JEE PYQ: Application Of Derivatives Question 6
Question 6 - 2021 (17 Mar Shift 2)
Consider the function $f : \mathbb{R} \to \mathbb{R}$ 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
(1) monotonic on $(-\infty, 0) \cup (0, \infty)$
(2) not monotonic on $(-\infty, 0)$ and $(0, \infty)$
(3) monotonic on $(0, \infty)$ only
(4) monotonic on $(-\infty, 0)$ only
Show Answer
Answer: (2)
Solution
For $x > 0$: $f’(x) = 2 - \sin\frac{1}{x} + \frac{1}{x}\cos\frac{1}{x}$. The term $\frac{1}{x}\cos\frac{1}{x}$ oscillates, so $f’(x)$ is an oscillating function which is non-monotonic on $(-\infty, 0)$ and $(0, \infty)$.
BETA
