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

$f(x)= \begin{cases}\frac{1}{2}-x, & x<\frac{1}{2} \\ \left(\frac{1}{2}-x\right)^2, & x \geq \frac{1}{2}\end{cases}$

For the function $f(x)$ given in option (a), we have

(LHD at x = $\frac{1}{2}$) = -1 and, (RHD at x = $\frac{1}{2}$) = 0

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

Hence, Lagrange's mean value theorem is not applicable.