Answer:
Option D
Explanation:
$\frac{3\pi}{2}<5<\frac{5\pi}{2}$
$\Rightarrow \sin^{-1}(\sin5)=5-2\pi$
$Given \sin^{-1}(\sin5)>x^{2}-4x$
$\Rightarrow x^{2}-4x+4<9-2\pi$
$\Rightarrow (x-2)^{2}<9-2\pi$
$\Rightarrow -\sqrt{9-2\pi}<x-2<\sqrt{9-2\pi}$
$\Rightarrow 2-\sqrt{9-2\pi}<x<2+\sqrt{9-2\pi}$
