JEE Advance - Mathematics (2025 - Paper 1 Online - No. 3)

Let $\mathbb{R}$ denote the set of all real numbers. Define the function $f : \mathbb{R} \to \mathbb{R}$ by

$f(x)=\left\{\begin{array}{cc}2-2 x^2-x^2 \sin \frac{1}{x} & \text { if } x \neq 0, \\ 2 & \text { if } x=0 .\end{array}\right.$

Then which one of the following statements is TRUE?

The function $f$ is NOT differentiable at $x = 0$

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

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

$x = 0$ is a point of local minima of $f$

The function $ f : \mathbb{R} \to \mathbb{R} $ is defined by:

$ f(x) = \begin{cases} 2 - 2x^2 - x^2 \sin \frac{1}{x}, & \text{if } x \neq 0, \\ 2, & \text{if } x = 0. \end{cases} $

Explanation:

(A) To determine differentiability at $ x = 0 $, consider the right-hand derivative (RHD):

$ \lim\limits_{h \rightarrow 0} \frac{(2 - 2h^2 - h^2 \sin \frac{1}{h}) - 2}{h} = 0 $

Similarly, for the left-hand derivative (LHD) at $ x = 0 $, we also find it equals 0. Therefore, $ f $ is differentiable at $ x = 0 $.

(B) For $ x \in (0, \delta) $, the derivative of $ f $ is:

$ f^{\prime}(x) = -\left(4x + 2x \sin \frac{1}{x}\right) + \cos \frac{1}{x} $

Thus:

$ f^{\prime}(x) = -\left(2x\left(4 - \sin \frac{1}{x}\right)\right) + \cos \frac{1}{x} $

Since $\cos \frac{1}{x}$ oscillates, we cannot assert that $ f(x) $ is strictly decreasing on $ (0, \delta) $.

(C) For $ x \in (-\delta, 0) $ and any $ \delta > 0 $, $ f(x) $ is not increasing since $\cos \frac{1}{x}$ oscillates between -1 and 1.

(D) Evaluating at $ x = 0 $:

$ f(0) = 2 $

For small $ h $:

$ f(0 + h) < 2 \quad \text{and} \quad f(0 - h) < 2 $

Therefore, $ x = 0 $ is a local maximum, not a local minimum.