Correct answer: (B) 5/6 ft/sec

Let \(x\) = distance of the base from the wall, \(y\) = height of the top on the wall. Since the ladder length is fixed: \(x^2+y^2=169\).

Differentiate with respect to \(t\): \(2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}=0\).

When \(x=5\): \(y=\sqrt{169-25}=12\). Substitute \(x=5,\ y=12,\ \dfrac{dx}{dt}=2\):

\[10(2)+24\dfrac{dy}{dt}=0 \;\Rightarrow\; \dfrac{dy}{dt}=-\dfrac{20}{24}=-\dfrac{5}{6}\]

The negative sign means \(y\) is decreasing, so the top slides down at \(\dfrac{5}{6}\) ft/sec. Common error: forgetting to solve for \(y\) first (choice C swaps \(x\) and \(y\)); choice D wrongly assumes \(\frac{dy}{dt}=-\frac{dx}{dt}\).

Correct answer: (A) −8/5

Differentiate both terms with the product rule, treating \(y\) as a function of \(x\):

\[\dfrac{d}{dx}(xy^2)=y^2+2xy\,y', \qquad \dfrac{d}{dx}(x^2y)=2xy+x^2y'\]

So: \(y^2+2xy\,y'+2xy+x^2y'=0 \;\Rightarrow\; y'(2xy+x^2) = -(y^2+2xy)\)

At \((1,2)\): numerator \(=-(4+4)=-8\); denominator \(=(4+1)=5\). So \(y'=-\dfrac{8}{5}\).

Common error: dropping one product-rule term (e.g. forgetting the \(2xy\) piece from either derivative) produces the wrong denominator or sign, leading to (B), (C), or (D).

Correct answer: (C) 2

Direct substitution gives \(\frac{0}{0}\), so apply L'Hôpital's Rule. Differentiate top and bottom:

\[\lim_{x\to0}\dfrac{2e^{2x}-2}{2x}\]

This is still \(\frac{0}{0}\) at \(x=0\), so apply L'Hôpital again:

\[\lim_{x\to0}\dfrac{4e^{2x}}{2}=\dfrac{4}{2}=2\]

Common error: stopping after only one application (giving the indeterminate form \(0/0\) mistaken for an answer, or careless algebra giving (B) or (D)).

Correct answer: (C) 5/3

By the Fundamental Theorem of Calculus (Part 1), \(g'(x)=\dfrac{x^2-4}{x}\) directly — there is no need to integrate first.

\[g'(3)=\dfrac{9-4}{3}=\dfrac{5}{3}\]

Common error: integrating the expression first and then differentiating (wastes time and invites arithmetic mistakes), or forgetting to divide by \(x\) (giving choice A).

Correct answer: (A) 1/√π

Average value \(=\dfrac{1}{\sqrt\pi-0}\displaystyle\int_0^{\sqrt\pi} x\sin(x^2)\,dx\). Let \(u=x^2,\ du=2x\,dx\):

\[\int_0^{\sqrt\pi} x\sin(x^2)\,dx=\dfrac12\int_0^{\pi}\sin(u)\,du=\dfrac12\big[-\cos u\big]_0^{\pi}=\dfrac12(1+1)=1\]

So the average value is \(\dfrac{1}{\sqrt\pi}\cdot 1=\dfrac{1}{\sqrt\pi}\).

Common error: forgetting to divide by the interval length \(\sqrt\pi-0\) (giving choice B), which is the most frequent mistake on average-value problems.

Correct answer: (A) Speeding up; speed = 9

\(v(t)=s'(t)=3t^2-12t+9\), \(a(t)=v'(t)=6t-12\).

At \(t=4\): \(v(4)=48-48+9=9\), and \(a(4)=24-12=12\).

Since \(v(4)\) and \(a(4)\) have the same sign (both positive), the particle is speeding up. Speed \(=|v(4)|=9\).

Common error: confusing velocity with speed, or assuming positive acceleration always means "speeding up" without first checking the sign of velocity.

Correct answer: (A) 2√3/3

Average rate of change: \(\dfrac{f(2)-f(0)}{2-0}=\dfrac{6-0}{2}=3\).

\(f'(x)=3x^2-1\). Set \(f'(c)=3\): \(3c^2-1=3 \Rightarrow c^2=\dfrac43 \Rightarrow c=\dfrac{2}{\sqrt3}=\dfrac{2\sqrt3}{3}\) (the negative root is outside \([0,2]\)).

Common error: leaving the answer as \(\frac{2}{\sqrt3}\) and mismatching it against a rationalized answer choice, or an arithmetic slip when solving \(3c^2=4\).

Correct answer: (B) x = 4

Volume: \(x^2h=32 \Rightarrow h=\dfrac{32}{x^2}\). Surface area (no top): \(S=x^2+4xh=x^2+\dfrac{128}{x}\).

\[S'(x)=2x-\dfrac{128}{x^2}=0 \;\Rightarrow\; 2x^3=128 \;\Rightarrow\; x^3=64 \;\Rightarrow\; x=4\]

\(S''(x)=2+\dfrac{256}{x^3}>0\) for \(x>0\), confirming a minimum. At \(x=4\), \(h=2\) and \(S=48\ \text{ft}^2\).

Common error: optimizing volume instead of surface area, or forgetting the box has no top (using \(4xh\) instead of \(2x^2+4xh\)).

Correct answer: (B) 2

Separate variables: \(\dfrac{dy}{y^2}=x\,dx\). Integrate: \(-\dfrac{1}{y}=\dfrac{x^2}{2}+C\).

Use \(y(0)=1\): \(-1=0+C \Rightarrow C=-1\). So \(-\dfrac1y=\dfrac{x^2}{2}-1 \Rightarrow y=\dfrac{2}{2-x^2}\).

