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Calculus Honors FRQ practice

6 problems · Worked solutions · 9-point rubrics · Mathematics

Code Scholars practice problems. These are not official exam prompts.

Problem 1 · Limits and Continuity · 9 points

Repair a removable discontinuity

For x≠2\displaystyle x\ne2, let f(x)=x2−4x−2\displaystyle f(x)=\frac{x^2-4}{x-2}, and set f(2)=k\displaystyle f(2)=k.

Problem diagram
-11234512345678xylimit
  1. A. Evaluate the limit as x approaches 2.
  2. B. Find k for continuity and justify with the definition.
  3. C. Find the derivative at 2 after that choice, using a difference quotient.
Show worked solution and rubric
  1. A. For x≠2\displaystyle x\ne2, f(x)=x+2\displaystyle f(x)=x+2, so lim⁡x→2f(x)=4\displaystyle \lim_{x\to2}f(x)=4.
  2. B. Continuity requires the value to equal the limit, hence k=4\displaystyle k=4. With this choice the entire function equals x+2\displaystyle x+2.
  3. C. f′(2)=lim⁡h→0f(2+h)−f(2)h=lim⁡h→04+h−4h=1\displaystyle f\prime(2)=\lim_{h\to0}\frac{f(2+h)-f(2)}h=\lim_{h\to0}\frac{4+h-4}h=1.
Solution diagram
-11234512345678xyk=4

Scoring rubric

  • 3 points: For x≠2\displaystyle x\ne2, f(x)=x+2\displaystyle f(x)=x+2, so lim⁡x→2f(x)=4\displaystyle \lim_{x\to2}f(x)=4.
  • 3 points: Continuity requires the value to equal the limit, hence k=4\displaystyle k=4
  • 3 points: f′(2)=lim⁡h→0f(2+h)−f(2)h=lim⁡h→04+h−4h=1\displaystyle f\prime(2)=\lim_{h\to0}\frac{f(2+h)-f(2)}h=\lim_{h\to0}\frac{4+h-4}h=1.
Review this unit

Problem 2 · The Derivative and Differentiation Rules · 9 points

Differentiate a composite model

Let f(x)=x2e3x\displaystyle f(x)=x^2e^{3x}.

  1. A. Differentiate, naming the product and chain rules.
  2. B. Find the tangent line at x = 0.
  3. C. Find and classify the critical points using the derivative sign.
Show worked solution and rubric
  1. A. f′(x)=2xe3x+3x2e3x=e3xx(2+3x)\displaystyle f\prime(x)=2xe^{3x}+3x^2e^{3x}=e^{3x}x(2+3x). The chain rule supplies the factor 3.
  2. B. f(0)=0\displaystyle f(0)=0 and f′(0)=0\displaystyle f\prime(0)=0, so the tangent line is y=0\displaystyle y=0.
  3. C. The critical points are x=−2/3\displaystyle x=-2/3 and x=0\displaystyle x=0. Since e3x>0\displaystyle e^{3x}>0, the derivative signs are positive, negative, positive on the three intervals. Thus −2/3\displaystyle -2/3 is a local maximum and 0 is a local minimum.

Scoring rubric

  • 3 points: f′(x)=2xe3x+3x2e3x=e3xx(2+3x)\displaystyle f\prime(x)=2xe^{3x}+3x^2e^{3x}=e^{3x}x(2+3x)
  • 3 points: f(0)=0\displaystyle f(0)=0 and f′(0)=0\displaystyle f\prime(0)=0, so the tangent line is y=0\displaystyle y=0.
  • 3 points: The critical points are x=−2/3\displaystyle x=-2/3 and x=0\displaystyle x=0
Review this unit

Problem 3 · Applications of Differentiation · 9 points

Optimize a fenced rectangle

A rectangular pen borders a straight river. Only the other three sides need fencing. There are 60 meters of fence. Let x be the length of each side perpendicular to the river.

Problem diagram
5101520253050100150200250300350400450500xy
  1. A. Express the area in terms of x and state the feasible interval.
  2. B. Use derivatives to find the maximum area.
  3. C. Justify the global maximum using endpoints and interpret the dimensions.
Show worked solution and rubric
  1. A. The parallel side is 60−2x\displaystyle 60-2x, so A(x)=x(60−2x)\displaystyle A(x)=x(60-2x) for 0≤x≤30\displaystyle 0\leq x\leq30 (nondegenerate pens use the open interval).
  2. B. A′(x)=60−4x=0\displaystyle A\prime(x)=60-4x=0 gives x=15\displaystyle x=15. The parallel side is 30 meters and area is 450 square meters.
  3. C. A(0)=A(30)=0\displaystyle A(0)=A(30)=0 and A′′(x)=−4<0\displaystyle A\prime\prime(x)=-4<0. The only interior critical point gives the global maximum, with sides 15, 15, and 30 meters fenced.
Solution diagram
5101520253050100150200250300350400450500xy

Scoring rubric

  • 3 points: The parallel side is 60−2x\displaystyle 60-2x, so A(x)=x(60−2x)\displaystyle A(x)=x(60-2x) for 0≤x≤30\displaystyle 0\leq x\leq30 (nondegenerate pens use the open interval).
  • 3 points: A′(x)=60−4x=0\displaystyle A\prime(x)=60-4x=0 gives x=15\displaystyle x=15
  • 3 points: A(0)=A(30)=0\displaystyle A(0)=A(30)=0 and A′′(x)=−4<0\displaystyle A\prime\prime(x)=-4<0
Review this unit

Problem 4 · Integrals and the Fundamental Theorem · 9 points

Track accumulated change

Let F(x)=∫1x2(t3+1) dt\displaystyle F(x)=\int_1^{x^2}(t^3+1)\,dt for real x.

