Question

Consider the function $f : R \rightarrow R$ defined by $f(x)=\begin{cases}\left(2-\sin \left(\frac{1}{x}\right)\right)|x|, x \neq 0 \\ 0 , \,\,\,\, x=0\end{cases}$ Then $f$ is:

• A. monotonic on $(-\infty, 0) \cup(0, \infty)$
• B. not monotonic on $(-\infty, 0)$ and $(0, \infty)$
• C. monotonic on $(0, \infty)$ only
• D. monotonic on $(-\infty, 0)$ only

Answer 1

Answer: (B)

$f(x)=\begin{cases}-x\left(2-\sin \left(\frac{1}{x}\right)\right) & x<0 \\ 0 & x=0 \\ x\left(2-\sin \left(\frac{1}{x}\right)\right) & \end{cases}$
$f ^{\prime}( x )=\begin{cases}-\left(2-\sin \frac{1}{ x }\right)- x \left(-\cos \frac{1}{ x } \cdot\left(-\frac{1}{ x ^{2}}\right)\right) & x <0 \\ \left(2-\sin \frac{1}{ x }\right)+ x \left(-\cos \frac{1}{ x }\left(-\frac{1}{ x ^{2}}\right)\right) & x >0\end{cases}$
$f^{\prime}(x)=\begin{cases}-2+\sin \frac{1}{x}-\frac{1}{x} \cos \frac{1}{x} x<0 \\ 2-\sin \frac{1}{x}+\frac{1}{x} \cos \frac{1}{x} x>0\end{cases}$
$f ^{\prime}( x )$ is an oscillating function which is non-monotonic in $(-\infty, 0) \cup(0, \infty)$.