Question

If value of $x$ which satisfy equation: ${\sin }^{-1}\left[\sin \left(\frac{2x^2+4}{1+x^2}\right)\right]<\pi -3$ is $-a<x<b$.
Find the value of $a+b$.

• A. $-2$
• B. $-1$
• C. $0$
• D. $1$

Answer 1

The correct option is C $0$

Given, ${\sin }^{-1}\left[\sin \left(\frac{2x^2+4}{1+x^2}\right)\right]<\pi -3$ (i)
$\frac{2x^2+4}{1+x^2}=\frac{2x^2+2+2}{1+x^2}=2+\frac{2}{1+x^2}$
Now $2+\frac{2}{1+x^2}ϵ(2,4]$
Therefor, Eq. (i) reduces to
$\pi -\frac{2x^2+4}{1+x^2}<\pi -3$
$\frac{2x^2+4}{1+x^2}>3$
$2x^2+4>3+3x^2$
$x^2-1<0$
$-1<x<1$