Functions Question 337
Question: Let $ f(x)= \begin{cases} \sin x, & \text{for }x\ge 0 \\ 1-\cos x, & \text{for }x\le 0 \\ \end{cases} . $ and $ g(x)=e^{x} $ . Then $ (gof)’(0) $ is
[UPSEAT 2004]
Options:
A) 1
B) ?1
C) 0
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ (gof),(x)=g[f(x)] $ = $ g,[1-\cos x]={e^{1-\cos x}},\text{ for }x\le 0 $ $ (gof{)}’(x)={e^{1-\cos x}}.\sin x,,\text{for }x\le 0 $ $ (gof{)}’(0)=0 $ .
BETA
