Question

Consider, $f(x)=|x-1|+|2x-\pi |+|x-3|$ 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

Answer: 00012

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