Question

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

Answer 1

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

Domain : $\quad-1 \leq\left[x^2-\frac{1}{2}\right] \leq 1 \quad \Rightarrow \quad x \in\left(-\sqrt{\frac{5}{2}}, \sqrt{\frac{5}{2}}\right)$

and $\quad-1 \leq\left[x^2+\frac{1}{2}\right] \leq 1 \quad \Rightarrow \quad x \in\left(-\sqrt{\frac{3}{2}} \cdot \sqrt{\frac{3}{2}}\right)$

$\Rightarrow \quad$ domain is $\quad x \in\left(-\sqrt{\frac{3}{2}}, \sqrt{\frac{3}{2}}\right) \quad$ or $\quad x^2 \in\left[0, \frac{3}{2}\right)$

if (i) $\quad x^2 \in\left[0, \frac{1}{2}\right)$, then $\quad f(x)=\pi$

if (ii) $\quad x^2 \in\left[\frac{1}{2}, 1\right)$, then $\quad f(x)=\pi$

if (iii) $\quad x^2 \in\left[1, \frac{3}{2}\right)$, then $\quad f(x)=\pi \quad \Rightarrow \quad$ range $=\{\pi\}$