JEE Advanced 2020 — Paper 1 — Question 35

$\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}$

$f(x)=x^{3}-x^{2}+(x-1) \sin x$

$\mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}$

(A) If $g$ is continuous at $x=1$ then $f g$ is differentiable

Let $h(x)=f(x) \cdot g(x)$

RHD $=h^{\prime}\left(1^{+}\right)=\lim _{h \rightarrow 0} \frac{f(1+h) g(1+h)-f(1) g(1)}{h}$

$=\lim _{h \rightarrow 0} \frac{f(1+h) g(1+h)-0}{h}$

$=\lim _{h \rightarrow 0} g(1+h) \cdot \lim _{h \rightarrow 0} \frac{f(1+h)-f(0)}{h}$

$$\begin{aligned} & =g(1) f^{\prime}(1) \\& \text { LHD at } x=1 \\& =\lim _{h \rightarrow 0} \frac{f(1-h) g(1-h)-f(1) g(1)}{-h} \\& =\lim _{h \rightarrow 0} g(1-h) \cdot \lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}=g(1) f^{\prime}(1) \end{aligned}$$

So, $h(x)$ is differentiable at $ x=1 $$

(B) Given $h(x)=f(x) \cdot g(x)$ is differentiable

$$\lim _{h \rightarrow 0} \frac{f(1+h) g(1+h)-f(1) g(1)}{h}=\lim _{h \rightarrow 0} \frac{f(1-h) g(1-h)-f(1) g(1)}{-h}$$ $$\lim _{h \rightarrow 0} g(1+h) \cdot f^{\prime}(1)=\lim _{h \rightarrow 0} g(1-h) \cdot f^{\prime}(1)$$

$f^{\prime}(1) \neq 0$ and $g(1)$ is not define

So, can not comment over continuity and differentiability

(C) Given $\mathrm{g}(\mathrm{x})$ is differentiable

So, $h(x)=f(x) g(x)$

$h^{\prime}(x)=f^{\prime}(x) g(x)+g^{\prime}(x) f(x)$, as $g(x)$ is differentiable

$=f^{\prime}(1) g(1)+0$ will exist

(D) Same as for B

$\lim _{h \rightarrow 0} g(1+h)=\lim _{h \rightarrow 0} g(1-h)$

can not say about differentiability of the $g(x)$