Question

Let $f(x)=\begin{cases}\sin x, & \text { for } x \geq 0 \\ 1-\cos x, & \text { for } x \leq 0\end{cases}$ and $g(x)=e^{x}$. Then the value of $(gof)'$ $(0)$ is

• A. 1
• B. -1
• C. 0
• D. None of these

Answer 1

Answer: (D)

Given, $f(x)=\begin{cases}\sin x, & \text { for } x \geq 0 \\ 1-\cos x, & \text { for } x \leq 0\end{cases}$ and $g(x)=e^{x}$.
$\therefore $ gof $=\begin{cases} e ^{\sin x}, & x \geq 0 \\ e ^{1-\cos x}, & x \leq 0 \end{cases}$
$\therefore LHD =(\text { gof })'(0- h )=\displaystyle\lim _{ h \rightarrow 0} \frac{gof(0- h )-gof( h )}{- h }$
$=\displaystyle\lim _{h \rightarrow 0} \frac{ e ^{1-\cos (0- h )}- e ^{1-\cosh }}{- h }=0$
RHD $=(\text { gof })'(0+h)=\displaystyle\lim _{h \rightarrow 0} \frac{gof(0+h)-gof(h)}{h}$
$=\displaystyle\lim _{h \rightarrow 0} \frac{ e ^{\sinh }- e ^{\sinh }}{ h }=0 $
$\therefore RHD = LHD =0 $
$\Rightarrow (\text { gof })'(0)=0$