Question

Consider the function f : R → R defined by $f\left(x\right)=\left\{\begin{array}{l}\{2-\sin (1/x\left)\right\}\left|x\right|x\ne 0\\ 0x=0\end{array}\right.$

Then f is

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

Answer 1

The correct option is B

not monotonic on $(-\infty ,0)$ and$(0,\infty )$

Explanation for the correct option:

Step 1. Definition of modulus function :

$\left|x\right|=\left\{\begin{array}{l}-xifx<0\\ xifx>0\\ 0ifx=0\end{array}\right.$

Step 2. The function can be written as :

$f\left(x\right)=\left\{\begin{array}{c}-\{2-\sin (1/x\left)\right\}xx<0\\ 0x=0\\ \{2-\sin (1/x\left)\right\}xx>0\end{array}\right.$

Step 3. Differentiate it with respect to x :

$$\begin{gathered} f\text{'}\left(x\right)=\left\{\begin{array}{l}-x\left[-\cos \left(\frac{1}{x}\right)\left(\frac{-1}{x^2}\right)\right]-\left[2-\sin \left(\frac{1}{x}\right)\right]x<0\\ x\left[-\cos \left(\frac{1}{x}\right)\left(\frac{-1}{x^2}\right)\right]+\left[2-\sin \left(\frac{1}{x}\right)\right]x>0\end{array}\right. \\ f\text{'}\left(x\right)=\left\{\begin{array}{l}-\left(\frac{1}{x}\right)\cos \left(\frac{1}{x}\right)+\sin \left(\frac{1}{x}\right)-2x<0\\ \left(\frac{1}{x}\right)\cos \left(\frac{1}{x}\right)+\sin \left(\frac{1}{x}\right)+2x>0\end{array}\right. \end{gathered}$$

Thus, f is not monotonic on $(-\infty ,0)$ and$(0,\infty )$.

Hence, the correct option is B.