Functions Question 571
Question: If $ f(x)= \begin{cases} & 1,x<0 \\ & 1+\sin x,0\le x<\frac{\pi }{2} \\ \end{cases} $ then $ f’(0)= $
[MP PET 1994]
Options:
A) 1
B) 0
C) $ \infty $
D) Does not exist
Show Answer
Answer:
Correct Answer: D
Solution:
$ Rf’(0)=\underset{h\to 0}{\mathop{\lim }}\frac{f(0+h)-f(0)}{h}=\underset{h\to 0}{\mathop{\lim }}\frac{1+\sinh -1}{h}=1 $ $ f’(0)=\underset{h\to 0}{\mathop{\lim }}\frac{f(0-h)-f(0)}{-h}=\underset{h\to 0}{\mathop{\lim }}\frac{1-1}{-h}=0 $ Hence, $ f’(0) $ does not exist.
BETA
