Explanation

(d)
There is only one function in option (a), whose critical point $\frac{1}{2} \in (0,\ 1)$ but in other parts critical point$0 \notin (0,\ 1)$. Then, we can say that functions in options (b), (c) and (d) are continuous on [0, 1] and differentiable in (0, 1)
Now, for $f(x) = \left\{ \begin{array}{r}
\frac{1}{2} - x,\ \ \ x < \frac{1}{2} \\
\left( \frac{1}{2} - x \right)^{2},\ \ \ x \geq \frac{1}{2}
\end{array} \right.\$
Here, $Lf^{'}\left( \frac{1}{2} \right) = - 1$
And $Rf^{'}\left( \frac{1}{2} \right) = 2\left( \frac{1}{2} - \frac{1}{2} \right)( - 1) = 0$
$\therefore\ \ Lf^{'}\left( \frac{1}{2} \right) \neq Rf^{'}\left( \frac{1}{2} \right)$
$\Rightarrow f$ is non-differentiable at $x = \frac{1}{2} \in (0,\ 1)$
$\therefore$ LMVT is NOT applicable to $f(x)in\ \lbrack 0,\ 1\rbrack$