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AP Calculus BC 1.5 Determining Limits Using Algebraic Properties of Limits Exam Style Questions – FRQ

AP Calculus BC 1.5 Determining Limits Using Algebraic Properties of Limits Exam Style Questions - FRQ

Question:

The continuous function f is defined on the closed interval −6 ≤ x ≤ 5. The figure above shows a portion of the graph of f, consisting of two line segments and a quarter of a circle centered at the point (5, 3). It is known that the point \((3,3-\sqrt{5})\) is on the graph of f. 

(a) If \(\int_{-6}^{5}f(x)dx=7,\) find the value of \(\int_{-6}^{-2}f(x)dx.\) Show the work that leads to your answer
(b) Evaluate \(\int_{3}^{5}(2f'(x)+4)dx.\)
(c) The function g is given by \(g(x)=\int_{-2}^{x}f(t)dt.\) Find the absolute maximum value of g on the interval − 2 ≤ x ≤ 5. Justify your answer.
(d) Find \(\lim_{x\rightarrow 1}\frac{10^{x}-3f'(x)}{f(x)-arctan x}.\)

▶️Answer/Explanation

Ans:

(a)

\(\int_{-6}^{-2}f(x)dx=\int_{-6}^{5}f(x)dx-\int_{-2}^{5}f(x)dx\)

\(\int_{-6}^{-2}f(x)dx=7-\left [ \frac{1}{2}-\frac{1}{2}-\frac{1}{4}+\frac{9}{4}+\left ( 9-\frac{9\pi }{4} \right ) \right ]\)

(b)

\(\int_{3}^{5}2f'(x)+4dx=\int_{3}^{5}2(f'(x)+2)dx\)

\(=2\int_{3}^{5}f'(x)+2dx\)

\(=2\cdot \left [ f(x)+2x \right ]_{3}^{5}\)

\(=2\cdot \left [ \left [ f(5)+2(5) \right ]-\left [ f(3)+2(3) \right ] \right ]\)

(c)

g'(x) = f(x)

g'(x) = f(x) = 0

x = -1,    x = 1/2,   x = 5

Since f(x) is continuous on -2 ≤ x ≤ 5, then g(x) = \(\int_{-2}^{5}f(t)dt \) is also continuous on -2 ≤ x ≤ 5. Therefore, by the EVT, g(x) has an absolute max. on -2 ≤ x ≤ 5.

Candidates for abs. max : x = -2, -1, 1/2, 5

\(x = -2 : \int_{-2}^{-2}f(t)dt=0\)

\(x = -1 : \int_{-2}^{-1}f(t)dt=\frac{1}{2}\)

\(x = \frac{1}{2} : \int_{-2}^{1/2}f(t)dt=\frac{1}{2}-\frac{1}{2}-\frac{1}{4}=-\frac{1}{4}\)

\(x = 5 : \int_{-2}^{-5}f(t)dt=-\frac{1}{4}+\frac{9}{4}+\left ( 9-\frac{\pi }{4} \right )=2+9-\frac{9\pi }{4}=11-\frac{9\pi }{4}\)

The absolute maximum value of g is \(11-\frac{9\pi }{4}\) and the absolute maximum occurs at x = 5.

(d)

\(\lim_{x\rightarrow 1}10^{x}-3f'(x)=10-3f'(1)=10-3(2)=4\)

\(\lim_{x\rightarrow 1}f(x)-arctan x=1-\frac{\pi }{4}\)

\(arctan(1)=\frac{\pi }{4}\)

\(\lim_{x\rightarrow 1}\frac{10^{x}-3f'(x)}{f(x)-arctan x}=\frac{4}{1-\pi /4}\)

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