Question 3

First, check differentiability at $x=0$: $f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{(2 - 2h^2 - h^2 \sin \frac{1}{h}) - 2}{h} = \lim_{h \to 0} (-2h - h \sin \frac{1}{h}) = 0$.

So, $f$ is differentiable at $x=0$, making (A) false.

Now find $f'(x)$ for $x \neq 0$: $f'(x) = -4x - [2x \sin \frac{1}{x} + x^2 \cos \frac{1}{x} (-\frac{1}{x^2})] = -4x - 2x \sin \frac{1}{x} + \cos \frac{1}{x}$.

As $x \to 0$, the terms $-4x$ and $-2x \sin \frac{1}{x}$ approach $0$, but $\cos \frac{1}{x}$ oscillates between $-1$ and $1$ infinitely often in any neighborhood of $0$.

Thus, $f'(x)$ changes sign infinitely often in any interval $(-\delta, 0)$ or $(0, \delta)$. This implies $f$ is not monotonic (neither strictly increasing nor strictly decreasing) on any such interval. Hence, (B) is false and (C) is true.

Finally, for $x \neq 0$, $f(x) - f(0) = -2x^2 - x^2 \sin \frac{1}{x} = -x^2 (2 + \sin \frac{1}{x})$.

Since $|\sin \frac{1}{x}| \le 1$, we have $2 + \sin \frac{1}{x} \ge 1$. Thus $f(x) - f(0) < 0$ for all $x \neq 0$, meaning $x=0$ is a point of local maxima. (D) is false.