Continuity and Differentiability
PYP_JEE_ADV_2025_P1
Grade None
Question:
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 $f$
Step-by-Step Solution
Key Concept: Differentiability via limit definition; oscillatory derivative near origin prevents local monotonicity
At $x = 0$: $f'(0) = \lim_{x \to 0} \dfrac{f(x)-2}{x} = \lim_{x \to 0} \dfrac{-2x^2 - x^2\sin(1/x)}{x} = \lim_{x \to 0} x(-2 - \sin(1/x)) = 0$. So $f$ is differentiable at $x=0$, ruling out (A).
For $x \neq 0$: $f'(x) = -4x - 2x\sin(1/x) + \cos(1/x)$. Near $x=0$, the $\cos(1/x)$ term oscillates between $-1$ and $1$, so $f'(x)$ changes sign in every interval $(-\delta,0)$ and $(0,\delta)$. This means $f$ is neither increasing nor decreasing on any $(0,\delta)$ or $(-\delta,0)$, and $x=0$ is not a local extremum.
(B) is FALSE — $f$ is not monotonically decreasing on any $(0,\delta)$.
(C) is TRUE — $f$ cannot be increasing on $(-\delta,0)$ for any $\delta > 0$ since $f'$ oscillates.
(D) is FALSE — $x=0$ is not a local minimum.
Correct Answer: C