-0|+|0-4|+|20-0|=4+4+20=28\).

Why students miss this: Choice A (20) is the net displacement \(x(5)-x(0)\) — a very common confusion between "distance" and "displacement." You must always find where \(v(t)=0\) first and sum the absolute value of position changes on each sub-interval, not just look at the start and end points.

Differentiate implicitly: \(2xy+x^2y'+3y^2y'=0 \Rightarrow y'=\dfrac{-2xy}{x^2+3y^2}\). At \((1,2)\): \(y'=\dfrac{-4}{1+12}=-\dfrac{4}{13}\). Differentiate again: \(2y+2xy'+2xy'+x^2y''+6y(y')^2+3y^2y''=0\), i.e. \(2y+4xy'+6y(y')^2+(x^2+3y^2)y''=0\). Substituting \(x=1,y=2,y'=-\tfrac{4}{13}\): \[ 4+4\left(-\tfrac{4}{13}\right)+6(2)\left(\tfrac{16}{169}\right)+13y''=0 \] Solving gives \(y''=-\dfrac{660}{2197}\) (confirmed independently by numerically solving the implicit curve near \((1,2)\) with a finite-difference check).

Why students miss this: Choice A is actually the value of the first derivative \(y'\) — a very common mistake of stopping one differentiation short and submitting \(y'\) as if it were \(y''\). Choice B has a sign error. Always differentiate the first-derivative equation again — don't try to shortcut algebraically.

By similar triangles, \(\dfrac{r}{h}=\dfrac{4}{10}=\dfrac{2}{5}\), so \(r=\tfrac{2}{5}h\). Volume: \(V=\dfrac13\pi r^2h=\dfrac13\pi\left(\tfrac{2}{5}h\right)^2h=\dfrac{4\pi}{75}h^3\). \[ \frac{dV}{dt}=\frac{4\pi}{25}h^2\frac{dh}{dt} \] At \(h=5\): \(\dfrac{dV}{dt}=\dfrac{4\pi}{25}(25)\dfrac{dh}{dt}=4\pi\dfrac{dh}{dt}\). Given \(\dfrac{dV}{dt}=-3\) (draining): \[ \frac{dh}{dt}=\frac{-3}{4\pi}\text{ cm/min} \] Why students miss this: Choice A drops the negative sign (the water level is falling, not rising). Choices C and D come from substituting \(r\) and \(h\) as independent constants instead of first eliminating \(r\) via similar triangles — a classic cone-related-rates trap.

Start from the known series \(\ln(1+u)=u-\dfrac{u^2}{2}+\dfrac{u^3}{3}-\cdots\). Substitute \(u=x^2\): \[ \ln(1+x^2)=x^2-\frac{x^4}{2}+\frac{x^6}{3}-\frac{x^8}{4}+\cdots \] The \(x^6\) term has coefficient \(+\dfrac13\).

Why students miss this: The sign pattern of \(\ln(1+u)\) alternates by power of \(u\), not by power of \(x\). Since \(u=x^2\), the third term (\(u^3\), which becomes \(x^6\)) keeps the positive sign from the original series — students who track signs by the power of \(x\) instead of the power of \(u\) get Choice A instead.

Let \(I=\int e^x\sin x\,dx\). Applying integration by parts twice leads to the classic identity \[ I=e^x\sin x-e^x\cos x-I \;\Rightarrow\; 2I=e^x(\sin x-\cos x) \;\Rightarrow\; I=\frac{e^x(\sin x-\cos x)}{2}+C \] Evaluating from \(0\) to \(\pi\): \[ \left[\frac{e^{\pi}(0-(-1))}{2}\right]-\left[\frac{e^0(0-1)}{2}\right]=\frac{e^{\pi}}{2}+\frac{1}{2}=\frac{e^{\pi}+1}{2}\approx 12.070 \] Why students miss this: Choice C is what happens when a student forgets that \(2I=\cdots\) requires dividing by 2 at the very end — a near-universal slip in the "IBP twice, solve algebraically for I" technique. Choice A has a sign error at the lower bound.

Factor: \(x^2-x-2=(x+1)(x-2)\). Partial fractions: \(\dfrac{3x+5}{(x+1)(x-2)}=\dfrac{A}{x+1}+\dfrac{B}{x-2}\). Solving, \(A=-\tfrac23\), \(B=\tfrac{11}{3}\), so \[ \int_3^5\left(\frac{-2/3}{x+1}+\frac{11/3}{x-2}\right)dx=\left[-\frac23\ln|x+1|+\frac{11}{3}\ln|x-2|\right]_3^5 \] Evaluating (both \(x+1>0\) and \(x-2>0\) throughout \([3,5]\), so absolute values are not needed) gives \(\approx 3.758\).

Why students miss this: Sign or coefficient-swap errors in setting up \(A\) and \(B\) are the most common failure point here — always check your partial fraction decomposition by re-combining it over a common denominator before integrating. Small coefficient slips shift the final numeric answer just enough to match a nearby wrong choice.

For a convergent alternating series satisfying the Alternating Series Test, the error after summing \(n\) terms is bounded by the magnitude of the first omitted term. After 4 terms, the first omitted term is the \(n=5\) term: \[ |{\rm Error}|\le a_5=\frac{1}{5^3}=\frac{1}{125} \] Why students miss this: Choice A uses the 4th term (\(1/4^3=1/64\)) instead of the next, 5th term — a classic off-by-one error. The error bound always uses the first term you did not include, not the last one you did.

Speed is \(\displaystyle\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\), with \(\dfrac{dx}{dt}=2t-4\) and \(\dfrac{dy}{dt}=3t^2-6t\). At \(t=2\): \(\dfrac{dx}{dt}=2(2)-4=0\) and \(\dfrac{dy}{dt}=3(4)-6(2)=0\) — both velocity components vanish simultaneously, so the particle momentarily comes to a complete stop. Minimum speed \(=0\) at \(t=2\).

Why students miss this: Many students assume minimum speed must occur at an endpoint (Choice B is the speed at \(t=0\); Choice C is the speed at \(t=3\)) and never check whether \(v_x\) and \(v_y\) can be zero at the same instant. Always check interior critical points of \(v_x^2+v_y^2\) — speed can genuinely reach exactly zero when a particle reverses direction in both coordinates at once.

By the Fundamental Theorem of Calculus combined with the Chain Rule (since the upper limit is \(x^2\), not just \(x\)): \[ g'(x)=\sqrt{1+(x^2)^3}\cdot\frac{d}{dx}(x^2)=2x\sqrt{1+x^6} \] At \(x=2\): \(g'(2)=2(2)\sqrt{1+64}=4\sqrt{65}\approx 32.249\).

Why students miss this: Choice A is the single most common FTC error on BC exams — evaluating the integrand at the upper limit correctly but forgetting to multiply by the derivative of that upper limit (\(2x\)) from the chain rule. Choice D mistakenly substitutes \(x=2\) directly into \(t^3\) instead of \(x^6\) (using \(x\) instead of \(x^2\) inside the radical).