Score a 5.

A ✓ Factor: \( \frac{x^2-9}{x-3} = \frac{(x-3)(x+3)}{x-3} = x+3 \) for \( x \neq 3 \). As \( x \to 3 \), this approaches \( 6 \).

B ✗ The problem explicitly says \( f(3) = k \), so its value depends entirely on the choice of \( k \).

C ✓ Continuity at \( x=3 \) requires \( \lim_{x\to3}f(x) = f(3) \). Since the limit is \( 6 \), setting \( k=6 \) achieves this.

D ✗ A 0/0 indeterminate form does not mean the limit doesn't exist — it means we need algebra (factoring, L'Hôpital, etc.) to evaluate it.

Lower bound at \( x=1 \): \( 3(1)-1 = 2 \). Upper bound: \( 1^2+2 = 3 \).
Wait — actually both limits: \( \lim_{x\to1}(3x-1) = 2 \) and \( \lim_{x\to1}(x^2+2) = 3 \)…

Correction check: \( 3(1)-1=2 \), \( 1+2=3 \). These are not equal, so the Squeeze Theorem gives \( 2 \leq \lim g(x) \leq 3 \), not a single value. B is true only if both bounds yield the same value. Since lower → 2 and upper → 3 at \( x=1 \), the squeeze does NOT pin the limit.

However the bounds given force: lower = \(2\), upper = \(3\). So B is the intended trap answer — students must verify both bounds give the same limit. They don't here, so D remains true (limit could still exist, just not forced by squeeze). A ✗ (arithmetic error). C ✗ (squeeze applies whenever bounds are known). D ✓ (limits don't require the function to be defined at the point).

Chain Rule: \( f'(x) = \cos(3x^2) \cdot \frac{d}{dx}(3x^2) = \cos(3x^2) \cdot 6x \).

A ✓ Correct application of the chain rule.
B ✗ You cannot just substitute — the argument inside cos must remain \(3x^2\), not simplify to \(6x\).
C ✓ \( f'(0) = 6(0)\cos(0) = 0 \). Correct.
D ✗ At \( x = \sqrt{\pi/3} \): \( 3x^2 = 3(\pi/3) = \pi \), so \( \cos(\pi) = -1 \). Therefore \( f'(\sqrt{\pi/3}) = 6\sqrt{\pi/3} \cdot (-1) = -6\sqrt{\pi/3} \), not positive.

Differentiating \( x^2 + xy + y^2 = 7 \): \( 2x + (y + x y') + 2yy' = 0 \).
Solve: \( y'(x + 2y) = -(2x + y) \), so \( y' = -\dfrac{2x+y}{x+2y} \).
At \( (1,2) \): \( y' = -\dfrac{2+2}{1+4} = -\dfrac{4}{5} \).

A ✓ Correct product rule for \(xy\).
B ✓ Correct calculation.
C ✗ The slope is \(-4/5\), not \(5/4\). (The student confused reciprocal with negative reciprocal.)
D ✓ Point-slope form with slope \(-4/5\) through \((1,2)\) is correct.

\( f(0)=0,\ f(2)=8-6=2 \). Average rate \( = \frac{2-0}{2-0} = 1 \).
\( f'(x) = 3x^2-3 = 1 \Rightarrow x^2 = 4/3 \Rightarrow x = \pm 2/\sqrt{3} \approx \pm 1.155 \).

A ✓ Correct average rate calculation.
B ✗ MVT guarantees at least one such \(c\); here there are actually two candidates, though only one is in \((0,2)\).
C ✓ Algebra is correct.
D ✓ \(2/\sqrt{3} \approx 1.155 \in (0,2)\); \(-2/\sqrt{3}\) is negative, not in the interval.

By FTC Part 2, \( F'(x) = \frac{x^2-1}{x+1} \). Factor: \( \frac{(x-1)(x+1)}{x+1} = x-1 \) (for \( x \neq -1 \)).

A ✗ It can be simplified to \( x-1 \).
B ✓ Correct simplification.
C ✓ Any integral from \(a\) to \(a\) is 0, so \( F(1) = \int_1^1 \cdots = 0 \).
D ✗ \( F'(1) = 1-1 = 0 \), not 2.

\( u = \sin x \): when \( x=0, u=0 \); when \( x=\pi, u=0 \). So \(\int_0^0 u\,du = 0\).
Via identity: \( \int_0^\pi \frac{1}{2}\sin 2x\,dx = \left[-\frac{1}{4}\cos 2x\right]_0^\pi = -\frac{1}{4}(1) + \frac{1}{4}(1) = 0 \).

A ✓ Correct substitution; limits both become 0.
B ✗ The integrand is not always positive (\(\sin x \cos x < 0\) for \( x \in (\pi/2, \pi) \)), and the reasoning is flawed.
C ✓ Correct identity and evaluation.
D ✓ Correct explanation for why substitution gives 0.

Using \(u=x, dv=e^x dx\): \(v=e^x, du=dx\).
\( \int xe^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = e^x(x-1)+C \).

