Functions Question 86
Question: The function $ f(x)=|x| $ at $ x=0 $ is
[MP PET 1993]
Options:
A) Continuous but non-differentiable
B) Discontinuous and differentiable
C) Discontinuous and non-differentiable
D) Continuous and differentiable
Show Answer
Answer:
Correct Answer: A
Solution:
Since this function is continuous at $ x=0 $ Now for differentiability $ f(x)=,|x|=|0|=0 $ and $ f(0+h)=f(h)=|h| $
$ \therefore \underset{h\to 0-}{\mathop{\lim }}\frac{f(0+h)-f(0)}{h}=\underset{h\to 0-}{\mathop{\lim }}\frac{|h|}{h}=-1 $ and $ \underset{h\to 0+}{\mathop{\lim }}\frac{f(0+h)-f(0)}{h}=\underset{h\to 0+}{\mathop{\lim }}\frac{|h|}{h}=1 $ . Therefore it is continuous and non-differentiable.
BETA
