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

Let the function f:R→R be defined by f(x)=x³-x²+(x-1)sin x and let g:R→R be an arbitrary function. Let fg:R→R be the product function defined by (fg)(x)=f(x)g(x). Then which of the following statements is/are TRUE? $f(x) = x^3 - x^2 + (x-1)\sin x, \quad (fg)(x) = f(x)g(x)$ $g \text{ is an arbitrary function}$ $\text{Check which statements are TRUE}$ $f(x) = x^2(x-1) + (x-1)\sin x = (x-1)(x^2 + \sin x)$ $f(1) = 0 \Rightarrow (fg)(x) = (x-1)\,(x^2+\sin x)\,g(x)$ $(fg)'(1) = \lim_{x\to 1}\frac{(fg)(x) - (fg)(1)}{x-1} = \lim_{x\to 1}(x^2+\sin x)\,g(x)$ $1 + \sin 1 \neq 0 \Rightarrow \lim_{x\to 1}(x^2+\sin x)g(x) \text{ exists} \iff \lim_{x\to 1} g(x) \text{ exists}$ $\therefore fg \text{ differentiable at } 1 \iff \lim_{x\to 1} g(x) \text{ exists (finite)}$ $(1): g \text{ continuous at } 1 \Rightarrow \lim_{x\to1} g(x) \text{ exists} \Rightarrow fg \text{ differentiable} \ \checkmark$ $(2): fg \text{ differentiable} \not\Rightarrow g \text{ continuous at } 1$ $g(x) = \begin{cases} 0, & x \neq 1 \\ 5, & x = 1 \end{cases}: \ \lim g = 0 \neq g(1) = 5$ $(3): g \text{ differentiable at } 1 \Rightarrow g \text{ continuous} \Rightarrow \lim_{x\to1} g(x) \text{ exists} \Rightarrow fg \text{ differentiable} \ \checkmark$ $(4): fg \text{ differentiable} \not\Rightarrow g \text{ differentiable at } 1$ $g(x) = |x-1|: \ \lim_{x\to1} g(x) = 0, \text{ but } g \text{ has a corner at } 1$ $\text{Correct statements: } (1) \text{ and } (3)$