JEE Main 2022 — Inverse Trigonometric Functions Question with Solution

$$-1\le \frac{2{\sin }^{-1}\left(\frac{1}{4x^2-1}\right)}{\pi }\le 1$$

$$\Rightarrow -\frac{\pi }{2}\le {\sin }^{-1}\left(\frac{1}{4x^2-1}\right)\le \frac{\pi }{2}$$

$$\Rightarrow -1\le \frac{1}{4x^2-1}\le 1$$

$$\therefore$$ $$\frac{1}{4x^2-1}+1\ge 0$$

$$\Rightarrow \frac{1+4x^2-1}{4x^2-1}\ge 0$$

$$\Rightarrow \frac{4x^2}{4x^2-1}\ge 0$$

$$\Rightarrow$$ $$\frac{4x^2}{\left(2x+1\right)\left(2x-1\right)}\ge 0$$ ...... (1)

$$\therefore$$ $$x\in \left(-\alpha ,-\frac{1}{2}\right)\cup \{0\}\cup \left(\frac{1}{2},\alpha \right)$$ .....(2)

And $$\frac{1}{4x^2-1}-1\le 0$$

$$\Rightarrow \frac{1-4x^2+1}{4x^2-1}\le 0$$

$$\Rightarrow \frac{2-4x^2}{4x^2-1}\le 0$$

$$\Rightarrow \frac{2x^2-1}{4x^2-1}\ge 0$$

$$\Rightarrow$$ $$\frac{\left(\sqrt{2}x+1\right)\left(\sqrt{2}x-1\right)}{\left(2x+1\right)\left(2x-1\right)}\ge 0$$ ...... (3)

$$x\in \left(-\alpha ,-\frac{1}{\sqrt{2}}\right)\cup \left(-\frac{1}{2},\frac{1}{2}\right)\cup \left(\frac{1}{\sqrt{2}},\alpha \right)$$ .....(4)

From (3) and (4), we get

$$\therefore$$ $$x\in \left[-\alpha ,-\frac{1}{\sqrt{2}}\right)\cup \left[\frac{1}{\sqrt{2}},\alpha \right)\cup \{0\}$$