Question

If the function $f:R\to R$ 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
$f(x)=\{\begin{array}{ll}x(x-\sin x)& ,x\ge 0\\ -x(x-\sin x)& ,x<0\end{array}$
for $x\ge 0,f^{\prime }(x)=2x-\sin x-x\cos x$
$=x(1-\cos x)+(x-\sin x)\ge 0$
for $x<0,f(x)=-2x+\sin x+x\cos x$
$=x(\cos x-1)-(x-\sin x)>0$ as $x<0$
$\Rightarrow$ So $f(x)$ strictly increases in $(-\infty ,\infty )$
$\Rightarrow f(x)$ is one-one
$x\to \infty f(x)\to \infty$
$x\to -\infty f(x)\to -\infty$.
So $f(x)$ is onto