Question

Let $f(x)=\{\begin{array}{ll}\sin x,& \text{ for }x\ge 0\\ 1-\cos x,& \text{ for }x\le 0\end{array}$ and $g(x)=e^x$. Then the value of $(gof{)}^{\prime }$ $(0)$ is

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

Answer 1

Answer: (D)

Given, $f(x)=\{\begin{array}{ll}\sin x,& \text{ for }x\ge 0\\ 1-\cos x,& \text{ for }x\le 0\end{array}$ and $g(x)=e^x$.
$\therefore$ gof $=\{\begin{array}{ll}{e}^{\sin x},& x\ge 0\\ e^{1-\cos x},& x\le 0\end{array}$
$\therefore LHD=(\text{ gof }{)}^{\prime }(0-h)=\underset{h\to 0}{\lim }\frac{gof(0-h)-gof(h)}{-h}$
$=\underset{h\to 0}{\lim }\frac{e^{1-\cos (0-h)}-e^{1-\cosh }}{-h}=0$
RHD $=(\text{ gof }{)}^{\prime }(0+h)=\underset{h\to 0}{\lim }\frac{gof(0+h)-gof(h)}{h}$
$=\underset{h\to 0}{\lim }\frac{e^{\sinh }-e^{\sinh }}{h}=0$
$\therefore RHD=LHD=0$
$\Rightarrow (\text{ gof }{)}^{\prime }(0)=0$