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

(D) Not monotonic on $(-\infty, 0)$ and $(0, \infty)$

(A) Monotonic on $(-\infty, 0)$ only

(B) Monotonic on $(-\infty, 0) \cup(0, \infty)$

(C) Monotonic on $(0, \infty)$ only

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Application of Derivatives

Consider the function

f : R \to R

defined by

f (x) = {(2 - sin (\frac{1}{x})) 0 ∣ x ∣, x \neq = 0, x = 0

. Then

f

is:

Answer: (D)

For

x > 0

f (x) = x (2 - sin \frac{1}{x})

and

f^{'} (x) = 2 - sin \frac{1}{x} + \frac{1}{x} cos \frac{1}{x}

x = \frac{1}{2 nπ}

this gives

f^{'} (x) = 2 + 2 nπ > 0

, while at

x = \frac{1}{( 2 n + 1 ) π}

it gives

f^{'} (x) = 2 - (2 n + 1) π < 0

for every

n \in N

. The derivative therefore changes sign in every neighbourhood of the origin, so

f

is not monotonic on

(0, \infty)

.
For

x < 0

f (x) = - x (2 - sin \frac{1}{x})

and

f^{'} (x) = - 2 + sin \frac{1}{x} - \frac{1}{x} cos \frac{1}{x}

x = - \frac{1}{2 nπ}

this gives

f^{'} (x) = 2 nπ - 2 > 0

and at

x = - \frac{1}{( 2 n + 1 ) π}

it gives

f^{'} (x) = - 2 - (2 n + 1) π < 0

, so the sign changes arbitrarily close to the origin here as well.
Hence

f

is monotonic neither on

(- \infty, 0)

nor on

(0, \infty)

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Single correct option