JEE Advanced 2025 Paper 1 Q03 Mathematics Differentiation & Applications Monotonicity & Differentiability Hard

JEE Advanced 2025 Paper 1 · Q03 · Monotonicity & Differentiability

Define $f: \mathbb{R} \to \mathbb{R}$ by $f(x) = 2 - 2x^2 - x^2 \sin(1/x)$ for $x \ne 0$, and $f(0) = 2$. Then which of the following is TRUE?

  1. A. $f$ is NOT differentiable at $x = 0$
  2. B. There is $\delta > 0$ such that $f$ is decreasing on $(0, \delta)$
  3. C. For every $\delta > 0$, $f$ is NOT increasing on $(-\delta, 0)$
  4. D. $x = 0$ is a point of local minima of $f$
Reveal answer + step-by-step solution

Correct answer:C

Solution

Set $g(x) = f(x) - 2 = -2x^2 - x^2 \sin(1/x)$. Differentiability at $0$: $g(x)/x = -2x - x \sin(1/x) \to 0$, so $g'(0) = 0$ and $f'(0) = 0$; (A) is FALSE. For $x \ne 0$, $f'(x) = -4x - 2x \sin(1/x) + \cos(1/x)$. The $\cos(1/x)$ term oscillates between $-1$ and $1$ infinitely often as $x \to 0$, while $-4x - 2x \sin(1/x) \to 0$. So $f'$ takes both positive and negative values in any neighbourhood of $0$. Hence on no interval $(0, \delta)$ is $f$ monotonic — (B) is FALSE — and on no interval $(-\delta, 0)$ is $f$ increasing throughout: in any $(-\delta, 0)$ one can find points where $f' < 0$, so $f$ cannot be increasing on the whole sub-interval. (C) is TRUE. Since $f'$ takes both signs in every nbd of $0$ with $f'(0)=0$, $x=0$ is neither a local min nor local max. (D) is FALSE. Answer: (C).

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