Question

If $tan^{- 1}\frac{x}{\pi }\leq \frac{\pi }{6},$ and the maximum value of $x$ is $\frac{\pi }{\sqrt{k}}$ , then $k=$

Answer 1

We have $tan^{- 1}\frac{x}{\pi }\leq \frac{\pi }{6}$
$\Rightarrow \tan \left(\tan ^{-1} \frac{x}{\pi}\right) \leq \tan \frac{\pi}{6}$
$\Rightarrow \frac{x}{\pi }\leq \frac{1}{\sqrt{3}}\Rightarrow x\leq \frac{\pi }{\sqrt{3}}$
$\therefore $ The maximum value of $x$ is $\frac{\pi }{\sqrt{3}}$ , so $k=3$