Applications Of Derivatives Question 253
Question: In
[0, 1] Lagrange’s mean value theorem is NOT applicable to [IIT Screening 2003]
Options:
A) $ f(x)= \begin{cases} & \frac{1}{2}-x,,x<\frac{1}{2} \\ & {{( \frac{1}{2}-x )}^{2}},,x\ge \frac{1}{2} \\ \end{cases} . $
B) $ f(x)= \begin{cases} & \frac{\sin x}{x},,x\ne 0 \\ & ,1,,,x=0 \\ \end{cases} . $
C) $ f(x)=x|x| $
D) $ f(x)=|x| $
Show Answer
Answer:
Correct Answer: A
Solution:
The function defined in option is not differentiable at $ x=\frac{1}{2} $ .
BETA
