JEE Main 2021 — Functions Question with Solution

The domain of the function

$f\left(x\right)={\sin }^{-1}\left(\frac{3x^2+x-1}{(x-1{)}^2}\right)+{\cos }^{-1}\left(\frac{x-1}{x+1}\right)$ 

For, ${\cos }^{-1}\left(\frac{x-1}{x+1}\right)$

$-1\le \frac{x-1}{x+1}\le 1$

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

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

$\Rightarrow 0\le \frac{1}{x+1}\le 1$

$\Rightarrow x+1\in [1,\infty )$

$\Rightarrow x\in [0,\infty ) ....\left(i\right)$

For, ${\sin }^{-1}\left(\frac{3x^2+x-1}{(x-1{)}^2}\right)$

$-1\le \frac{3x^2+x-1}{{\left(x-1\right)}^2}\le 1$

$\Rightarrow -{\left(x-1\right)}^2\le 3x^2+x-1\le {\left(x-1\right)}^2, x\ne 1$

Now, $-{\left(x-1\right)}^2\le 3x^2+x-1, x\ne 1$

$\Rightarrow 4x^2-x\ge 0 , x\ne 1$

$\Rightarrow x\left(4x-1\right)\ge 0, x\ne 1$

$\Rightarrow x\in (-\infty , 0]\cup \left[\frac{1}{4}, \infty \right) -\left\{1\right\} ....\left(ii\right)$

And $3x^2+x-1\le {\left(x-1\right)}^2, x\ne 1$

$\Rightarrow 2x^2+3x-2\le 0, x\ne 1$$$

$\Rightarrow \left(x+2\right)\left(2x-1\right)\le 0, x\ne 1$

$\Rightarrow x\in \left[-2, \frac{1}{2}\right] ....\left(iii\right)$

Domain of the function ${\sin }^{-1}\left(\frac{3x^2+x-1}{(x-1{)}^2}\right)$ from the equations $\left(ii\right) \& \left(iii\right)$ is

$\Rightarrow x\in \left[-2, 0\right]\cup \left[\frac{1}{4}, \frac{1}{2}\right] ....\left(iv\right)$

Now the domain of the given function will be the intersection of the equation $\left(i\right) \& \left(iv\right)$

Hence, domain is $x\in \left[\frac{1}{4}, \frac{1}{2}\right]\cup \left\{0\right\}$