If the function $f : \mathbb{R} \to \mathbb{R}$ is defined by $f(x) = |x|(x - \sin x)$, then which of the following statements is TRUE?

$f:\mathbb{R}\to\mathbb{R}, \quad f(x)=|x|(x-\sin x)$ $\text{one-one? onto? } \ f:\mathbb{R}\to\mathbb{R}$ $(1)\ \text{one-one only}, \quad (2)\ \text{onto only}$ $(3)\ \text{both}, \quad (4)\ \text{neither}$ $1.\ \text{symmetry}; \quad 2.\ \text{sign of } f$ $3.\ f' \Rightarrow \text{monotonic} \Rightarrow \text{one-one}$ $4.\ \lim_{x\to\pm\infty}f \Rightarrow \text{onto}$ $|x| \text{ is even}, \quad x-\sin x \text{ is odd}$ $f(-x)=|x|(-x+\sin x)=-f(x) \Rightarrow f \text{ is odd}$ $x>0:\ x-\sin x>0 \Rightarrow f(x)>0$ $x<0:\ f(x)<0, \quad f(0)=0$ $y=|x|(x-\sin x)$ $x>0:\ f(x)=x^{2}-x\sin x$ $f'(x)=2x-\sin x-x\cos x=x(2-\cos x)-\sin x$ $0<x\le\pi:\ f''(x)=2(1-\cos x)+x\sin x\ge 0$ $f'(0)=0 \Rightarrow f'(x)>0$ $x\ge\pi:\ f'(x)\ge x-1>0$ $\Rightarrow f'(x)>0 \ \forall x>0$ $f \text{ odd} + \text{increasing on } (0,\infty) \Rightarrow f \text{ strictly increasing}$ $\Rightarrow f \text{ is one-one}$ $\lim_{x\to+\infty}f(x)=+\infty, \quad \lim_{x\to-\infty}f(x)=-\infty$ $f \text{ continuous} \Rightarrow \text{range}(f)=\mathbb{R} \Rightarrow f \text{ is onto}$ $\boxed{f \text{ is both one-one and onto} \ \Rightarrow\ (3)}$