Math JEE Adv. 2025 paper 1

 To determine which statement about the function 

$f(x)$ is true, let's analyze each option step-by-step.

Given Function:

$$f(x)=\{\begin{array}{ll}2-2x^2-x^2\sin \frac{1}{x}& \text{if }x\mathrm{\ne }0,\\ 2& \text{if }x=0.\end{array}$$

Step 1: Differentiability at $x=0$

To check differentiability at $x=0$, compute the derivative using the limit definition:

$$f^{\mathrm{\prime }}(0)=\underset{h\to 0}{\lim }\frac{f(h)-f(0)}{h}=\underset{h\to 0}{\lim }\frac{2-2h^2-h^2\sin \frac{1}{h}-2}{h}=\underset{h\to 0}{\lim }(-2h-h\sin \frac{1}{h})\mathrm{.}$$

Since $\mid \sin \frac{1}{h}\mid \le 1$, the term $h\sin \frac{1}{h}$ approaches 0 as $h\to 0$. Thus:

$$f^{\mathrm{\prime }}(0)=0.$$

The function is differentiable at $x=0$, so Option (A) is false.

Step 2: Behavior on $(0,\delta )$

Compute the derivative for $x\mathrm{\ne }0$:

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

For small $x>0$, the term $\cos \frac{1}{x}$ oscillates between $-1$ and $1$, making $f^{\mathrm{\prime }}(x)$ change sign infinitely often near $x=0$. Therefore, $f$ is not decreasing on any interval $(0,\delta )$, so Option (B) is false.

Step 3: Behavior on $(-\delta ,0)$

For $x<0$, the derivative $f^{\mathrm{\prime }}(x)$ also oscillates due to the $\cos \frac{1}{x}$ term. This means $f$ is not increasing on any interval $(-\delta ,0)$, making Option (C) true.

Step 4: Local Minima at $x=0$

For $x\mathrm{\ne }0$, $f(x)=2-2x^2-x^2\sin \frac{1}{x}$. Since $\mid \sin \frac{1}{x}\mid \le 1$, we have:

$$f(x)\ge 2-2x^2-x^2=2-3x^2\mathrm{.}$$

However, for $x$ approaching 0, $f(x)$ can be less than 2 (e.g., when $\sin \frac{1}{x}=1$, $f(x)=2-3x^2$, which is less than 2 for $x\mathrm{\ne }0$). Thus, $x=0$ is not a local minima, so Option (D) is false.

Conclusion:

The only true statement is Option (C).

Final Answer: $\boxed{{\displaystyle C}}$