JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Sets Relations and Functions

$f(x)=|x|(x-\sin x)$ is a non-periodic, continuous, and odd function.

For $x\ge 0$: $f(x)=x^2-x\sin x$

For $x<0$: $f(x)=-x^2+x\sin x$

As $x\to \infty$, $f(x)=x^2(1-\frac{\sin x}{x})\to \infty$

As $x\to -\infty$, $f(x)=-x^2(1-\frac{\sin x}{x})\to -\infty$

Therefore, the range of $f(x)$ is $\mathbb{R}$, so $f$ is onto.

For $x\ge 0$: $f^{\prime }(x)=2x-\sin x-x\cos x=(x-\sin x)+x(1-\cos x)$

Both terms are non-negative (and zero only at $x=0$), so $f^{\prime }(x)\ge 0$ for all $x\in \mathbb{R}$ with equality only at $x=0$.

Therefore, $f$ is strictly increasing, hence one-one.

Thus, $f$ is both one-one and onto.