Let $\mathbb{R}$ denote the set of all real numbers. Define the function $f: \mathbb{R} \rightarrow \mathbb{R}$ by \[ f(\mathrm{x})=\left\{\begin{array}{cc} 2-2 x^{2}-x^{2} \sin \frac{1}{x} & \text { if } x \neq 0 \\ 2 & \text { if } x=0 \end{array}\right. \] Then which one of the following statements is TRUE ?

(C) For any positive real number $\delta$, the function $f$ is NOT an increasing function on the interval $(-\delta, 0)$

(A) The function $f$ is NOT differentiable at $x=0$

(B) There is a positive real number $\delta$, such that $f$ is a decreasing function on the interval $(0, \delta)$

(D) $x=0$ is a point of local minima of $f$

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Application of Derivatives

Let

R

denote the set of all real numbers. Define the function

f : R \to R

f (x) = {2 - 2 x^{2} - x^{2} sin \frac{1}{x} 2 if x \neq = 0 if x = 0

Then which one of the following statements is TRUE ?

(A)The function $f$ is NOT differentiable at $x = 0$
(B)There is a positive real number $δ$ , such that $f$ is a decreasing function on the interval $(0, δ)$
(C)For any positive real number $δ$ , the function $f$ is NOT an increasing function on the interval $(- δ, 0)$ ✓ Correct
(D) $x = 0$ is a point of local minima of $f$

Answer: (C)

First, check differentiability at

x = 0

h \to 0 lim \frac{f ( h ) - f ( 0 )}{h} = h \to 0 lim \frac{- 2 h ^{2} - h ^{2} sin \frac{1}{h}}{h} = 0

f

is differentiable at

x = 0

.

For

x \neq = 0

f^{'} (x) = - 4 x - 2 x sin \frac{1}{x} + cos \frac{1}{x}

For

x \in (0, δ)

, the term

cos \frac{1}{x}

oscillates between -1 and 1, so we cannot guarantee

f^{'} (x) < 0

throughout. Option (B) is false.

For

x \in (- δ, 0)

, the oscillation of

cos \frac{1}{x}

again prevents

f

from being monotone increasing. Option (C) is true.

Since

f (0) = 2

and

f (0 \pm h) < 2

for small

h > 0

x = 0

is a local maximum, not minimum. Option (D) is false.

JEE Advanced · 2025 · 18 May 2025 · Paper 1 · Single correct option