QJEEM22Ma5b66: JEE Main 2022 Mathematics - Class 11 JEEM22Ma5b66 Solution: Let f : R R be defined as f (x) = x 1 and g : R {1, 1} R be…

$f:R \to R$ defined as

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

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

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

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

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

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

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

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

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

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