JEE Advanced 2018 — Paper 1 — Question 36

$$\begin{aligned} & \sin ^{-1}\left(\frac{x^{2}}{1-x}-\frac{x\left(\frac{x}{2}\right)}{1-\frac{x}{2}}\right)=\frac{\pi}{2}-\cos ^{-1}\left(\frac{-x / 2}{1+\frac{x}{2}}-\frac{(-x)}{1+x}\right) \\& \Rightarrow \sin ^{-1}\left(x^{2}\left(\frac{1}{1-x}-\frac{1}{2-x}\right)\right)=\frac{\pi}{2}-\cos ^{-1}\left(x\left(\frac{1}{1+x}-\frac{1}{2+x}\right)\right) \\& \sin ^{-1}\left[\frac{x^{2}}{(1-x)(2-x)}\right]=\frac{\pi}{2}-\cos ^{-1}\left[\frac{x}{(1+x)(2+x)}\right]=\sin ^{-1}\left[\frac{x}{(1+x)(2+x)}\right] \\&\end{aligned}$$

$\Rightarrow x\left[\frac{x}{(1-x)(2-x)}-\frac{1}{(1+x)(2+x)}\right]=0$

$\Rightarrow \ x^{3}+2 \text { or }^{3}+5 x^{3}+2 x^{2}+2 x=x^{2}-3 x+2$

increasing function $\forall x$

$f(0)=-2, f(1 / 2)>0$

$\Rightarrow$ one root between $\left(0, \frac{1}{2}\right)$

$\Rightarrow$ total number of solutions $=2$