AP Calculus BC FRQ Practice

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.

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.

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.

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

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

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.

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.

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

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

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.

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.

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.

$\displaystyle dV/dt=(\pi h^2/9)dh/dt$. 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.

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

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

$\displaystyle f'=3(x^2-1)$ 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.

$\displaystyle f''=6x$ 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.

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.

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.

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.

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

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

Choose $\displaystyle u=x$, $\displaystyle dv=e^x\,dx$, so $\displaystyle du=dx$ and $\displaystyle v=e^x$. The integral is $\displaystyle [xe^x-e^x]_0^1=0-(-1)=1$. The lower endpoint contributes $\displaystyle -1$ to the antiderivative evaluation.

Scoring: 1 point for a correct integration-by-parts setup; 1 point for the antiderivative evaluation and value.

The identity $\displaystyle 2x+3=(x+2)+(x+1)$ gives the decomposition $\displaystyle 1/(x+1)+1/(x+2)$. An antiderivative is $\displaystyle \ln(x+1)+\ln(x+2)$ on this interval. Evaluating yields $\displaystyle \ln2+\ln3-(\ln1+\ln2)=\ln3$.

Scoring: 1 point for the correct partial-fraction decomposition; 1 point for integration and the exact value.

Write $\displaystyle \lim_{b\to\infty}\int_0^b(1+x)^{-2}\,dx$. The finite integral is $\displaystyle [-1/(1+x)]_0^b=1-1/(1+b)$. Its limit is $\displaystyle 1$, so the improper integral converges to $\displaystyle 1$.

Scoring: 1 point for the improper-limit setup; 1 point for evaluation and a justified convergence conclusion.

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.

$\displaystyle y(1)=3e$ 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.

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.

At $\displaystyle (0,2)$, the slope is $\displaystyle 0-2=-2$. The tangent line is $\displaystyle L(x)=2-2x$, so the first Euler value is $\displaystyle y_1=L(1/2)=1$.

Scoring: 1 point for the slope and tangent line; 1 point for the first Euler estimate.

The second step starts at the computed point $\displaystyle (1/2,1)$. Its slope is $\displaystyle 1/2-1=-1/2$. Thus $\displaystyle y_2=1+(1/2)(-1/2)=3/4$, the two-step estimate of $\displaystyle y(1)$.

Scoring: 1 point for recomputing the slope at the first Euler point; 1 point for the second update and value.

Differentiating the exact solution twice gives $\displaystyle y''=3e^{-x}>0$, so the solution is concave up. The tangent line at $\displaystyle 0$ lies below this curve for positive $\displaystyle x$. Therefore the first Euler estimate $\displaystyle 1$ underestimates the true value $\displaystyle y(1/2)$.

Scoring: 1 point for a positive second derivative and concavity; 1 point for the tangent-line error interpretation.

Setting the growth rate to zero gives equilibria $\displaystyle P=0$ and $\displaystyle P=100$. For $\displaystyle 0<P<100$, $\displaystyle P'>0$; above $\displaystyle 100$, $\displaystyle P'<0$. Thus $\displaystyle 100$ organisms is the carrying capacity, and the solution starting at $\displaystyle 20$ increases toward $\displaystyle 100$.

Scoring: 1 point for both equilibria and the carrying capacity; 1 point for sign reasoning and the long-term conclusion.

The growth-rate function is the downward-opening quadratic $\displaystyle g(P)=P-0.01P^2$. Its vertex is at $\displaystyle P=50$, where $\displaystyle g(50)=0.01(50)(50)=25$ organisms per day.

Scoring: 1 point for maximizing at half the carrying capacity with reasoning; 1 point for the rate and units.

Rewrite the equation as $\displaystyle P'=P(1-P/100)$, so $\displaystyle K=100$ and $\displaystyle r=1$. The initial condition gives $\displaystyle 20=100/(1+A)$, hence $\displaystyle A=4$ and $\displaystyle P(t)=100/(1+4e^{-t})$. Setting $\displaystyle P=50$ gives $\displaystyle 4e^{-t}=1$, so $\displaystyle t=\ln4$ days.

Scoring: 1 point for the particular logistic solution; 1 point for solving the half-capacity equation and reporting the time.

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.

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.

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.

Since $\displaystyle f'(x)=x$, the length is $\displaystyle L=\int_0^2\sqrt{1+x^2}\,dx\approx2.958$. The integral uses the derivative of the function inside the length formula.

Scoring: 1 point for the correct arc-length integral; 1 point for its numerical evaluation.

The disk radius is $\displaystyle f(x)=x^2/2$, not $\displaystyle f'(x)$. Therefore $\displaystyle V=\pi\int_0^2(x^2/2)^2\,dx=\frac\pi4[x^5/5]_0^2=8\pi/5\approx5.027$.

Scoring: 1 point for the disk integral with the correct radius; 1 point for its value.

Divide the integral of the function by the interval length $\displaystyle 2-0=2$. Thus $\displaystyle f_{\mathrm{avg}}=\frac12\int_0^2x^2/2\,dx=2/3\approx0.667$. This average height is distinct from both the arc length and the volume.

Scoring: 1 point for averaging over the interval length; 1 point for evaluation and interpretation of the factor.

