JEE Advanced 2025 · Paper 1 · Mathematics Q3

MathematicsSingle correct+3 / -1Question 3 of 48

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

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

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

Answer and worked solution

Official answer: C

(A) False. $\dfrac{f(x)-f(0)}{x}=-2x-x\sin\frac1x\to0$ , so $f'(0)=0$ exists.
(B) False. For $x\ne0$ , $f'(x)=-4x-2x\sin\frac1x+\cos\frac1x$ . As $x\to0$ the term $\cos\frac1x$ dominates and takes values near $\pm1$ in every interval $(0,\delta)$ , so $f'$ changes sign there and $f$ is not decreasing on any $(0,\delta)$ .
(C) True. The same holds on $(-\delta,0)$ , so $f$ is not increasing on any such interval.
(D) False. $f(x)-f(0)=-x^2\left(2+\sin\frac1x\right)<0$ for $x\ne0$ , so $x=0$ is a local maximum.

JEE Advanced 2025 · Paper 1, Mathematics question 3