AP Calculus BC 3.1 The Chain Rule - FRQs - Exam Style Questions

(a) Slope \(=f'(0)=6\) ⇒ solve for \(k\) (\(k>0\))
\(f(x)=(x^{2}-2x+k)^{-1}\). By chain/power rule, \[ f'(x)=-1\cdot (x^{2}-2x+k)^{-2}\cdot(2x-2)=\frac{-(2x-2)}{(x^{2}-2x+k)^{2}}. \] Evaluate at \(x=0\): \[ f'(0)=\frac{-(-2)}{k^{2}}=\frac{2}{k^{2}}. \] Set \(f'(0)=6\Rightarrow \dfrac{2}{k^{2}}=6\Rightarrow k^{2}=\dfrac{1}{3}\Rightarrow \boxed{k=\dfrac{1}{\sqrt{3}}}\ (\text{since }k>0). \)

(b) \(\displaystyle \int_{0}^{1} f(x)\,dx\) for \(k=-8\)
Then \(f(x)=\dfrac{1}{x^{2}-2x-8}=\dfrac{1}{(x-4)(x+2)}\). Use linear partial fractions (BC): find \(A,B\) with \[ \frac{1}{(x-4)(x+2)}=\frac{A}{x-4}+\frac{B}{x+2} \ \Rightarrow\ 1=A(x+2)+B(x-4). \] Solve: set \(x=4\Rightarrow 1=6A\Rightarrow A=\dfrac{1}{6}\); set \(x=-2\Rightarrow 1=-6B\Rightarrow B=-\dfrac{1}{6}\). Hence \[ \int_{0}^{1}\!f(x)\,dx =\int_{0}^{1}\!\left(\frac{1}{6}\frac{1}{x-4}-\frac{1}{6}\frac{1}{x+2}\right)\!dx =\left[\frac{1}{6}\ln|x-4|-\frac{1}{6}\ln|x+2|\right]_{0}^{1}. \] Evaluate: \[ =\left(\tfrac{1}{6}\ln 3-\tfrac{1}{6}\ln 3\right)-\left(\tfrac{1}{6}\ln 4-\tfrac{1}{6}\ln 2\right) =\boxed{\tfrac{1}{6}\ln 2}. \] 

(c) \(\displaystyle \int_{0}^{2} f(x)\,dx\) for \(k=1\) (improper)
Now \(f(x)=\dfrac{1}{x^{2}-2x+1}=\dfrac{1}{(x-1)^{2}}\), which is unbounded at \(x=1\). Split and evaluate as limits: \[ \int_{0}^{2}\frac{1}{(x-1)^{2}}\,dx =\lim_{b\to 1^{-}}\int_{0}^{b}\frac{1}{(x-1)^{2}}\,dx +\lim_{b\to 1^{+}}\int_{b}^{2}\frac{1}{(x-1)^{2}}\,dx. \] Antiderivative: \(\displaystyle \int \frac{1}{(x-1)^{2}}dx=-\,\frac{1}{x-1}+C\). Thus \[ \lim_{b\to 1^{-}}\!\Big[-\tfrac{1}{x-1}\Big]_{0}^{b} =\lim_{b\to 1^{-}}\!\left(-\frac{1}{b-1}+1\right)=+\infty, \quad \lim_{b\to 1^{+}}\!\Big[-\tfrac{1}{x-1}\Big]_{b}^{2} =\lim_{b\to 1^{+}}\!\left(1+\frac{1}{b-1}\right)=+\infty. \] Since at least one (indeed both) limits diverge, the integral \[ \boxed{\text{diverges (does not exist as a finite value).}} \]