Let f, g : R → R be two functions defined by $$f(x) = \begin{cases} x\sin\left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$$ and $$g(x) = xf(x)$$. Statement I: f is a continuous function at x = 0. Statement II: g is a differentiable function at x = 0.

We are given two functions, $$f$$ and $$g$$, both mapping real numbers to real numbers. The function $$f(x)$$ is defined as:

$$f(x) = \begin{cases} x \sin\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}$$

and $$g(x)$$ is defined as $$g(x) = x \cdot f(x)$$. We need to verify two statements: Statement I claims that $$f$$ is continuous at $$x = 0$$, and Statement II claims that $$g$$ is differentiable at $$x = 0$$.

First, let's check Statement I: continuity of $$f$$ at $$x = 0$$. For $$f$$ to be continuous at $$x = 0$$, the limit of $$f(x)$$ as $$x$$ approaches 0 must equal $$f(0)$$. We know $$f(0) = 0$$. So, we need to compute:

$$\lim_{x \to 0} f(x) = \lim_{x \to 0} x \sin\left(\frac{1}{x}\right)$$

Since $$|\sin(\theta)| \leq 1$$ for any real $$\theta$$, we have:

$$\left| x \sin\left(\frac{1}{x}\right) \right| \leq |x| \cdot 1 = |x|$$

This implies:

$$-|x| \leq x \sin\left(\frac{1}{x}\right) \leq |x|$$

As $$x$$ approaches 0, both $$-|x|$$ and $$|x|$$ approach 0. By the Squeeze Theorem,

$$\lim_{x \to 0} x \sin\left(\frac{1}{x}\right) = 0$$

Thus, $$\lim_{x \to 0} f(x) = 0 = f(0)$$, so $$f$$ is continuous at $$x = 0$$. Statement I is true.

Next, check Statement II: differentiability of $$g$$ at $$x = 0$$. First, express $$g(x)$$ explicitly. Since $$g(x) = x \cdot f(x)$$, and using the definition of $$f(x)$$:

For $$x \neq 0$$, $$f(x) = x \sin\left(\frac{1}{x}\right)$$, so:

$$g(x) = x \cdot \left( x \sin\left(\frac{1}{x}\right) \right) = x^2 \sin\left(\frac{1}{x}\right)$$

At $$x = 0$$, $$g(0) = 0 \cdot f(0) = 0 \cdot 0 = 0$$. Thus,

$$g(x) = \begin{cases} x^2 \sin\left(\frac{1}{x}\right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}$$

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

$$g'(0) = \lim_{h \to 0} \frac{g(0 + h) - g(0)}{h} = \lim_{h \to 0} \frac{g(h) - 0}{h} = \lim_{h \to 0} \frac{g(h)}{h}$$

For $$h \neq 0$$, $$g(h) = h^2 \sin\left(\frac{1}{h}\right)$$, so:

$$\frac{g(h)}{h} = \frac{h^2 \sin\left(\frac{1}{h}\right)}{h} = h \sin\left(\frac{1}{h}\right)$$

Thus,

$$g'(0) = \lim_{h \to 0} h \sin\left(\frac{1}{h}\right)$$

Again, $$|\sin(\theta)| \leq 1$$, so:

$$\left| h \sin\left(\frac{1}{h}\right) \right| \leq |h| \cdot 1 = |h|$$

This implies:

$$-|h| \leq h \sin\left(\frac{1}{h}\right) \leq |h|$$

As $$h$$ approaches 0, both $$-|h|$$ and $$|h|$$ approach 0. By the Squeeze Theorem,

$$\lim_{h \to 0} h \sin\left(\frac{1}{h}\right) = 0$$

Therefore, $$g'(0)$$ exists and equals 0, so $$g$$ is differentiable at $$x = 0$$. Statement II is true.

Both statements are true. Now, reviewing the options:

A. Both statement I and II are false.

B. Both statement I and II are true.

C. Statement I is true, statement II is false.

D. Statement I is false, statement II is true.

Hence, the correct answer is Option B.