Question

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

• A. If $g$ is continuous at $x=1$, then $fg$ is differentiable at $x=1$
• B. If $fg$ is differentiable at $x=1$, then $g$ is continuous at $x=1$
• C. If $g$ is differentiable at $x=1$, then $fg$ is differentiable at $x=1$
• D. If $fg$ is differentiable at $x=1$, then $g$ is differentiable at $x=1$

Answer 1

Answer: (A), (C)

If $fg$ is differentiable at $x=1$, then
$\underset{h\to 0}{\lim }\frac{fg(1+h)-fg(1)}{h}$ exists finitely
$\Rightarrow \underset{h\to 0}{\lim }\frac{h\left[(1+h{)}^2+\sin (1+h)g(1+h)\right]-0}{h}$ exists finitely
$\Rightarrow \underset{h\to 0}{\lim }(1+\sin 1)g(1+h)$ exists finitely
So $\underset{h\to 0^{-}}{\lim }g(1+h)=\underset{h\to 0^{-}}{\lim }g(1+h)$
$\Rightarrow$ it does not mean that
$g(x)$ is continuous or differentiable at $x=1$
But if $g(x)$ is continuous or differentiable at $x=1$
then $\underset{h\to 0^{-}}{\lim }g(1+h)=\underset{h\to 0^{-}}{\lim }g(1+h)$,
hence $fg(x)$ will be differentiable