Question

If the domain of the function $f\left(x\right)=\sqrt{3{\cos }^{-1}\left(4x\right)-\pi }$ is $[a,b],$ then the value of $(4a+64b)$ is

Answer 1

For $f\left(x\right)$ to be defined,
$-1\le 4x\le 1$
$\Rightarrow \frac{-1}{4}\le x\le \frac{1}{4}\dots \left(1\right)$
and
$3{\cos }^{-1}\left(4x\right)-\pi \ge 0$
$\Rightarrow 3{\cos }^{-1}\left(4x\right)\ge \pi$
$\Rightarrow {\cos }^{-1}\left(4x\right)\ge \frac{\pi }{3}$
$\Rightarrow 4x\le \frac{1}{2}$
$\Rightarrow x\le \frac{1}{8}\dots \left(2\right)$

Hence, domain of $f\left(x\right)$ is $\left(1\right)\cap \left(2\right)$
$\Rightarrow x\in [\frac{-1}{4},\frac{1}{8}]$
$\Rightarrow a=-\frac{1}{4},b=\frac{1}{8}$
$\therefore 4a+64b=-1+8=7$