If the function f: R → R is defined by f(x) = |x|(x - x) , then...

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

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

$⇒ {f}^{'}\left(x\right) = 2x - xcosx - \sin x$

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

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

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