Differentiability and Monotonicity of Piecewise Oscillatory Function

To determine the correct statement, we analyze the differentiability, extrema, and derivative behavior of $f(x)$ around $x = 0$.

1. Differentiability at $x = 0$: $f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} = \lim_{x \to 0} \frac{-2x^2 - x^2 \sin \frac{1}{x}}{x} = \lim_{x \to 0} \left( -2x - x \sin \frac{1}{x} \right) = 0$ Thus, $f$ is differentiable at $x = 0$, which makes option (A) false.

2. Local Extrema at $x = 0$: For all $x \neq 0$, since $-1 \le \sin \frac{1}{x} \le 1$, we have $2 + \sin \frac{1}{x} \ge 1 > 0$. Therefore, $f(x) - f(0) = -x^2 \left( 2 + \sin \frac{1}{x} \right) < 0$ This implies that $f(x) < f(0)$ for all $x \neq 0$, so $x = 0$ is a point of strict local maximum, making option (D) false.

3. Monotonicity on Intervals Near $x = 0$: For $x \neq 0$, the derivative is given by: $f'(x) = -4x - 2x \sin \frac{1}{x} + \cos \frac{1}{x}$ As $x \to 0$, the term $\cos \frac{1}{x}$ oscillates infinitely often between $-1$ and $1$, while $-4x - 2x \sin \frac{1}{x} \to 0$. Hence, in every open interval of the form $(-\delta, 0)$ and $(0, \delta)$ for any $\delta > 0$, $f'(x)$ changes sign infinitely many times.

   As a result, $f$ cannot be monotonically increasing or decreasing on any such interval $(-\delta, 0)$ or $(0, \delta)$. Therefore, option (B) is false and option (C) is true.

Correct Answer: C