Question

The domain of $sin^{-1} [2x^2 - 5]$, where $[\cdot]$ represent greatest integer function, is

• A. $\left[-\sqrt{\frac{7}{2}}, -\sqrt{2}\right]$
• B. $\left[\sqrt{2}, \frac{7}{2}\right]$
• C. $\left[-\sqrt{\frac{7}{2}}, -\sqrt{2}\right]\cup\left[\sqrt{2} , \sqrt{\frac{7}{2}}\right]$
• D. None of these

Answer 1

Answer: (D)

$\frac{\pi}{2} \le sin^{-1} t \le \frac{\pi}{2}$
$\therefore -1 \le t \le 1, t = [2x^2 - 5]$
$\Rightarrow -1 \le [2x^2 - 5] \le 1$
$\Rightarrow -1 \le 2x^2 - 5 < 2$
$\Rightarrow 2 \le x^2 < \frac{7}{2}$
$\Rightarrow \sqrt{2} \le |x| < \sqrt{\frac{7}{2}}$
$\therefore |x| \ge \sqrt{2}$ and $|x| < \sqrt{\frac{7}{2}}$
$\Rightarrow x \le - \sqrt{2} \cup x \ge \sqrt{2}$ and $-\sqrt{\frac{7}{2}} < x < \sqrt{\frac{7}{2}}$