Home / AP Calculus BC: 6.12 Using Linear Partial  Fractions bc only – Exam Style questions with Answer- FRQ

AP Calculus BC: 6.12 Using Linear Partial  Fractions bc only – Exam Style questions with Answer- FRQ

Question

Consider the family of functions \(f(x)=\frac{1}{x^{2}-2x+k},\) where k is a constant.
(a) Find the value of k, for k > 0, such that the slope of the line tangent to the graph of f at x = 0 equals 6.
(b) For k = −8, find the value of \(\int_{0}^{1}f(x)dx.\)
(c) For k = 1, find the value of \(\int_{0}^{1}f(x)dx\) or show that it diverges.

Answer/Explanation

Ans:

(a) \(f'(x)=\frac{-(2x-2)}{\left ( x^{2}-2x+k \right )^{2}}\)

\(f'(0)=\frac{2}{k^{2}}=6\Rightarrow k^{2}=\frac{1}{3}\Rightarrow k = \frac{1}{\sqrt{3}}\)

(b) \(\frac{1}{x^{2}-2x-8}=\frac{1}{(x-4)(x+2)}=\frac{A}{x-4}+\frac{B}{x+2}\)

\(\Rightarrow 1 = A(x+2)+B(x-4)\)

\(\Rightarrow A = \frac{1}{6}, B=-\frac{1}{6}\)

\(\int_{0}^{1}f(x)dx=\int_{0}^{1}\left ( \frac{\frac{1}{6}}{x-4} -\frac{\frac{1}{6}}{x+2}\right )dx\)

\(=\left [\frac{1}{6}In|x-4| -\frac{1}{6} In|x+2| \right ]_{x=0}^{x=1}\)

\(=\left ( \frac{1}{6}In3-\frac{1}{6}In3 \right )-\left ( \frac{1}{6}In4-\frac{1}{6}In2 \right )=-\frac{1}{6}In2\)

(c) \(\int_{0}^{2}\frac{1}{x^{2}-2x+1}dx=\int_{0}^{2}\frac{1}{(x-1)^{2}}dx=\int_{0}^{1}\frac{1}{(x-1)^{2}}dx+\int_{1}^{2}\frac{1}{(x-1)^{2}}dx\)

\(=\lim_{b\rightarrow 1^{-}}\int_{0}^{b}\frac{1}{(x-1)^{2}}dx+\lim_{b\rightarrow 1^{+}}\int_{b}^{2}\frac{1}{(x-1)^{2}}dx\)

\(=\lim_{b\rightarrow 1^{-}}\left ( -\frac{1}{x-1|_{x=0}^{x=b}} \right )+\lim_{b\rightarrow 1^{+}}\left ( -\frac{1}{x-1|_{x=b}^{x=2}} \right )\)

\(=\lim_{b\rightarrow 1^{-}}\left ( -\frac{1}{b-1}-1 \right )+\lim_{b\rightarrow 1^{+}}\left ( -1+\frac{1}{b-1} \right )\)

Because \(=\lim_{b\rightarrow 1^{-}}\left ( -\frac{1}{b-1} \right )\) does not exist, the integral diverges.

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