Question

Consider, $f ( x )=| x -1|+|2 x -\pi|+| x -3|$ and $g ( x )=\sin ^{-1} x +\tan ^{-1} x$. The value of $x$ for which $f ( g ( x ))$ is minimum is $\sqrt{ k \sin \frac{\pi}{\lambda}}$ where $k$ and $\lambda \in N$ then find the value of $( k +\lambda)$.

Answer 1

$ f(x)$ will be minimum at $x=\frac{\pi}{2} \Rightarrow g(x)=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1} x+\tan ^{-1} x=\frac{\pi}{2} \Rightarrow \sin ^{-1} x=\cot ^{-1} x \Rightarrow \sin ^{-1} x=\sin ^{-1} \frac{1}{\sqrt{1+x^2}} $
$\Rightarrow x =\frac{1}{\sqrt{1+ x ^2}} \Rightarrow x ^2\left(1+ x ^2\right)=1 \Rightarrow x ^4+ x ^2-1=0 $
$\Rightarrow x ^2=\frac{-1 \pm \sqrt{1+4}}{2}=\frac{-1 \pm \sqrt{5}}{2}=2\left(\frac{\sqrt{5}-1}{4}\right)=2 \sin \left(\frac{\pi}{10}\right) \Rightarrow x =\sqrt{2 \sin \left(\frac{\pi}{10}\right)}$
$\therefore \lambda+ k =12 $