AP Calculus AB FRQ Practice | Code Scholars Tutoring

Unit 1 · No calculator · 6 instructional points

Repairing a hole and proving a root

Let $\displaystyle f(x)=(x^2-9)/(x-3)$ for $\displaystyle x\ne3$ and $\displaystyle f(3)=k$ . Also let $\displaystyle g(x)=x^3+x-1$ . No calculator is needed.

Part a · 2 points

Find $\displaystyle \lim_{x\to3}f(x)$ and show the algebra.

Show solution and scoring notes

For $\displaystyle x\ne3$ , $\displaystyle f(x)=x+3$ , so the limit is $\displaystyle 6$ .

Scoring: 1 point for valid factorization; 1 for the limit.

Part b · 2 points

Find $\displaystyle k$ that makes $\displaystyle f$ continuous at $\displaystyle 3$ . Justify.

Show solution and scoring notes

The limit is $\displaystyle 6$ . Continuity requires $\displaystyle f(3)$ to equal that limit, so $\displaystyle k=6$ .

Scoring: 1 point for k; 1 for the continuity condition.

Part c · 2 points

Prove $\displaystyle g$ has a zero in $\displaystyle (0,1)$ .

Show solution and scoring notes

As a polynomial, $\displaystyle g$ is continuous on $\displaystyle [0,1]$ . Since $\displaystyle g(0)=-1$ and $\displaystyle g(1)=1$ , the IVT guarantees a zero strictly between them.

Scoring: 1 point for continuity; 1 for endpoint signs and the IVT conclusion.

Review Unit 1

Unit 2 · No calculator · 6 instructional points

Definition, tangent line, and a corner

Let $\displaystyle f(x)=x^2+1$ and $\displaystyle h(x)=|x-2|$ . No calculator is needed.

Part a · 2 points

Use the limit definition to find $\displaystyle f'(2)$ .

Show solution and scoring notes

$\displaystyle [f(2+s)-f(2)]/s=(4s+s^2)/s=4+s$ for $\displaystyle s\ne0$ . The limit is $\displaystyle 4$ .

Scoring: 1 point for the correct quotient and simplification; 1 for the limit.

Part b · 2 points

Write the tangent line to $\displaystyle f$ at $\displaystyle x=2$ .

Show solution and scoring notes

The point is $\displaystyle (2,5)$ and slope is $\displaystyle 4$ , giving $\displaystyle y-5=4(x-2)$ .

Scoring: 1 point for the point; 1 for the tangent equation.

Part c · 2 points

Explain why $\displaystyle h$ is continuous but not differentiable at $\displaystyle 2$ .

Show solution and scoring notes

The two function limits equal $\displaystyle h(2)=0$ , so continuity holds. The left and right difference-quotient limits are $\displaystyle -1$ and $\displaystyle 1$ , so the derivative does not exist.

Scoring: 1 point for continuity reasoning; 1 for unequal one-sided slopes.

Review Unit 2

Unit 3 · No calculator · 6 instructional points

An implicit curve and its inverse slope

Consider the upper semicircle $\displaystyle y=f(x)=\sqrt{25-x^2}$ near $\displaystyle (3,4)$ , restricted to $\displaystyle 0<x<5$ so $\displaystyle f$ is one-to-one. No calculator is needed.

Part a · 2 points

Use $\displaystyle x^2+y^2=25$ to find $\displaystyle dy/dx$ and its value at $\displaystyle (3,4)$ .

Show solution and scoring notes

$\displaystyle 2x+2yy'=0$ , so $\displaystyle y'=-x/y$ and the slope is $\displaystyle -3/4$ .

Scoring: 1 point for the differentiated equation; 1 for the slope.

Part b · 2 points

Find $\displaystyle y''$ at $\displaystyle (3,4)$ .

Show solution and scoring notes

Differentiate again: $\displaystyle 2+2(y')^2+2yy''=0$ . Therefore $\displaystyle y''=-[1+9/16]/4=-25/64$ .

Scoring: 1 point for a valid second derivative equation; 1 for the value.

Part c · 2 points

Find $\displaystyle (f^{-1})'(4)$ . Explain which original point is used.

