JEE Advanced 2025 — Paper 1 — MATH-I solution

JEE Advanced2025Paper 1MATH-I

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

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

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

Solution

Examine each option in turn.

Differentiability at 0. For $h \neq = 0$ , $\frac{f ( h ) - f ( 0 )}{h} = \frac{- 2 h ^{2} - h ^{2} sin ( 1/ h )}{h} = - 2 h - h sin (1/ h) \to 0$ as $h \to 0$ . So $f$ is differentiable at $0$ with $f^{'} (0) = 0$ . Option (A) is false.

Derivative for $x \neq = 0$ . Differentiating, $f^{'} (x) = - 4 x - 2 x sin \frac{1}{x} + cos \frac{1}{x} .$

As $x \to 0$ , the first two terms vanish, while $cos (1/ x)$ oscillates between $- 1$ and $1$ . Therefore in every neighbourhood of $0$ on either side, $f^{'}$ takes both positive and negative values.

Option (B): On any interval $(0, δ)$ , $f^{'}$ changes sign repeatedly, so $f$ is not monotonic there. False.

Option (C): By the same oscillation argument, $f^{'}$ is not non-negative throughout any interval $(- δ, 0)$ , so $f$ is not increasing on $(- δ, 0)$ for any $δ > 0$ . True.

Option (D): For small $h$ , $f (0 \pm h) = 2 - 2 h^{2} - h^{2} sin (1/ (\pm h)) < 2 = f (0)$ when $2 + sin (1/ h) > 0$ , which holds. So $f$ has a local maximum at $0$ , not a minimum. False.

$∴$ The correct option is (C).

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