Question

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 $f g: R \rightarrow R$ 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 $f g$ is differentiable at $x=1$
• B. If $f g$ is differentiable at $x=1$, then $g$ is continuous at $x=1$
• C. If $g$ is differentiable at $x=1$, then $f g$ is differentiable at $x=1$
• D. If $f g$ is differentiable at $x=1$, then $g$ is differentiable at $x=1$

Answer 1

Answer: (A), (C)

If $f g$ is differentiable at $x=1$, then
$\displaystyle\lim _{h \rightarrow 0} \frac{f g(1+h)-f g(1)}{h}$ exists finitely
$\Rightarrow \displaystyle\lim _{h \rightarrow 0} \frac{h\left[(1+h)^{2}+\sin (1+h) g(1+h)\right]-0}{h}$ exists finitely
$\Rightarrow \displaystyle\lim _{h \rightarrow 0}(1+\sin 1) g(1+h)$ exists finitely
So $\displaystyle\lim _{h \rightarrow 0^{-}} g(1+h)=\displaystyle\lim _{h \rightarrow 0^{-}} 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 $\displaystyle\lim_{h \rightarrow 0^{-}} g(1+h)=\displaystyle\lim _{h \rightarrow 0^{-}} g(1+h)$,
hence $f g(x)$ will be differentiable