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

Given interval range is $\left[0,1\right]$.

Lagrange's mean value theorem: If $f\left(x\right)$ is representing a polynomial function in $x$ and the two roots of the equation $f\left(x\right)=0$ are $x=a$ and $x=b$, then there exists at                                                         least one root of the equation$f\text{'}\left(x\right)=0$ lying between these values.

                                                      $f\text{'}\left(x\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$f$f\text{'}\left({\frac{1}{2}}^{+}\right)=\frac{{\left({\displaystyle \frac{1}{2}}-1\right)}^2-{\displaystyle \frac{1}{2}}}{1}=0$

The function $f\left(x\right)=\left\{\begin{array}{c}\frac{1}{2}-x,x<\frac{1}{2}\\ {\left(\frac{1}{2}-x\right)}^2,x\ge \frac{1}{2}\end{array}\right.$

The left hand derivative $f\text{'}\left({\frac{1}{2}}^{-}\right)=\frac{\left({\displaystyle \frac{1}{2}}-1\right)-{\displaystyle \frac{1}{2}}}{1-0}=-1$.

The right had derivative 

For the function $f\left(x\right)$ given in option $1$, we have (LHD at $x=\frac{1}{2})=-1$ and $($ RHD at $x=\frac{1}{2})=0$.

So, it is not differentiable at $x=\frac{1}{2}\in (0,1)$.

$\therefore$ Lagrange's mean value theorem is not applicable.

Hence, option $1$ is the correct option.