  1. A. Differentiate F using the Fundamental Theorem and chain rule.
  2. B. Evaluate F(2) exactly.
  3. C. Find intervals where F is increasing or decreasing and explain why the lower limit contributes no derivative term.
Show worked solution and rubric
  1. A. F′(x)=((x2)3+1)(2x)=2x(x6+1)\displaystyle F\prime(x)=((x^2)^3+1)(2x)=2x(x^6+1).
  2. B. F(2)=[t44+t]14=68−54=2674\displaystyle F(2)=[\frac{t^4}{4}+t]_1^4=68-\frac54=\frac{267}4.
  3. C. x6+1>0\displaystyle x^6+1>0 for every real x, so F decreases on (−∞,0)\displaystyle (-\infty,0) and increases on (0,∞)\displaystyle (0,\infty). The lower limit is constant, so its evaluated antiderivative is constant.

Scoring rubric

  • 3 points: F′(x)=((x2)3+1)(2x)=2x(x6+1)\displaystyle F\prime(x)=((x^2)^3+1)(2x)=2x(x^6+1).
  • 3 points: F(2)=[t44+t]14=68−54=2674\displaystyle F(2)=[\frac{t^4}{4}+t]_1^4=68-\frac54=\frac{267}4.
  • 3 points: x6+1>0\displaystyle x^6+1>0 for every real x, so F decreases on (−∞,0)\displaystyle (-\infty,0) and increases on (0,∞)\displaystyle (0,\infty)
Review this unit

Problem 5 · Applications of Integration · 9 points

Compare distance and displacement

A particle has velocity v(t)=t−2\displaystyle v(t)=t-2 meters per second on 0≤t≤5\displaystyle 0\leq t\leq5. Its initial position is 3 meters.

Problem diagram
0.511.522.533.544.55-3-2-11234xy
  1. A. Find displacement and final position.
  2. B. Find total distance traveled by splitting at the direction change.
  3. C. Explain why total distance differs from displacement.
Show worked solution and rubric
  1. A. Displacement is ∫05(t−2) dt=[t2/2−2t]05=5/2\displaystyle \int_0^5(t-2)\,dt=[t^2/2-2t]_0^5=5/2 meters. Final position is 3+5/2=11/2\displaystyle 3+5/2=11/2 meters.
  2. B. Velocity changes sign at t=2\displaystyle t=2. Distance is −∫02(t−2) dt+∫25(t−2) dt=2+9/2=13/2\displaystyle -\int_0^2(t-2)\,dt+\int_2^5(t-2)\,dt=2+9/2=13/2 meters.
  3. C. Displacement is signed change; the first two seconds contribute −2\displaystyle -2 meters. Distance counts their magnitude positively. Integrating ∣v∣\displaystyle |v| counts both directions.
Solution diagram
0.511.522.533.544.55-3-2-11234xy

Scoring rubric

  • 3 points: Displacement is ∫05(t−2) dt=[t2/2−2t]05=5/2\displaystyle \int_0^5(t-2)\,dt=[t^2/2-2t]_0^5=5/2 meters
  • 3 points: Velocity changes sign at t=2\displaystyle t=2
  • 3 points: Displacement is signed change; the first two seconds contribute −2\displaystyle -2 meters
Review this unit

Problem 6 · Techniques of Integration and Differential Equations · 9 points

Solve exponential decay

A quantity y satisfies dydt=−0.4y\displaystyle \frac{dy}{dt}=-0.4y and y(0)=50\displaystyle y(0)=50.

  1. A. Solve the initial value problem by separation.
  2. B. Find the half-life exactly.
  3. C. Evaluate the total accumulated quantity over t ≥ 0 as an improper integral and give its units when y is in mg and t in hours.
Show worked solution and rubric
  1. A. For the positive solution, dy/y=−0.4 dt\displaystyle dy/y=-0.4\,dt, so ln⁡y=−0.4t+C\displaystyle \ln y=-0.4t+C and y=50e−0.4t\displaystyle y=50e^{-0.4t}. The initial value fixes the constant.
  2. B. 25=50e−0.4t\displaystyle 25=50e^{-0.4t} gives t=ln⁡20.4\displaystyle t=\frac{\ln2}{0.4} hours, approximately 1.733 hours.
  3. C. ∫0∞50e−0.4t dt=lim⁡b→∞125(1−e−0.4b)=125\displaystyle \int_0^\infty50e^{-0.4t}\,dt=\lim_{b\to\infty}125(1-e^{-0.4b})=125 mg-hours. This is cumulative exposure, not mass at one moment.

Scoring rubric

  • 3 points: For the positive solution, dy/y=−0.4 dt\displaystyle dy/y=-0.4\,dt, so ln⁡y=−0.4t+C\displaystyle \ln y=-0.4t+C and y=50e−0.4t\displaystyle y=50e^{-0.4t}
  • 3 points: 25=50e−0.4t\displaystyle 25=50e^{-0.4t} gives t=ln⁡20.4\displaystyle t=\frac{\ln2}{0.4} hours, approximately 1.733 hours.
  • 3 points: ∫0∞50e−0.4t dt=lim⁡b→∞125(1−e−0.4b)=125\displaystyle \int_0^\infty50e^{-0.4t}\,dt=\lim_{b\to\infty}125(1-e^{-0.4b})=125 mg-hours
Review this unit

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