Question

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

Answer 1

Answer: (A), (C)

If $g$ is differentiable at $x=1$, then $fg$ is differentiable at $x=1$

Explanation for the correct options:

Finding the value of the function at $x=1$:

Given that,

$f:R\to R$

$$\begin{gathered} f\left(x\right)=x^3–x^2+(x–1)\sin \left(x\right)\mathrm{and}g:R\to R \\ \mathrm{Also} \\ h\left(x\right)=f\left(x\right).g\left(x\right)=x^3–x^2+(x–1)\sin x·g\left(x\right) \end{gathered}$$

Substituting $x=1+h$ in the above function and taking limits

$$\begin{gathered} \mathrm{First}\mathrm{we}\mathrm{find}h\text{'}\left(1^{+}\right) \\ h\text{'}\left(1^{+}\right)=\underset{h\to 0}{\lim }\frac{\{{(1+h)}^3-{(1+h)}^2+h.\sin \left(1+h\right)\}g(1+h)}{h} \\ =\underset{h\to 0}{\lim }\frac{{(1+h}^3+3h+3h^2-{1-h}^2-2h+h.\sin \left(1+h\right)\left)g\right(1+h)}{h} \\ =\underset{h\to 0}{\lim }\frac{{(h}^3+2h^2+h+h.\sin \left(1+h\right)\left)g\right(1+h)}{h} \\ =\underset{h\to 0}{\lim }(1+\sin \left(1+h\right))g(1+h) \\ =\left(1+\sin \left(1\right)\right)g\left(1\right) \\ \mathrm{Now}h\text{'}\left(1^{-}\right) \\ h\text{'}\left(1^{-}\right)=\underset{h\to 0}{\lim }\frac{\{{(1-h)}^3-{(1-h)}^2+(-h).\sin \left(1-h\right)\}g(1-h)}{-h} \\ =\underset{h\to 0}{\lim }\frac{{(-h}^3-3h+3h^2-h^2+2h-h.\sin \left(1-h\right)\left)g\right(1-h)}{-h} \\ =\underset{h\to 0}{\lim }(1+\sin \left(1-h\right))g(1-h) \\ =\left(1+\sin \left(1\right)\right)g\left(1\right) \end{gathered}$$

Therefore, $h\text{'}\left(1^{+}\right)=h\text{'}\left(1^{-}\right)$

As $g\left(x\right)$ is constant at $x=1$

$$\begin{gathered} g(1+h)=g(1–h)=g\left(1\right) \\ \mathrm{And} \\ h\text{'}(1+)=h\text{'}(1–)=\left(1+\sin \left(1\right)\right)g\left(1\right) \end{gathered}$$

Therefore, the correct answers are options (A) and (C).