Question

The range of $f\left(x\right)={\sin }^{-1}\left[\begin{array}{c}\frac{1}{2}+x^2\end{array}\right]$ $\left(\right[\cdot ]$ denotes greatest integer function) is

• A. $\{-\frac{\pi }{2},0,\frac{\pi }{2}\}$
• B. $[\frac{\pi }{6},\frac{\pi }{2}]$
• C. $\left\{\frac{\pi }{2}\right\}$
• D. $\{0,\pi \}$

Answer 1

The correct option is B $[\frac{\pi }{6},\frac{\pi }{2}]$

$\Rightarrow$$f\left(x\right)={\sin }^{-1}[\frac{1}{2}+x^2]$

The sin functions range is $[-1,1]$ so,

$\Rightarrow$$-1⩽\frac{1}{2}+x^2⩽1$

Subtracting all by $\frac{1}{2}$ we get,

$\Rightarrow$$\frac{-3}{2}⩽x^2⩽\frac{1}{2}$

$\Rightarrow$$0⩽x^2⩽\frac{1}{2}$

$\Rightarrow$$0⩽x⩽\frac{1}{\sqrt{2}}$

Minimum value when x=0

$={\sin }^{-1}[\frac{1}{2}+0]$

$={\sin }^{-1}\left[\frac{1}{2}\right]$

$=\frac{\pi }{6}$

Maximum value when $x=$$\frac{1}{\sqrt{2}}$

$={\sin }^{-1}[\frac{1}{2}+\frac{1}{2}]$

$={\sin }^{-1}\left[1\right]$

$=\frac{\pi }{2}$

So the range of f(x) is $[\frac{\pi }{6},\frac{\pi }{2}]$