If the function f: R rightarrow R is defined | vedclass

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(C) Given $f(x) = |x|(x - \sin x)$.Since $f(-x) = |-x|(-x - \sin(-x)) = |x|(-x + \sin x) = -|x|(x - \sin x) = -f(x)$,the function is an odd function.F

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$f$ is $BOTH$ one-one and onto

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(C) Given $f(x) = |x|(x - \sin x)$.Since $f(-x) = |-x|(-x - \sin(-x)) = |x|(-x + \sin x) = -|x|(x - \sin x) = -f(x)$,the function is an odd function.For $x \geq 0$,$f(x) = x^2 - x \sin x$. For $x As $x ightarrow \infty$,$f(x) = x^2(1 - \frac{\sin x}{x}) ightarrow \infty$. As $x ightarrow -\infty$,$f(x) ightarrow -\infty$. Since $f$ is continuous,the range is $R$,so $f$ is onto.For $x > 0$,$f'(x) = 2x - \sin x - x \cos x = x(1 - \cos x) + (x - \sin x)$. Since $x > \sin x$ and $1 - \cos x \geq 0$ for $x > 0$,$f'(x) > 0$.For $x  0$ (by symmetry of the odd function).Since $f'(x) > 0$ for all $x 
eq 0$ and $f$ is continuous,$f$ is strictly increasing on $R$.Thus,$f$ is one-one.Therefore,$f$ is both one-one and onto.

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

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