Show solution and scoring notes

The preimage of $\displaystyle 4$ is $\displaystyle 3$ . Thus $\displaystyle (f^{-1})'(4)=1/f'(3)=-4/3$ .

Scoring: 1 point for identifying the preimage; 1 for the reciprocal derivative.

Review Unit 3

Unit 4 · No calculator · 6 instructional points

A draining conical tank

A conical tank has height $\displaystyle 12$ ft and top radius $\displaystyle 4$ ft. Water drains at $\displaystyle 2$ cubic feet per minute. At an instant the water depth is $\displaystyle 6$ ft. Use $\displaystyle V=\pi r^2h/3$ . No calculator is needed.

Part a · 2 points

Express water volume as a function of depth alone.

Show solution and scoring notes

Similar triangles give $\displaystyle r/h=4/12=1/3$ . Thus $\displaystyle V=\pi(h/3)^2h/3=\pi h^3/27$ .

Scoring: 1 point for the radius-depth relationship; 1 for volume in h.

Part b · 2 points

Find $\displaystyle dh/dt$ when $\displaystyle h=6$ . Include units.

Show solution and scoring notes

$\displaystyle dV/dt=(\pi h^2/9)dh/dt$ . At this instant $\displaystyle -2=4\pi\,dh/dt$ , so $\displaystyle dh/dt=-1/(2\pi)$ feet per minute.

Scoring: 1 point for implicit time differentiation; 1 for the signed rate with units.

Part c · 2 points

Find a linear estimate of the depth $\displaystyle 0.1$ minute later.

Show solution and scoring notes

$\displaystyle h(t+0.1)\approx6-0.1/(2\pi)=6-1/(20\pi)$ feet, using the instantaneous depth rate.

Scoring: 1 point for a linearization setup; 1 for the estimate.

Review Unit 4

Unit 5 · No calculator · 6 instructional points

Extrema, concavity, and a theorem

Let $\displaystyle f(x)=x^3-3x$ on $\displaystyle [-2,2]$ . No calculator is needed.

Part a · 2 points

Find all absolute extrema, showing the candidate comparison.

Show solution and scoring notes

$\displaystyle f'=3(x^2-1)$ vanishes at $\displaystyle -1,1$ . Candidate values are $\displaystyle f(-2)=-2$ , $\displaystyle f(-1)=2$ , $\displaystyle f(1)=-2$ , $\displaystyle f(2)=2$ . Maximum value $\displaystyle 2$ occurs at $\displaystyle -1,2$ ; minimum value $\displaystyle -2$ occurs at $\displaystyle -2,1$ .

Scoring: 1 point for all candidates; 1 for correct comparisons and extrema.

Part b · 2 points

Find the inflection point and justify it.

Show solution and scoring notes

$\displaystyle f''=6x$ changes from negative to positive at $\displaystyle 0$ . Since $\displaystyle f(0)=0$ , the inflection point is $\displaystyle (0,0)$ .

Scoring: 1 point for second derivative evidence; 1 for a concavity-change justification.

Part c · 2 points

Find all $\displaystyle c\in(-2,2)$ whose tangent slope equals the secant slope over $\displaystyle [-2,2]$ . State why such a point must exist.

Show solution and scoring notes

The polynomial is continuous on the closed interval and differentiable on its interior, so the MVT applies. The secant slope is $\displaystyle (2-(-2))/4=1$ . Solving $\displaystyle 3c^2-3=1$ gives $\displaystyle c=\pm2/\sqrt3$ , both in the interior.

Scoring: 1 point for theorem hypotheses; 1 for the equation and both solutions.

Review Unit 5

Unit 6 · No calculator · 6 instructional points

Tabular accumulation and a composite bound

A continuous rate $\displaystyle r$ has values $\displaystyle r(0)=2$ , $\displaystyle r(1)=4$ , and $\displaystyle r(3)=8$ . Define $\displaystyle A(x)=5+\int_0^x r(t)\,dt$ and $\displaystyle B(x)=\int_0^{x^2}r(t)\,dt$ . No calculator is needed.

