Question

$si{n}^{-1}\left(sin5\right)>x^2-4x$ 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)

$\frac{3\pi }{2}<5<\frac{5\pi }{2}$
$\Rightarrow si{n}^{-1}\left(sin5\right)=5-2\pi$
Given $si{n}^{-1}\left(sin5\right)>x^2-4x$
$\Rightarrow x^2-4x+4<9-2\pi$
$\Rightarrow {\left(x-2\right)}^2<9-2\pi$
$\Rightarrow -\sqrt{9-2\pi }<x-2<\sqrt{9-2\pi }$
$\Rightarrow 2-\sqrt{9-2\pi }<x<2+\sqrt{9-2\pi }$