JEE Advanced 2020 — Paper 1 — Question 30

$f(x)$ is odd, continuous function

$f(x)= \begin{cases}x(x-\sin x) & , \quad x \geq 0\\ -x(x-\sin x) & , x<0\end{cases}$

for $x \geq 0, f^{\prime}(x)=2 x-\sin x-x \cos x=x(1-\cos x)+(x-\sin x) \geq 0$

for $x<0, f^{\prime}(x)=-2 x+\sin x+x \cos x=x(\cos x-1)-(x-\sin x)>0$ as $x<0$

$\Rightarrow$ So $f(x)$ strictly increases in $(-\infty, \infty) \Rightarrow f(x)$ is one-one

$x \rightarrow \infty ,\mathrm{f}(\mathrm{x}) \rightarrow \infty$

$x \rightarrow-\infty, f(x) \rightarrow-\infty$. So $f(x)$ is onto