Home / AP Calculus AB 5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points – MCQs

AP Calculus AB 5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points - MCQs - Exam Style Questions

AP Calculus AB-MCQs and Answers - All Topics

No-Calc Question

Let \(f(x)=k\sqrt{x}+\dfrac{1}{x}\) for \(x>0\), where \(k\) is a constant. If \(f\) has a relative extremum at \(x=9\), what is \(k\)?
(A) \(\dfrac{2}{3}\)
(B) \(\dfrac{2}{27}\)
(C) \(\dfrac{2}{81}\)
(D) \(\dfrac{2}{243}\)
▶️ Answer/Explanation
Set \(f'(9)=0\). Compute \(f'(x)=\dfrac{k}{2\sqrt{x}}-\dfrac{1}{x^{2}}\).
Then \(0=f'(9)=\dfrac{k}{2\cdot 3}-\dfrac{1}{9^{2}}=\dfrac{k}{6}-\dfrac{1}{81}\).
Solve: \(\dfrac{k}{6}=\dfrac{1}{81}\Rightarrow k=\dfrac{2}{27}\).
✅ Answer: (B)

Calc-Ok Question

The function \(f\) is \(f(x)=3x-4\cos(2x+1)\) with derivative \(f'(x)=3+8\sin(2x+1)\). What values of \(x\) satisfy the Mean Value Theorem on \([-1,2]\)?

(A) \(-0.692\) and \(1.263\)
(B) \(-0.479\) and \(1.049\)
(C) \(0.285\)
(D) \(0.517\)
(E) \(1.578\)

▶️ Answer/Explanation

MVT gives \(f'(c)=\dfrac{f(2)-f(-1)}{2-(-1)}=\dfrac{9-4\cos 5+4\cos 1}{3}\approx 3.3422\).
Solve \(3+8\sin(2c+1)=3.3422 \Rightarrow \sin(2c+1)\approx 0.042773\).
Principal solutions: \(2c+1=\sin^{-1}(0.042773)\) or \(\pi-\sin^{-1}(0.042773)\).
Thus \(c=\dfrac{\sin^{-1}(0.042773)-1}{2}\approx -0.479\) and \(c=\dfrac{\pi-\sin^{-1}(0.042773)-1}{2}\approx 1.049\), both in \([-1,2]\).
✅ Answer: (B) \(-0.479\) and \(1.049\)

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