Question

$si{n}^{-1}\left(sin\left(\frac{2x^2+4}{1+x^2}\right)\right)<\pi -3$ if

• A. $-1\le x\le 0$
• B. $0\le x\le 1$
• C. $-1<x<1$
• D. $x>1$

Answer 1

The correct option is C $-1<x<1$

Let $t=\frac{2x^2+4}{1+x^2}\Rightarrow x^2=\frac{4-t}{t-2}$
Since $x^2>0$ we get $2<t\le 4$
So we have to solve $si{n}^{-1}\left(sint\right)<\pi -3$ and $2<t\le 4$ simultaneously.
Now $\frac{\pi }{2}<2<t\le 4<\frac{3\pi }{2}$
So $si{n}^{-1}\left(sint\right)=\pi -t$
and $\pi -t<\pi -3$
$\Rightarrow t>3$
$\therefore$ We get $3<t\le 4$
$\Rightarrow 3<\frac{2x^2+4}{x^2+1}\le 4$
$\Rightarrow -1<x<1$ which is the required solution