Consider the function $$f : R \to R$$ defined by $$f(x) = \begin{cases} \left(2 - \sin\left(\frac{1}{x}\right)\right)|x|, & x \neq 0 \\ 0, & x = 0 \end{cases}$$. Then $$f$$ is:

We have $$f(x) = \begin{cases} \left(2 - \sin\left(\frac{1}{x}\right)\right)|x|, & x \neq 0 \\ 0, & x = 0 \end{cases}$$.

For $$x > 0$$, $$|x| = x$$, so $$f(x) = \left(2 - \sin\left(\frac{1}{x}\right)\right)x = 2x - x\sin\left(\frac{1}{x}\right)$$.

Differentiating using the product rule: $$f'(x) = 2 - \sin\left(\frac{1}{x}\right) - x \cdot \cos\left(\frac{1}{x}\right) \cdot \left(-\frac{1}{x^2}\right) = 2 - \sin\left(\frac{1}{x}\right) + \frac{1}{x}\cos\left(\frac{1}{x}\right)$$.

For $$f$$ to be monotonically increasing on $$(0, \infty)$$, we need $$f'(x) \geq 0$$ for all $$x > 0$$. The term $$2 - \sin\left(\frac{1}{x}\right)$$ is always at least $$1$$ (since $$-1 \leq \sin \leq 1$$). However, the term $$\frac{1}{x}\cos\left(\frac{1}{x}\right)$$ oscillates and becomes unbounded near $$x = 0$$.

To see this, consider $$x_n = \frac{1}{(2n+1)\pi}$$ for large positive integers $$n$$. At these points, $$\frac{1}{x_n} = (2n+1)\pi$$, so $$\cos\left(\frac{1}{x_n}\right) = \cos((2n+1)\pi) = -1$$ and $$\sin\left(\frac{1}{x_n}\right) = 0$$. Then $$f'(x_n) = 2 - 0 + (2n+1)\pi \cdot (-1) = 2 - (2n+1)\pi$$. For $$n \geq 1$$, this gives $$f'(x_n) = 2 - 3\pi < 0$$.

Similarly, at $$x_n = \frac{1}{2n\pi}$$, we have $$\cos\left(\frac{1}{x_n}\right) = 1$$ and $$\sin\left(\frac{1}{x_n}\right) = 0$$, so $$f'(x_n) = 2 + 2n\pi > 0$$.

Since $$f'(x)$$ takes both positive and negative values for $$x$$ near $$0^+$$, the function is not monotonic on $$(0, \infty)$$.

For $$x < 0$$, $$|x| = -x$$, so $$f(x) = -\left(2 - \sin\left(\frac{1}{x}\right)\right)x$$. Differentiating: $$f'(x) = -2 + \sin\left(\frac{1}{x}\right) - \frac{1}{x}\cos\left(\frac{1}{x}\right)$$.

By the same argument, at $$x_n = -\frac{1}{2n\pi}$$ (approaching $$0$$ from the left), $$\cos\left(\frac{1}{x_n}\right) = 1$$ and $$\frac{1}{x_n} = -2n\pi$$, so $$f'(x_n) = -2 + 0 - (-2n\pi)(1) = -2 + 2n\pi > 0$$ for $$n \geq 1$$. At $$x_n = -\frac{1}{(2n+1)\pi}$$, $$\cos\left(\frac{1}{x_n}\right) = -1$$ and $$\frac{1}{x_n} = -(2n+1)\pi$$, so $$f'(x_n) = -2 + 0 - (-(2n+1)\pi)(-1) = -2 - (2n+1)\pi < 0$$.

So $$f'(x)$$ also changes sign near $$0^-$$, meaning $$f$$ is not monotonic on $$(-\infty, 0)$$ either.

Therefore, $$f$$ is not monotonic on $$(-\infty, 0)$$ and $$(0, \infty)$$, which is Option B.