Differentiate both velocity components: $\displaystyle \mathbf a(t)=\langle-2t\sin(t^2),-e^{-t}\rangle$. Thus $\displaystyle \mathbf a(1)=\langle-2\sin1,-e^{-1}\rangle\approx\langle-1.683,-0.368\rangle$ meters per second squared.

Scoring: 1 point for component derivatives with the chain rule; 1 point for the evaluated vector with units.

Position is initial position plus component-wise displacement: $\displaystyle \mathbf r(1)=\langle1+\int_0^1\cos(t^2)\,dt,-1+\int_0^1e^{-t}\,dt\rangle$. The second component simplifies to $\displaystyle -e^{-1}$, so $\displaystyle \mathbf r(1)\approx\langle1.905,-0.368\rangle$ meters.

Scoring: 1 point for both displacement integrals plus the initial position; 1 point for the final vector.

Integrate speed, the magnitude of velocity: $\displaystyle D=\int_0^1\sqrt{\cos^2(t^2)+e^{-2t}}\,dt\approx1.108$ meters. The magnitude of the net displacement generally differs from this path length.

Scoring: 1 point for the speed integral; 1 point for its numerical evaluation with distance units.

Solve $\displaystyle 2\cos\theta=1$ to obtain $\displaystyle \theta=\pm\pi/3$. The first radius is at least $\displaystyle 1$ exactly for $\displaystyle -\pi/3\le\theta\le\pi/3$, so those are the rays containing the region between the curves.

Scoring: 1 point for both angles; 1 point for the radius comparison and angular interval.

At this angle, $\displaystyle r=\sqrt3$ and $\displaystyle r'=-1$. Using $\displaystyle x'=r'\cos\theta-r\sin\theta$ and $\displaystyle y'=r'\sin\theta+r\cos\theta$ gives $\displaystyle x'=-\sqrt3$ and $\displaystyle y'=1$. Hence $\displaystyle dy/dx=-1/\sqrt3$.

Scoring: 1 point for Cartesian parametric derivatives from the polar equation; 1 point for their evaluated ratio.

The outer radius is $\displaystyle 2\cos\theta$ and the inner radius is $\displaystyle 1$. Thus $\displaystyle A=\frac12\int_{-\pi/3}^{\pi/3}(4\cos^2\theta-1)\,d\theta$. Using $\displaystyle 4\cos^2\theta-1=1+2\cos(2\theta)$ gives $\displaystyle A=[\theta/2+\sin(2\theta)/2]_{-\pi/3}^{\pi/3}=\pi/3+\sqrt3/2$.

Scoring: 1 point for a correct polar-area integral with bounds; 1 point for the exact evaluation.

The absolute ratio of successive terms is $\displaystyle |x-1|n/(n+1)$, which tends to $\displaystyle |x-1|$. The ratio test gives absolute convergence for $\displaystyle |x-1|<1$ and divergence when $\displaystyle |x-1|>1$. The radius is $\displaystyle R=1$. At the center, every term is zero.

Scoring: 1 point for the ratio limit; 1 point for the radius and interior convergence condition.

At $\displaystyle x=0$, each term is $\displaystyle -1/n$, so the series diverges. At $\displaystyle x=2$, it becomes the alternating harmonic series, which converges by the alternating series test but not absolutely. Thus the interval is $\displaystyle (0,2]$, with conditional convergence at $\displaystyle 2$.

Scoring: 1 point for the two justified endpoint tests; 1 point for the interval and conditional-convergence classification.

The three-term estimate is $\displaystyle 1/2-(1/2)^2/2+(1/2)^3/3=1/2-1/8+1/24=5/12$. The positive term magnitudes decrease to zero, so the error is at most the first omitted magnitude $\displaystyle (1/2)^4/4=1/64$.

Scoring: 1 point for the three-term value; 1 point for a valid first-omitted-term error bound with its conditions.

The first four derivative values at $\displaystyle 0$ are $\displaystyle 1,2,4,8$. Divide each by its factorial: $\displaystyle P_3(x)=1+2x+4x^2/2!+8x^3/3!=1+2x+2x^2+\frac43x^3$.

Scoring: 1 point for the derivative values and factorial coefficients; 1 point for the polynomial.

Substitution gives $\displaystyle 1+1/2+1/8+1/48=79/48$. The Taylor remainder is $\displaystyle f^{(4)}(c)(1/4)^4/4!$ for some $\displaystyle c$ between $\displaystyle 0$ and $\displaystyle 1/4$. Since $\displaystyle f^{(4)}(c)=16e^{2c}>0$, the approximation is below the true value.

Scoring: 1 point for the exact estimate; 1 point for the positive-remainder argument and underestimate conclusion.

On $\displaystyle [0,1/4]$, $\displaystyle |f^{(4)}(x)|=16e^{2x}\le16e^{1/2}<32$, so $\displaystyle M=32$ is valid. The error is at most $\displaystyle 32(1/4)^4/4!=32/(256\cdot24)=1/192$. This is a guaranteed upper bound, not an exact error.

Scoring: 1 point for bounding the fourth derivative on the full interval; 1 point for the correctly evaluated Lagrange bound.