in` [0,1] `, lagrange mean value theorem is NOT applicable to

Options

A. `f(x)={(1/2-x,x lt 1/2),((1/2-x)^(2),xge1/2):}`
B. `f(x)={((sinx)/x,x!=0),(1,x=0):}`
C. `f(x)=x|x|`
D. `f(x)=|x|`

Answer

A

Explanation

There is only one function in option a whose critical point `1/2 epsilon(0,1)` but in other parts critical point `0 !in(0,1)`.
Then, we can say that functions is options b, c and d are continuous on `[0,1]` and differentiable in (0,1).
Nor for `f(x)={(1/2-x, xlt 1/2),((1/2-x)^(2),x ge1/2):}`
Here `Lf'(1/2)=-1` and `Rf'(1/2)=2(1/2-1/2)(-1)=0`
`:.Lf'(1/2)!=Rf'(1/2)`
`implies f` is non-differentiable at `x=1/2 epsilon(0,1)`.
`:.` LMVT is NOT applicable to `f(x)` in `[0,1]`.