Let mathbb R denote the set of all real numbers. Define the function f: mathbb R rightarrow mathbb R by f( x )= \ begin array cc 2-2 x 2 -x 2 sin 1/x & if x neq 0 \\ 2 & if x=0 end array . Then which one of the following statements is TRUE ?

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

The function f is NOT differentiable at x=0

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

x=0 is a point of local minima of f

Mathematics · Limits, Continuity and Differentiability

JEE Advanced 2025 — Paper 1 — Question 17

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 ?

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

(A) $\quad R H D$ at $x=0: \lim _{h \rightarrow 0} \frac{\left(2-2 h^{2}-h^{2} \sin \frac{1}{h}\right)-2}{h}=0$

Similarly LHD at $\mathrm{x}=0$ is also equal to 0 . $\therefore$ Differentiable at $\mathrm{x}=0$

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

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

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

for $\mathrm{x} \in(0, \delta)$

$\Rightarrow$ We can't say $f(x)$ is decreasing on $(0, \delta)$ as $\cos \frac{1}{x}$ oscillates.

$\Rightarrow \mathrm{f}(\mathrm{x})$ is not increasing on $(-\delta, 0)$ as $\cos \frac{1}{\mathrm{x}}$ oscillates from -1 to 1 .

(D) $\mathrm{f}(0)=2$

$\mathrm{f}(0+\mathrm{h})<2$

$\mathrm{f}(0-\mathrm{h})<2$

$\therefore \mathrm{x}=0$ is local maxima

Answer key and solution verified before publishing.

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