Question

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

Answer 1

Answer: 3

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