Question

The soultion satisfying the equation $3{\cos }^{-1}x={\sin }^{-1}\left[\left(\sqrt{1-x^2}\right)\left(4x^2-1\right)\right]$ is $[a,b]$ then $[a+b]$ equals ([.] denotes G.I.E)

Answer 1

Answer: 1

$3{\cos }^{-1}x={\sin }^{-1}\left[\left(\sqrt{1-x^2}\left(4x^2-1\right)\right)\right]$
put $x=\cos \theta ,3{\cos }^{-1}(\cos \theta )$
$={\sin }^{-1}(\sin \theta \left(4{\cos }^2\theta -1\right)$
$={\sin }^{-1}(\sin \theta \left(4\left(1-{\sin }^2\theta \right)-1\right)$
$={\sin }^{-1}\left(3\sin \theta -4{\sin }^3\theta \right)$
$={\sin }^{-1}(\sin 3\theta )$
equality holds when $0\le \theta \le {\pi }^{\wedge }-\frac{\pi }{2}\le 3\theta \le \frac{\pi }{2}$
$\Rightarrow 0\le \theta \le \frac{\pi }{6}$
$\therefore \frac{\sqrt{3}}{2}\le \cos \theta \le 1$
$\therefore x\in \left[\frac{\sqrt{3}}{2},1\right]$