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

For $x \leq 0$,
$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, $\left(\frac{1}{ a }+\frac{1}{ b }\right) \pi=\left(\frac{4}{\pi}+\frac{2}{\pi}\right) \times \pi=6$.