Let the function $$f: R \rightarrow R$$ be defined by $$f(x) = x^{3} − x^{2} + (x − 1) \sin x$$ and let $$g: R \rightarrow R$$ be an arbitrary function. Let $$fg: R \rightarrow R$$ be the product function defined by $$(fg)(x) = f(x)g(x)$$. Then which of the following statements is/are TRUE?

First write $$f(x)$$ in a factored form about $$x = 1$$ because that is the point of interest.

Since $$f(1)=1^3-1^2+(1-1)\sin 1=0$$, factor out $$(x-1)$$:

$$f(x)= (x-1)\,k(x)$$
where $$k(x)=\dfrac{x^{3}-x^{2}+(x-1)\sin x}{x-1}\qquad(x\neq 1).$$

The limit of $$k(x)$$ as $$x\to 1$$ is the derivative of $$f$$ at $$1$$:

$$k(1)=f'(1)=\left[3x^{2}-2x+\sin x+(x-1)\cos x\right]_{x=1}=1+\sin 1\neq 0.$$

Thus, in a neighbourhood of $$1$$ we can write

$$f(x)=(x-1)k(x)\quad\text{with}\quad k(x)\to k(1)=1+\sin 1.$$

Now examine each statement.

Case A: $$g$$ is continuous at $$1$$.
Then $$\displaystyle\lim_{x\to 1}\dfrac{f(x)g(x)-f(1)g(1)}{x-1} =\lim_{x\to 1}\dfrac{(x-1)k(x)g(x)}{x-1} =\lim_{x\to 1}k(x)g(x)=k(1)g(1).$$ The limit exists, so $$(fg)(x)$$ is differentiable at $$x=1$$. Statement A is TRUE.

Case B: $$(fg)(x)$$ is differentiable at $$1$$.
Take a counter-example: $$g(x)=\begin{cases}1,&x\neq 1\\ 2,&x=1\end{cases}.$$ Here $$g$$ is not continuous at $$1$$. For $$x\neq 1,\; (fg)(x)=f(x).$$ Define $$(fg)(1)=f(1)g(1)=0.$$ Because $$f$$ itself is differentiable at $$1$$, the product $$fg$$ is differentiable at $$1$$ even though $$g$$ is not continuous. Therefore differentiability of $$fg$$ does not force continuity of $$g$$. Statement B is FALSE.

Case C: $$g$$ is differentiable at $$1$$.
Differentiability implies continuity, and from Case A continuity of $$g$$ already guarantees differentiability of $$fg$$. Hence $$fg$$ is differentiable at $$1$$. Statement C is TRUE.

Case D: $$(fg)(x)$$ is differentiable at $$1$$ implies $$g$$ is differentiable at $$1$$.
Using the same counter-example given in Case B, $$fg$$ is differentiable but $$g$$ is not even continuous, let alone differentiable. So the implication fails. Statement D is FALSE.

Therefore the correct options are:
Option A (true) and Option C (true).

Final Answer: Option A and Option C are correct.