Question

The range of function $f\left(x\right)={\sin }^{-1}[x^2+\frac{1}{2}]+{\cos }^{-1}[x^2-\frac{1}{2}]$, where [.] is the greatest integer function is

• A. $\{\frac{\pi }{2},\pi \}$
• B. $\{0,\frac{-1}{2}\}$
• C. $\left\{\pi \right\}$
• D. $\left(0.\frac{\pi }{2}\right)$

Answer 1

The correct option is C $\left\{\pi \right\}$

$f\left(x\right)={\sin }^{-1}[x^2+\frac{1}{2}]+{\cos }^{-1}[x^2-\frac{1}{2}]$
$={\sin }^{-1}[x^2+\frac{1}{2}]+{\cos }^{-1}[x^2+\frac{1}{2}-1]$
$={\sin }^{-1}[x^2+\frac{1}{2}]+{\cos }^{-1}([x^2+\frac{1}{2}]-1)$
Since, $x^2+\frac{1}{2}\ge \frac{1}{2}$
So, $[x^2+\frac{1}{2}]$ is defined only for the two values.
$[x^2+\frac{1}{2}]=0\Rightarrow f\left(x\right)={\sin }^{-1}0+{\cos }^{-1}(-1)=\pi$
$[x^2+\frac{1}{2}]=1\Rightarrow f\left(x\right)={\sin }^{-1}1+{\cos }^{-1}0=\pi$.
So, range of $f\left(x\right)=\pi$