In `[0, 1]` Lagrange's mean value theorem is not applicable to

Options

A. `f(x){{:((1)/(2)-x,xlt(1)/(2)),(((1)/(2)-x)^(2),xge(1)/(2)):}`
B. `f(x)={{:((sinx)/(x),xne0),(1,x=0`):}
C. f(x)=x|x|
D. f(x)=|x|

Answer

A

Explanation

For the function f(x) given in option (a), we have
`(LHD at x (1)/(2))=-1 and (RHD at x =(1)/(2))=0`
So, it is not differentiable at `x=(1)/(2)= ein (0,1)`