Part a · 2 points

Estimate $\displaystyle A(3)$ with a trapezoidal sum on the given partition.

Show solution and scoring notes

The integral estimate is $\displaystyle (2+4)/2\cdot1+(4+8)/2\cdot2=3+12=15$ . Add the initial value to get $\displaystyle A(3)\approx20$ .

Scoring: 1 point for the nonuniform trapezoidal sum; 1 for adding the initial value.

Part b · 2 points

Find $\displaystyle A'(1)$ and explain why the derivative exists.

Show solution and scoring notes

The integrand is continuous, so the FTC gives $\displaystyle A'(x)=r(x)$ . Consequently $\displaystyle A'(1)=4$ .

Scoring: 1 point for use of continuity and FTC; 1 for the value.

Part c · 2 points

Find $\displaystyle B'(1)$ using the FTC and chain rule.

Show solution and scoring notes

$\displaystyle B'(x)=2x\,r(x^2)$ , hence $\displaystyle B'(1)=2r(1)=8$ .

Scoring: 1 point for the composite-bound derivative; 1 for evaluation.

Review Unit 6

Unit 7 · No calculator · 6 instructional points

An initial-value problem and its behavior

A quantity satisfies $\displaystyle dy/dx=2xy$ and $\displaystyle y(0)=3$ . No calculator is needed.

Part a · 2 points

Solve explicitly for $\displaystyle y$ and justify the initial condition.

Show solution and scoring notes

Separate $\displaystyle dy/y=2x\,dx$ . Integration gives $\displaystyle \ln|y|=x^2+C$ , hence $\displaystyle y=Ae^{x^2}$ . At $\displaystyle 0$ , $\displaystyle A=3$ , so $\displaystyle y=3e^{x^2}$ .

Scoring: 1 point for separation/integration; 1 for applying the initial condition.

Part b · 2 points

Find the equation of the tangent line at $\displaystyle x=1$ .

Show solution and scoring notes

$\displaystyle y(1)=3e$ and $\displaystyle y'(1)=2(1)(3e)=6e$ . Thus $\displaystyle y-3e=6e(x-1)$ .

Scoring: 1 point for point and slope; 1 for the tangent equation.

Part c · 2 points

Determine where this solution increases and identify its minimum value. Justify.

Show solution and scoring notes

Since $\displaystyle y>0$ , the derivative $\displaystyle 2xy$ has the sign of $\displaystyle x$ . The solution decreases on $\displaystyle (-\infty,0)$ and increases on $\displaystyle (0,\infty)$ . It attains its absolute minimum $\displaystyle 3$ at $\displaystyle x=0$ .

Scoring: 1 point for sign reasoning; 1 for intervals and the minimum.

Review Unit 7

Unit 8 · No calculator · 6 instructional points

Area, rotation, and average value

The region $\displaystyle R$ lies between $\displaystyle y=x$ and $\displaystyle y=x^2$ for $\displaystyle 0\le x\le1$ . No calculator is needed.

Part a · 2 points

Find the area of $\displaystyle R$ , showing which curve is above the other.

Show solution and scoring notes

On $\displaystyle [0,1]$ , $\displaystyle x\ge x^2$ . The area is $\displaystyle \int_0^1(x-x^2)\,dx=1/6$ .

Scoring: 1 point for top-minus-bottom setup; 1 for evaluation.

Part b · 2 points

Find the volume when $\displaystyle R$ rotates about the x-axis.

Show solution and scoring notes

Outer radius is $\displaystyle x$ , inner radius is $\displaystyle x^2$ . Volume is $\displaystyle \pi\int_0^1(x^2-x^4)\,dx=2\pi/15$ .

Scoring: 1 point for washer radii and integral; 1 for volume.

Part c · 2 points

Find the average vertical width of $\displaystyle R$ on $\displaystyle [0,1]$ .

Show solution and scoring notes

The width function is $\displaystyle w(x)=x-x^2$ . Its average is $\displaystyle \frac1{1-0}\int_0^1w(x)\,dx=1/6$ . The equality to the area occurs because the interval length is one.

Scoring: 1 point for an average-value setup; 1 for the value and interpretation.

Review Unit 8