Question

If ${\tan }^{-1}\left(x^2+3|x|-4\right)+{\cot }^{-1}\left(4\pi +{\sin }^{-1}(\sin 14)\right)=\frac{\pi }{2}$, then the value of ${\sin }^{-1}(\sin 2|x|)$ is equal to

• A. $6-2\pi$
• B. $2\pi -6$
• C. $\pi -3$
• D. $3-\pi$

Answer 1

Answer: (D)

${\sin }^{-1}(\sin 14)=(14-4\pi )$
$\text{ So, }{x}^2+3|x|-4=4\pi +(14-4\pi )$
$\Rightarrow x^2+3|x|-18=0$
$\Rightarrow (|x|+6)(|x|-3)=0$
$\therefore |x|=3$
So, ${\sin }^{-1}(\sin 2|x|)={\sin }^{-1}(\sin 6)=(6-2\pi )$