Mathematics - Functions Question with Solution | TestHub

$f\left(x\right)=\left|x\right|\left(x-\sin x\right)$ is odd function $\left[\because f\left(-x\right)=-f\left(x\right)\right]$

Now, $f\left(x\right)=x^2-x\sin x\forall x\ge 0$

$\Rightarrow f^{\text{'}}\left(x\right)=2x-x\cos x-\sin x$

$\Rightarrow f^{\text{'}}\left(x\right)=\left(x-\sin x\right)+x\left(1-\cos x\right)>0$

$\therefore$Graph of $y=f\left(x\right)$ will be

From graph, it's one-one and onto both.