Question

Find the range of

$f\left(x\right)={\sin }^{-1}[x^2+\frac{1}{2}]+{\cos }^{-1}[x^2-\frac{1}{2}]$ where $[\cdot ]$denotes the greatest integer function.

• A. $\left\{\pi \right\}$
• B. $\left\{2\pi \right\}$
• C. $\left\{3\pi \right\}$
• D. $\left\{7\pi \right\}$

Answer 1

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

Given function is, $f\left(x\right)={\cos }^{-1}[x^2-\frac{1}{2}]+{\sin }^{-1}[x^2+\frac{1}{2}]$

For this function to be defined $-1\le [x^2+\frac{1}{2}]\le 1$ and $-1\le [x^2-\frac{1}{2}]\le 1$

or $-1\le x^2+\frac{1}{2}<2$ and $-1\le x^2-\frac{1}{2}<2$

or $-\frac{3}{2}\le x^2<\frac{3}{2}$ and $-\frac{1}{2}\le x^2<\frac{5}{2}$ but $x^2\ge 0$

Taking intersection of above, domain of $f\left(x\right)$ is $x^2\in [0,\frac{3}{2})$

Now taking different cases,

case 1. $x^2\in [0,\frac{1}{2})$

$f\left(x\right)={\cos }^{-1}[x^2-\frac{1}{2}]+{\sin }^{-1}[x^2+\frac{1}{2}]={\cos }^{-1}(-1)+{\sin }^{-1}\left(0\right)=\pi -0=\pi$

case 2. $x^2\in [\frac{1}{2},\frac{3}{2})$

$f\left(x\right)={\cos }^{-1}[x^2+\frac{1}{2}]+{\sin }^{-1}[x^2-\frac{1}{2}]={\cos }^{-1}\left(0\right)+{\sin }^{-1}\left(1\right)=\frac{\pi }{2}+\frac{\pi }{2}=\pi$

Hence range of $f\left(x\right)$ is only singleton set $\left\{\pi \right\}$