Mathematics - Continuity - Differentiability Question with Solution | TestHub

Differentiability of $fg$ at $x=1$

Left-hand derivative :

${\left(fg\right)}^{\text{'}}\left(1^{-}\right)=\underset{h\to 0}{\lim }\frac{fg\left(1-h\right)-fg\left(1\right)}{-h}$

$=\underset{h\to 0}{\lim }\frac{\left\{{\left(1-h\right)}^3-{\left(1-h\right)}^2-h\sin \left(1-h\right)\right\}g\left(1-h\right)-0}{-h}$

$=\underset{h\to 0}{\lim }\left\{{\left(1-h\right)}^2+\sin \left(1-h\right)\right\}g\left(1-h\right) .......\left(i\right)$

Right-hand derivative :

${\left(fg\right)}^{\text{'}}\left(1^{+}\right)=\underset{h\to 0}{\lim }\frac{fg\left(1+h\right)-fg\left(1\right)}{h}$

$=\underset{h\to 0}{\lim }\frac{\left\{{\left(1+h\right)}^3-{\left(1+h\right)}^2+h\sin \left(1+h\right)\right\}g\left(1+h\right)-0}{h}$

$=\underset{h\to 0}{\lim }\left\{{\left(1+h\right)}^2+\sin \left(1+h\right)\right\}g\left(1+h\right) .......\left(ii\right)$

If $g$ is continuous at $x=1$, then

$\underset{h\to 0}{\lim }g\left(1+h\right)=\underset{h\to 0}{\lim }g\left(1-h\right)=g\left(1\right) .........\left(iii\right)$

From equations $\left(i\right), \left(ii\right) \& \left(iii\right)$, we get

$\underset{h\to 0}{\lim }{\left(fg\right)}^{\text{'}}\left(1^{+}\right)=\underset{h\to 0}{\lim }{\left(fg\right)}^{\text{'}}\left(1^{-}\right)=\left(1+\sin 1\right)g\left(1\right)$

$\therefore fg$ is differentiable at $x=1$.

So, option $\left(A\right)$ is correct.

Now, from equations $\left(i\right) \& \left(ii\right)$, we can say that for $fg$ to be differentiable, we need only $g\left(1+h\right)=g\left(1-h\right)$.

But for $g$ to be continuous, we need $g\left(1+h\right)=g\left(1-h\right)=g\left(1\right)$.

So, option $\left(B\right)$ is incorrect.

Now, if $g$ is differentiable at $x=1$ and $f$ is already differentiable at $x=1$ as $f^{\text{'}}\left(1^{-}\right)=f^{\text{'}}\left(1^{+}\right)=1+\sin 1$, so product of two differentiable functions is also differentiable.

$\therefore fg$ is differentiable at $x=1$.

So, option $\left(C\right)$ is correct.

Now, from option $\left(B\right)$, if $fg$ is differentiable at $x=1$, we cannot guarantee $g$ to be continuous at $x=1$.

So, we also cannot guarantee $g$ to be differentiable at $x=1$.

So, option $\left(D\right)$ is incorrect.