A ✓ LIATE: Log, Inverse trig, Algebraic, Trig, Exponential — pick leftmost as \(u\).
B ✓ Correct final answer.
C ✗ Setting \(u=e^x, dv=x\,dx\) gives a more complicated integral \(\int \frac{x^2}{2}e^x dx\), which is harder.
D ✓ Factoring out \(e^x\): \(xe^x - e^x = e^x(x-1)\). Equivalent to B.

Separate: \( y\,dy = 2x\,dx \). Integrate: \( \frac{y^2}{2} = x^2 + C \).
At \( x=0, y=3 \): \( \frac{9}{2} = 0 + C \Rightarrow C = \frac{9}{2} \).
So \( y^2 = 2x^2 + 9 \Rightarrow y = \sqrt{2x^2+9} \).

A ✓ Correct separation and integration.
B ✗ \( C = 9/2 \neq 0 \). Common error: forgetting to plug in.
C ✓ Correct initial condition application.
D ✓ Correct particular solution; positive root chosen since \( y(0)=3>0 \).

Step 1: \(y' = 0+1=1\), \(y_1 = 1 + 0.5(1) = 1.5\).
Step 2: \(y' = 0.5+1.5 = 2\), \(y_2 = 1.5+0.5(2) = 2.5\).
True solution: \(y = 2e^x - x - 1\), \(y(1) = 2e-2 \approx 3.44\).

A ✓ Correct first step.
B ✓ Correct second step.
C ✗ Euler's method is never exact (unless the slope is constant); it always introduces local truncation error.
D ✓ \(y'' = 1 + y' > 0\), so the curve is concave up. Euler uses tangent lines, which lie below a concave-up curve → underestimate.

\( \frac{a_{n+1}}{a_n} = \frac{n+1}{3^{n+1}} \cdot \frac{3^n}{n} = \frac{n+1}{3n} \).
Limit: \( \lim_{n\to\infty} \frac{n+1}{3n} = \lim_{n\to\infty}\frac{1 + 1/n}{3} = \frac{1}{3} \).

A ✓ Correct ratio setup.
B ✗ The limit is \(1/3\), not \(1/2\).
C ✓ Correct limit value.
D ✓ Ratio Test: \(L < 1\) implies absolute convergence.

Substitute \(-x^2\) for \(x\) in the \(e^x\) series: \(e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} - \frac{x^6}{3!}+\cdots\).
Integrate term by term from 0 to 1: \(1 - \frac{1}{3} + \frac{1}{10} - \frac{1}{42} + \cdots\)
(Check: \(\int_0^1 x^{2k} dx = \frac{1}{2k+1}\).)

A ✓ Correct substitution into Maclaurin series.
B ✓ Correct term-by-term integration.
C ✗ The series for \(e^x\) (and hence \(e^{-x^2}\)) converges for all real \(x\) — infinite radius of convergence.
D ✓ The error function \(\text{erf}(x)\) has no elementary antiderivative; series is the standard approach.

Set equal: \(x^2 = 2x \Rightarrow x(x-2)=0 \Rightarrow x=0,2\).
On \([0,2]\): \(2x \geq x^2\) (check \(x=1\): \(2>1\)). So top = \(2x\), bottom = \(x^2\).
Area \(= \int_0^2(2x-x^2)dx = [x^2 - x^3/3]_0^2 = 4 - 8/3 = 4/3\).

A ✓ Correct intersection points.
B ✗ Wrong order — this gives a negative value. Area must use (top − bottom).
C ✓ Correct integrand.
D ✓ Correct numerical answer.

On \([0,1]\), \(x \geq x^2\). Revolving creates a washer with outer radius \(R=x\), inner radius \(r=x^2\).
\(V = \pi\int_0^1(x^2 - x^4)dx = \pi[x^3/3 - x^5/5]_0^1 = \pi(1/3-1/5) = \pi \cdot 2/15 = 2\pi/15\).

A ✗ There IS a hole (the inner \(x^2\) boundary), so we need the washer method, not disk.
B ✓ Correct identification of outer and inner radii.
C ✓ \(V = \pi\int(R^2-r^2)dx = \pi\int_0^1(x^2-x^4)dx\). Correct.
D ✓ Correct evaluation.

\( v(t) = t^2-4t+3 = (t-1)(t-3) \). Zeros at \(t=1,3\).
\( a(t) = v'(t) = 2t-4 \). At \(t=2\): \(a(2)=0\).
On \((1,3)\): \(v(t) < 0\), \(a(t) = 2t-4\): at \(t=1.5\), \(a=-1<0\), same sign → speeding up. But at \(t=2.5\), \(a=1>0\), opposite sign → slowing down. So NOT uniformly speeding up on entire \((1,3)\).

A ✓ Direction changes when \(v=0\) and \(v\) changes sign: at \(t=1\) and \(t=3\).
B ✗ Speeding up requires velocity and acceleration to have the same sign. This fails at \(t=2.5 \in (1,3)\).
C ✓ Total distance (not displacement) always equals \(\int|v|\,dt\).
D ✓ \(a(2) = 2(2)-4 = 0\).

