JEE Main 2021 — Inverse Trigonometric Functions Question with Solution

Given equation is

${\sin }^{-1}\left[x^2+\frac{1}{3}\right]+{\cos }^{-1}\left[x^2-\frac{2}{3}\right]=x^2$

Now, ${\sin }^{-1}\left[x^2+\frac{1}{3}\right]$ is defined if $-1\le \left[x^2+\frac{1}{3}\right]\le 1$

$\Rightarrow -1\le x^2+\frac{1}{3}<2$

$\Rightarrow \frac{-4}{3}\le x^2<\frac{5}{3}$

$\Rightarrow 0\le x^2<\frac{5}{3} ...\left(1\right)$

Also, and ${\cos }^{-1}\left[{\mathrm{x}}^2-\frac{2}{3}\right]$ is defined if $-1\le \left[x^2-\frac{2}{3}\right]\le 1$

 $\Rightarrow -1\le x^2-\frac{2}{3}<2$

$\Rightarrow \frac{-1}{3}\le x^2<\frac{8}{3}$

$\Rightarrow 0\le x^2<\frac{8}{3} ...\left(2\right)$

So, from $\left(1\right)$ and $\left(2\right)$ we can conclude $0\le x^2<\frac{5}{3}$

Case -$I:$ If $0\le x^2<\frac{2}{3}$

$\Rightarrow {\sin }^{-1}\left(0\right)+{\cos }^{-1}(-1)=x^2$

$\Rightarrow 0+\pi =x^2$

$\Rightarrow x^2=\pi$ but $\pi \notin \left[0,\frac{2}{3}\right)$

$\Rightarrow$ No value of $\text{'}x\text{'}$

Case - $II:$ If $\frac{2}{3}\le x^2<\frac{5}{3}$

$\Rightarrow {\sin }^{-1}\left(1\right)+{\cos }^{-1}\left(0\right)=x^2$

$\Rightarrow \frac{\pi }{2}+\frac{\pi }{2}={\mathrm{x}}^2$

$\Rightarrow x^2=\pi$ but $\pi \notin \left[\frac{2}{3},\frac{5}{3}\right)$

$\Rightarrow$ No value of $\text{'}x\text{'}$

So, number of solutions of the equation is zero.