Answer:
Option D
Explanation:
We have,
$f(x)= \frac{x}{1+x^{2}}$ $x \in R$
$g(x)=\frac{x^{2}}{1+x^{2}},x\in R$
$f'(x)= \frac{1+x^{2}-2x^{2}}{(1+x^{2})^{2}}$
$f'(x)= \frac{1-x^{2}}{(1+x^{2})^{2}}$
Clearly f'(x) is not monotonic
$\therefore$ f(x) is not one-one function range of f(x) , is $\left[ -\frac{1}{2},\frac{1}{2}\right]$
$\therefore$ f(x) is not onto
Clearly g(x) is even function
$\therefore$ g(x) is not one-one function
Range of g(x) is [0,1]
$\therefore$ g(x) is also not onto . Here , f(x) and g(x) both are neither one-one not onto.