Question

$\mathrm{The}\mathrm{domain}\mathrm{of}f\left(x\right)={\sin }^{-1}\left(2x^2-3\right],\mathrm{where}[.]\mathrm{denotes}\mathrm{the}\mathrm{greatest}\mathrm{integer}\mathrm{function},\mathrm{is}$

• A. $\left(-\sqrt{\frac{3}{2}},\sqrt{\frac{3}{2}}\right)$
• B. $(-\sqrt{\frac{3}{2}},-1]\cup \left(-\sqrt{\frac{5}{2}},\sqrt{\frac{5}{2}}\right)$
• C. $\left(-\sqrt{\frac{5}{2}},\sqrt{\frac{5}{2}}\right)$
• D. $(-\sqrt{\frac{5}{2}},-1]\cup [1,\sqrt{\frac{5}{2}})$

Answer 1

The correct option is D $(-\sqrt{\frac{5}{2}},-1]\cup [1,\sqrt{\frac{5}{2}})$

$\mathrm{Given}\mathrm{function},f\left(x\right)={\sin }^{-1}\left(2x^2-3\right],\mathrm{where}(.]\mathrm{denotes}\mathrm{the}\mathrm{greatest}\mathrm{integer}\mathrm{function}.\mathrm{We}\mathrm{must}\mathrm{have}-1\le \left(2x^2-3\right]\le 1\Rightarrow -1\le 2x^2-3<2\Rightarrow 2\le 2x^2<5\Rightarrow 1\le x^2<\frac{5}{2}\mathrm{or}x\in (-\sqrt{\frac{5}{2}},-1]\cup [1,\sqrt{\frac{5}{2}})$