Let $$\mathbb{R}$$ denote the set of all real numbers. Define the function $$f: \mathbb{R} \to \mathbb{R}$$ by

$$f(x) = \begin{cases} 2 - 2x^2 - x^2 \sin \frac{1}{x} & \text{if } x \neq 0, \\ 2 & \text{if } x = 0. \end{cases}$$

Then which one of the following statements is TRUE?

The function is defined by
$$f(x)=\begin{cases}2-2x^{2}-x^{2}\sin\frac1x,&x\neq0\\[4pt]2,&x=0\end{cases}$$

1. Continuity at $$x=0$$
$$\displaystyle\lim_{x\to0}\bigl(2-2x^{2}-x^{2}\sin\tfrac1x\bigr)=2-0-0=2=f(0)$$
Hence $$f$$ is continuous at the origin.

2. Differentiability at $$x=0$$
Use the definition of the derivative:

$$\begin{aligned} f'(0)&=\lim_{h\to0}\frac{f(h)-f(0)}{h} =\lim_{h\to0}\frac{-2h^{2}-h^{2}\sin\tfrac1h}{h}\\[4pt] &=\lim_{h\to0}\bigl(-h(2+\sin\tfrac1h)\bigr)=0. \end{aligned}$$
The limit exists, so $$f$$ is differentiable at $$x=0$$ with $$f'(0)=0$$. Therefore Option A is false.

3. Derivative for $$x\neq0$$
Differentiate term-wise:

$$\begin{aligned} f'(x)&=\frac{d}{dx}\Bigl(-2x^{2}-x^{2}\sin\tfrac1x\Bigr)\\[4pt] &=-4x-\Bigl[2x\sin\tfrac1x-x^{2}\cos\tfrac1x\cdot\frac1{x^{2}}\Bigr]\\[4pt] &=-4x-2x\sin\tfrac1x+\cos\tfrac1x\\[4pt] &=\cos\tfrac1x-4x-2x\sin\tfrac1x.\qquad(x\neq0) \end{aligned}$$

Near the origin the terms $$-4x$$ and $$-2x\sin\tfrac1x$$ are of order $$x$$ and tend to $$0$$, while $$\cos\tfrac1x$$ oscillates between $$-1$$ and $$1$$. Hence the sign of $$f'(x)$$ changes infinitely often in every punctured neighbourhood of $$0$$.

4. Testing the options

Option B: To be decreasing on $$(0,\delta)$$ we would need $$f'(x)\le0$$ for all $$x\in(0,\delta)$$. But pick a sequence $$x_n=\tfrac1{2n\pi}$$; then $$\cos\tfrac1{x_n}=\cos(2n\pi)=1$$ and

$$f'(x_n)=1-4x_n-2x_n\sin(2n\pi)=1-4x_n>0$$

for all sufficiently large $$n$$. Thus $$f$$ is not always decreasing on any interval $$(0,\delta)$$, so Option B is false.

Option C: To be increasing on $$(-\delta,0)$$ we would need $$f'(x)\ge0$$ there. Take the sequence $$y_n=-\tfrac1{(2n+1)\pi}$$; then $$\cos\tfrac1{y_n}=\cos\bigl(-(2n+1)\pi\bigr)=-1$$ and

$$f'(y_n)=-1-4y_n-2y_n\sin\bigl(-(2n+1)\pi\bigr)=-1-4y_n<0$$

for all large $$n$$ (since $$y_n\to0^-$$). Therefore, no matter how small $$\delta$$ is chosen, $$f'(x)$$ takes negative values in $$(-\delta,0)$$, so $$f$$ cannot be increasing there. Hence Option C is true.

Option D: For $$x\neq0$$ we have
$$f(x)=2-x^{2}\bigl(2+\sin\tfrac1x\bigr), \qquad 1\le2+\sin\tfrac1x\le3,$$
so $$f(x)\le2-x^{2}<2$$ for every non-zero $$x$$. Thus $$f(0)=2$$ is a strict maximum, not a minimum. Therefore Option D is false.

Conclusion
The only correct statement is Option C: “For any positive real number $$\delta$$, the function $$f$$ is NOT an increasing function on the interval $$(-\delta,0)$$.”