Question

If the range of values of $k$ for which the equation $f(x)=k$ has exactly two solutions is $[a,b)$ then find the value of $\left(\frac{1}{a}+\frac{1}{b}\right)\pi$.

Answer 1

Answer: 6

For $x\le 0$,
$f(x)={\cos }^{-1}\left(\frac{1+x}{\sqrt{2\left(1+x^2\right)}}\right)\text{ put }x=\tan \theta \Rightarrow \theta ={\tan }^{-1}x;\theta \in \left(\frac{-\pi }{2},0\right]$
$={\cos }^{-1}\left(\frac{1+\tan \theta }{\sqrt{2{\sec }^2\theta }}\right)={\cos }^{-1}\left(\frac{1}{\sqrt{2}}(\cos \theta +\sin \theta )\right)={\cos }^{-1}\left(\cos \left(\theta -\frac{\pi }{4}\right)\right);\theta -\frac{\pi }{4}\in \left(\frac{-3\pi }{4},\frac{-\pi }{4}\right]$
$=-\left(\theta -\frac{\pi }{4}\right)=\frac{\pi }{4}-{\tan }^{-1}x$


From the graph it is clear that equation $f(x)=k$ has exactly two roots then $k\in \left[\frac{\pi }{4},\frac{\pi }{2}\right)\equiv [a,b)$
Hence, $\left(\frac{1}{a}+\frac{1}{b}\right)\pi =\left(\frac{4}{\pi }+\frac{2}{\pi }\right)\times \pi =6$.