If the function $$f: R \rightarrow R$$ is defined by $$f(x) = |x|(x − \sin x)$$, then which of the following statements is TRUE?

For $$x \in \mathbb{R}$$ the function is defined as $$f(x)=|x|\,(x-\sin x)$$.

Write it piece-wise:
  • If $$x \ge 0$$, then $$|x| = x$$, so $$f(x)=x^2-x\sin x \quad -(1)$$
  • If $$x \lt 0$$, then $$|x| = -x$$, so $$f(x)=-x^2+x\sin x \quad -(2)$$

Step 1 : Monotonicity (injectivity)

Differentiate each branch.

For $$x \ge 0$$, using $$(1)$$:
$$f'(x)=\frac{d}{dx}\bigl[x^2-x\sin x\bigr]=2x-\sin x-x\cos x$$
Factor as $$f'(x)=x(2-\cos x)-\sin x$$.

Because $$\sin x \le x$$ for $$x \ge 0$$,
$$f'(x) \ge x(2-\cos x)-x = x(1-\cos x) \ge 0,$$
and for every $$x \gt 0$$ we have $$1-\cos x \gt 0$$, giving $$f'(x) \gt 0$$. Thus $$f(x)$$ is strictly increasing on $$[0,\infty)$$.

For $$x \lt 0$$, using $$(2)$$:
$$f'(x)=\frac{d}{dx}\bigl[-x^2+x\sin x\bigr]=-2x+\sin x+x\cos x$$
Rewrite as $$f'(x)=-x(2-\cos x)+\sin x$$.

For negative $$x$$ we have $$\sin x \ge x$$, hence
$$f'(x) \ge -x(2-\cos x)+x = -x(1-\cos x).$$
Now $$x \lt 0 \;\Rightarrow\; -x \gt 0$$ and $$1-\cos x \ge 0$$, so $$-x(1-\cos x) \gt 0$$ whenever $$x \ne 0$$. Therefore $$f'(x) \gt 0$$ for all $$x \lt 0$$, making $$f(x)$$ strictly increasing on $$(-\infty,0)$$ as well.

Combining both intervals, $$f'(x) \gt 0$$ for every $$x \ne 0$$ and $$f'(0)=0$$, so the entire function is strictly increasing on $$\mathbb{R}$$. Hence $$f$$ is one-one.

Step 2 : Range (surjectivity)

Find the limits at the ends of the domain.

As $$x \to -\infty$$ use $$(2)$$:
$$f(x)=-x^2+x\sin x = -x^2 \bigl(1-\tfrac{\sin x}{x}\bigr).$$
Since $$\Bigl|\tfrac{\sin x}{x}\Bigr| \le 1/|x| \to 0$$, the quadratic term dominates and $$f(x) \to -\infty.$$}

As $$x \to +\infty$$ use $$(1)$$:
$$f(x)=x^2-x\sin x = x^2\bigl(1-\tfrac{\sin x}{x}\bigr) \to +\infty.$$}

The function is continuous (both pieces and their first derivatives meet at $$x=0$$), strictly increasing, and its values run from $$-\infty$$ to $$+\infty$$. By the Intermediate Value Theorem the range equals $$\mathbb{R}$$, so $$f$$ is onto.

Step 3 : Conclusion

$$f$$ is both one-one and onto. Hence the true statement is:
Option C which is: f is BOTH one-one and onto.