At \(x=1\): \(y(1)=\dfrac{2}{2-1}=2\).

Common error: sign errors when integrating \(1/y^2\), or forgetting to solve explicitly for \(y\) before substituting \(x=1\).

Correct answer: (A)

This requires the chain rule three layers deep. Let \(u=\sin(2x^2+1)\), so \(f=u^3\).

\[f'(x)=3u^2\cdot u' = 3\sin^2(2x^2+1)\cdot\cos(2x^2+1)\cdot(4x)\]

\[=12x\sin^2(2x^2+1)\cos(2x^2+1)\]

Common error: stopping after only two layers of the chain rule and forgetting the derivative of the innermost \(2x^2+1\) (giving choice B), or dropping the outer \(\cos\) factor (choice C).

Correct answer: (D)

Critical points occur where \(f'(x)=0\): \(x=1\) and \(x=-2\).

Near \(x=-2\), the factor \((x+2)^2\) is never negative, so the sign of \(f'(x)\) is controlled entirely by \((x-1)\), which is negative on both sides of \(x=-2\). Since \(f'\) does not change sign at \(x=-2\), it is not a local extremum (it's a stationary/inflection-type point).

Near \(x=1\): for \(x<1\), \(f'(x)<0\); for \(x>1\), \(f'(x)>0\). The sign changes from negative to positive, so \(x=1\) is a local minimum.

Common error: assuming every root of \(f'\) is an extremum, without checking whether the sign actually changes — squared factors are the classic trap.

Correct answer: (A)

Let \(u=x^2+5,\ du=2x\,dx\). When \(x=0,\ u=5\); when \(x=2,\ u=9\).

\[\int_5^9 \sqrt u\cdot\dfrac{du}{2}=\dfrac12\cdot\dfrac23 u^{3/2}\Big|_5^9=\dfrac13\left(9^{3/2}-5^{3/2}\right)=\dfrac{27-5\sqrt5}{3}\]

Common error: forgetting the factor of \(\tfrac12\) from \(du=2x\,dx\) (giving choice B or C), or forgetting to shift the bounds by 5 when substituting (giving choice D).

Correct answer: (A) a = 2, b = 0

Continuity at \(x=1\) requires: \(1^2+1=a(1)+b \Rightarrow a+b=2\).

Differentiability requires the one-sided derivatives to match. Since \(\frac{d}{dx}(x^2+1)=2x\), the left derivative at \(x=1\) is \(2\). The right derivative is simply \(a\). So \(a=2\).

Substituting into \(a+b=2\): \(b=0\).

Common error: checking only continuity (which alone permits infinitely many \((a,b)\) pairs) and forgetting that differentiability adds the second, stronger condition that pins down both constants uniquely.

Correct answer: (C) 1/4

Multiply numerator and denominator by the conjugate \(\sqrt{x}+2\):

\[\dfrac{(\sqrt x-2)(\sqrt x+2)}{(x-4)(\sqrt x+2)}=\dfrac{x-4}{(x-4)(\sqrt x+2)}=\dfrac{1}{\sqrt x+2}\]

As \(x\to4\): \(\dfrac{1}{\sqrt4+2}=\dfrac{1}{4}\).

Common error: substituting \(x=4\) directly into the original expression (gives the undefined form \(0/0\) and tempts students to guess \(0\)), or an arithmetic slip after rationalizing.

Correct answer: (A) 1/(3π) m/min

By similar triangles, \(\dfrac{r}{h}=\dfrac{5}{10}=\dfrac12 \Rightarrow r=\dfrac h2\).

\[V=\dfrac13\pi r^2h=\dfrac13\pi\left(\dfrac h2\right)^2h=\dfrac{\pi h^3}{12}\]

\[\dfrac{dV}{dt}=\dfrac{\pi h^2}{4}\dfrac{dh}{dt}\]

At \(h=6\), with \(\dfrac{dV}{dt}=-3\): \(-3=\dfrac{\pi(36)}{4}\dfrac{dh}{dt}=9\pi\dfrac{dh}{dt} \Rightarrow \dfrac{dh}{dt}=-\dfrac{1}{3\pi}\).

Common error: using the radius directly without substituting \(r=h/2\) first, which leaves two variables in the volume equation and makes the related-rates equation unsolvable.

Correct answer: (A) 83

The subintervals are unequal width, so each trapezoid must be computed separately: \(\frac{\Delta x}{2}\big(f(\text{left})+f(\text{right})\big)\).

\([0,2]:\ \tfrac22(3+5)=8 \quad [2,5]:\ \tfrac32(5+8)=19.5 \quad [5,7]:\ \tfrac22(8+10)=18 \quad [7,10]:\ \tfrac32(10+15)=37.5\)

Total: \(8+19.5+18+37.5=83\).

Common error: using only left endpoints (choice B, \(=67\)) or only right endpoints (choice C, \(=99\)) instead of averaging both — this is the single most common mistake on non-uniform table data. Note that the trapezoidal estimate is always the average of the left and right sums: \((67+99)/2=83\), which is a good way to double-check your work.

Correct answer: (A)

Both bounds are functions of \(x\), so this needs the generalized Fundamental Theorem of Calculus (Leibniz rule):

\[g'(x)=\sqrt{1+(x^3)^2}\cdot\dfrac{d}{dx}(x^3)-\sqrt{1+(x^2)^2}\cdot\dfrac{d}{dx}(x^2)\]

\[=3x^2\sqrt{1+x^6}-2x\sqrt{1+x^4}\]

Common error: forgetting to multiply each term by the derivative of its bound (choice B), or making a sign error and adding instead of subtracting (choice C) — remember the lower-bound term is always subtracted.