Question

The range of $f\left(x\right)={\sin }^{-1}\left(\frac{x^2+1}{x^2+2}\right)$ is

• A. $[0,\frac{\pi }{2}]$
• B. $(0,\frac{\pi }{6})$
• C. $[\frac{\pi }{6},\frac{\pi }{2})$
• D. None of these

Answer 1

The correct option is C $[\frac{\pi }{6},\frac{\pi }{2})$

Here, $\frac{x^2+1}{x^2+2}=1-\frac{1}{x^2+2}$
Now, $2\le x^2+2<\infty$ for all $x\in R$
or, $\frac{1}{2}\ge \frac{1}{x^2+2}>0$
or, $-\frac{1}{2}\le \frac{-1}{x^2+2}<0$
or, $\frac{1}{2}\le 1-\frac{1}{x^2+2}<1$
or, $\frac{\pi }{6}\le {\sin }^{-1}(1-\frac{1}{x^2+2})<\frac{\pi }{2}$