Question

Let the function $f:R\to R$ be defined by $f\left(x\right)=x^3-x^2+(x-1)\sin x$ and let $g:R\to R$ be an arbitrary function. Let $fg:R\to R$ be the product function defined by $\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right).$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 $g$ is differentiable at $x=1,$ then $fg$ is differentiable at $x=1$

$f:R\to R$
$\left(A\right):f\left(x\right)=x^3-x^2+(x-1)\sin x;g:R\to R$
$h\left(x\right)=f\left(x\right)\cdot g\left(x\right)=[x^3-x^2+(x-1\left)\sin x\right]\cdot g\left(x\right)$
$h^{\prime }\left(1^{+}\right)=\underset{h\to 0}{lim}\frac{\left[\right(1+h{)}^3-(1+h{)}^2+h\cdot \sin (1+h\left)\right]g(1+h)}{h}$
$=\underset{h\to 0}{lim}\frac{(1+h^3+3h+3h^2-1-h^2-2h+h\sin (1+h\left)\right)g(1+h)}{h}$
$=\underset{h\to 0}{lim}\frac{(h^3+2h^2+h+h\sin (1+h\left)\right)g(1+h)}{h}$
$=\underset{h\to 0}{lim}(1+\sin (1+h\left)\right)g(1+h)$
$h^{\prime }\left(1^{-}\right)=\underset{h\to 0}{lim}\frac{\left[\right(1-h{)}^3-(1-h{)}^2+(-h)\cdot \sin (1-h\left)\right]g(1-h)}{-h}$
$=\underset{h\to 0}{lim}\frac{(1-h^3-3h+3h^2-1-h^2+2h-h\sin (1-h\left)\right)g(1-h)}{-h}$
$=\underset{h\to 0}{lim}(1+\sin (1-h\left)\right)g(1-h)$

as $g\left(x\right)$ is continuous at $x=1$
$\therefore g(1+h)=g(1-h)=g\left(1\right)$
$h^{\prime }\left(1^{+}\right)=h^{\prime }\left(1^{-}\right)=(1+\sin 1)g\left(1\right)$
${}^{\prime }{A}^{\prime }$ is correct

$C$ is always true as
$h\left(x\right)=f\left(x\right)\cdot g\left(x\right)\Rightarrow h^{\prime }\left(x\right)=f^{\prime }\left(x\right)\cdot g\left(x\right)+f\left(x\right)\cdot g^{\prime }\left(x\right)$