Question

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

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

Answer 1

The correct option is B $\left\{0\right\}$

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 $x\in [0,\frac{3}{2})$

Now taking different cases,

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

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

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

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

Hence range of $f\left(x\right)$ is only singleton set $\left\{0\right\}$