\(\frac{dy}{dt} = 3t^2-3\), \(\frac{dx}{dt} = 2t\). So \(\frac{dy}{dx} = \frac{3t^2-3}{2t}\).
Horizontal tangent: \(dy/dt = 0 \Rightarrow 3t^2-3=0 \Rightarrow t=\pm1\) (and \(dx/dt \neq 0\) there: \(2(\pm1)\neq0\) ✓).
Vertical tangent: \(dx/dt = 0 \Rightarrow 2t=0 \Rightarrow t=0\) (and \(dy/dt \neq 0\) there: \(-3\neq0\) ✓).

A ✓ Correct formula.
B ✓ Horizontal when numerator = 0, denominator ≠ 0.
C ✗ \(t=\pm1\) give horizontal, not vertical tangents.
D ✓ Vertical when denominator = 0, numerator ≠ 0: at \(t=0\).

Convert: \(r = 2\cos\theta \Rightarrow r^2 = 2r\cos\theta \Rightarrow x^2+y^2 = 2x \Rightarrow (x-1)^2+y^2=1\). Circle center \((1,0)\), radius 1.
The full circle traces once for \(\theta \in [0,\pi]\) (for \(\theta \in (\pi/2,\pi)\), \(r<0\) but still traces the same circle).
Area: \(\frac{1}{2}\int_0^\pi 4\cos^2\theta\,d\theta = 2\int_0^\pi\frac{1+\cos2\theta}{2}d\theta = \int_0^\pi(1+\cos2\theta)d\theta = \pi\). ✓ Also, circle of radius 1: area = \(\pi(1)^2 = \pi\). ✓

A ✓ Correct; full circle in \([0,\pi]\).
B ✗ Using \([0,2\pi]\) would double-count the circle, giving area \(2\pi\) instead of \(\pi\).
C ✓ Correct integral and answer.
D ✓ Correct Cartesian form.

\( f^{(4)}(x) = \frac{6}{(1+x)^4} \). On \([0, 0.5]\), max is at \(x=0\): \(f^{(4)}(0) = 6\).
Lagrange: \(|R_3| \leq \frac{6}{4!}(0.5)^4 = \frac{6}{24}\cdot\frac{1}{16} = \frac{6}{384} = \frac{1}{64}\).
ASET: Next term is \(-\frac{(0.5)^4}{4} = -\frac{1}{64}\), so error \(\leq \frac{1}{64}\).

A ✓ Correct ASET bound.
B ✗ The calculation has an error: \(\frac{6(0.5)^4}{24} = \frac{6/16}{24} = \frac{6}{384} = \frac{1}{64}\), not \(\frac{1}{16}\). The student forgot to divide by \(4! = 24\).
C ✓ The series \(\sum(-1)^{n+1}\frac{x^n}{n}\) alternates and terms decrease to 0 for \(0 < x \leq 1\).
D ✓ Correct Lagrange bound calculation.

Ratio Test: \(\left|\frac{(x-2)^{n+1}}{(n+1)3^{n+1}} \cdot \frac{n\cdot3^n}{(x-2)^n}\right| = \frac{|x-2|}{3}\cdot\frac{n}{n+1} \to \frac{|x-2|}{3}\).
Converges when \(|x-2|<3\), i.e., \(-1 < x < 5\).

At \(x=-1\): \((x-2)^n = (-3)^n\), so term = \(\frac{(-3)^n}{n\cdot3^n} = \frac{(-1)^n}{n}\). This is \(\sum\frac{(-1)^n}{n}\), converges by AST.
At \(x=5\): \((x-2)^n = 3^n\), so term = \(\frac{3^n}{n\cdot3^n} = \frac{1}{n}\). Harmonic series — diverges.

A ✓ Correct radius and center of convergence.
B ✓ Correct endpoint check for \(x=-1\).
C ✗ At \(x=5\), the \(3^n\) factors cancel, leaving \(\sum 1/n\) — the harmonic series, which diverges.
D ✓ Correct; interval of convergence is \([-1, 5)\).

\(\frac{dx}{dt} = e^t\cos t - e^t\sin t = e^t(\cos t - \sin t)\).
\(\frac{dy}{dt} = e^t\sin t + e^t\cos t = e^t(\sin t + \cos t)\).

\(\left(\frac{dx}{dt}\right)^2 = e^{2t}(\cos t-\sin t)^2 = e^{2t}(1 - \sin 2t)\).
\(\left(\frac{dy}{dt}\right)^2 = e^{2t}(\sin t+\cos t)^2 = e^{2t}(1 + \sin 2t)\).
Sum: \(e^{2t}(2) = 2e^{2t}\).

Arc length: \(\int_0^\pi\sqrt{2e^{2t}}\,dt = \int_0^\pi\sqrt{2}\,e^t\,dt = \sqrt{2}[e^t]_0^\pi = \sqrt{2}(e^\pi-1)\).

A ✓ Correct derivatives by product rule.
B ✗ The \(\pm\sin 2t\) terms cancel when summed, leaving \(2e^{2t}\), not \(2e^{2t}\sin 2t\).
C ✓ Correct sum after cancellation.
D ✓ Correct final arc length.