Question

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

• A. $f$ is one-one, but NOT onto
• B. $f$ is onto, but NOT one-one
• C. $f$ is BOTH one-one and onto
• D. $f$ is NEITHER one-one NOR onto

Answer 1

Answer: (C)

$f(x)$ is odd, continuous function
$f(x)= \begin{cases}x(x-\sin x) & , x \geq 0 \\ -x(x-\sin x) & , x < 0\end{cases}$
for $x \geq 0, f^{\prime}(x)=2 x-\sin x-x \cos x$
$=x(1-\cos x)+(x-\sin x) \geq 0$
for $x < 0, f(x)=-2 x+\sin x+x \cos x$
$=x(\cos x-1)-(x-\sin x)>0$ as $x < 0$
$\Rightarrow $ So $f(x)$ strictly increases in $(-\infty, \infty)$
$ \Rightarrow f(x)$ is one-one
$x \rightarrow \infty f ( x ) \rightarrow \infty$
$x \rightarrow-\infty f(x) \rightarrow-\infty$.
So $f(x)$ is onto