JEE Main 2022 — Functions Question with Solution

$$f:R\to R$$ defined as

$$f(x)=x-1$$ and $$g:R\to \{1,-1\}\to R,g(x)=\frac{x^2}{x^2-1}$$

Now $$fog(x)=\frac{x^2}{x^2-1}-1=\frac{1}{x^2-1}$$

$$\therefore$$ Domain of $$fog(x)=R-\{-1,1\}$$

And range of $$fog(x)=(-\infty ,-1]\cup (0,\infty )$$

Now, $$\frac{d}{dx}(fog(x))=\frac{-1}{x^2-1}.2x=\frac{2x}{1-x^2}$$

$$\therefore$$ $$\frac{d}{dx}(fog(x))>0$$ for $$\frac{2x}{(1-x)(1+x)}>0$$

$$\Rightarrow \frac{x}{(x-1)(x+1)}<0$$

$$\therefore$$ $$x\in (-\infty ,-1)\cup (0,1)$$

and $$\frac{d}{dx}(fog(x))<0$$ for $$x\in (-1,0)\cup (1,\infty )$$

$$\therefore$$ $$fog(x)$$ is neither one-one nor onto.