Question

The inequality $\sin ^{-1}(\sin 5)>x^{2}-4 x$ holds if

• A. $x=2-\sqrt{9-2\pi}$
• B. $x=2+\sqrt{9-2\pi}$
• C. $x>\, 2+\sqrt{9-2\pi}$
• D. $x \in\left(2-\sqrt{9-2\pi,} 2+\sqrt{9-2\pi}\right)$

Answer 1

Answer: (D)

$\frac{3 \pi}{2}<5< \frac{5 \pi}{2}$
$\Rightarrow \text{Sin}^{-1}(\text{Sin} 5)=5-2 \pi$
Given $\sin ^{-1}(\text{Sin} 5) >x^{2}-4 x$
$\Rightarrow x^{2}-4 x+4< 9-2 \pi$
$\Rightarrow (x-2)^{2}< 9-2 \pi$
$\Rightarrow (x-2)^{2}< 9-2 \pi$
$\Rightarrow -\sqrt{9-2 \pi} < x-2 < \sqrt{9-2 \pi}$
$\Rightarrow 2-\sqrt{9-2 \pi} < x < 2+\sqrt{9-2 \pi}$