JEE Advanced 2020 · Paper 1, Mathematics question 2

Odd. $f(-x)=|x|(-x+\sin x)=-f(x)$.

Increasing for $x>0$. There $f(x)=x^2-x\sin x$ and
$$f'(x)=x(1-\cos x)+(x-\sin x)>0.$$
So $f$ is strictly increasing on $[0,\infty)$, and by oddness on all of $\mathbb R$. Hence it is one-one.

Onto. $f$ is continuous, with $f(x)\ge x^2-x\to\infty$ as $x\to\infty$ and, by oddness, $f\to-\infty$ as $x\to-\infty$. So it takes every real value.

(C)