Lagrange’s Theorem cannot be applied for:
A) $f(x) = \log x$ in $[1,e]$
B) $f(x) = x - \dfrac{1}{x}$ in $[1,3]$
C) $f(x) = \sqrt {{x^2} - 4} $ in $[2,4]$
D) $f(x) = |x|$ in $[ - 1,2]$

Hint: First of all, we will write the conditions required for applying the Lagrange’s Theorem. Then, we will check these conditions for every option and see which does not fall under such conditions.

Complete step-by-step answer:
For applying the Lagrange’s Theorem, we need the function to be continuous in {a, b} and differentiable in (a, b).
Let us check every option one by one.
Option (A): $f(x) = \log x$ in $[1,e]$
We here have, a = 1 and b = e.
Let us find $f'(x)$. We know that if $f(x) = \log x$, then $f'(x) = \dfrac{1}{x}$.
Since the derivative is defined for all values from (1, e). Hence, the function is differentiable everywhere.
Now, we will use the fact that: “Every differentiable function is continuous.”
Therefore, the function is continuous as well.
Therefore, Lagrange’s Theorem can be applied to this.
Option (B): $f(x) = x - \dfrac{1}{x}$ in $[1,3]$
We here have, a = 1 and b = 3 .
Let us find $f'(x)$.
We know that if $f(x) = {x^n}$, then $f'(x) = {x^{n - 1}}$.
So, here $f(x) = {x^1} - {x^{ - 1}}$ and therefore, $f'(x) = {x^0} - {x^{ - 2}} = 1 - \dfrac{1}{{{x^2}}}$
Since the derivative is defined for all values from (1, 3). Hence, the function is differentiable everywhere.
Now, we will use the fact that: “Every differentiable function is continuous.”
Therefore, the function is continuous as well.
Therefore, Lagrange’s Theorem can be applied to this.
Option (C): $f(x) = \sqrt {{x^2} - 4} $ in $[2,4]$
We here have, a = 2 and b = 4 .$f(x) = |x|$
Let us find $f'(x)$.
We know that if $f(x) = {x^n}$, then $f'(x) = {x^{n - 1}}$.
So, here $f(x) = {\left( {{x^2} - 4} \right)^{\dfrac{1}{2}}}$ and therefore, $f'(x) = \dfrac{1}{2}{\left( {{x^2} - 4} \right)^{ - \dfrac{1}{2}}} \times \dfrac{d}{{dx}}({x^2} - 4)$ (Chain Rule)
Hence, \[f'(x) = \dfrac{1}{2}{\left( {{x^2} - 4} \right)^{ - \dfrac{1}{2}}} \times 2x = \dfrac{{2x{{({x^2} - 4)}^{ - \dfrac{1}{2}}}}}{2} = \dfrac{x}{{\sqrt {{x^2} - 4} }}\]
Since the derivative is defined for all values from (2, 4). Hence, the function is differentiable everywhere.
Now, we will use the fact that: “Every differentiable function is continuous.”
Therefore, the function is continuous as well.
Therefore, Lagrange’s Theorem can be applied to this.
Option (D): in $[ - 1,2]$
We here have, a = -1 and b = 2.
Redefining the function as:-
$f(x) = \left\{ {\begin{array}{*{20}{c}}
  { - x, - 1 \leqslant x \leqslant 0} \\
  {x,0 < x \leqslant 2}
\end{array}} \right.$
Let us find $f'(x)$.
$f'(x) = \left\{ {\begin{array}{*{20}{c}}
  { - 1, - 1 \leqslant x \leqslant 0} \\
  {1,0 < x \leqslant 2}
\end{array}} \right.$
Since, the derivative is different for $x = 0$ when it approaches from left and right that is:-
$\mathop {\lim }\limits_{x \to {0^ - }} f'(x) = \mathop {\lim }\limits_{x \to {0^ - }} ( - 1) = - 1$
$\mathop {\lim }\limits_{x \to {0^ + }} f'(x) = \mathop {\lim }\limits_{x \to {0^ + }} (1) = 1$
Hence, the function is not differentiable at $x = 0$.
Therefore, we cannot apply Lagrange’s Theorem to this.

Option D is the correct answer.

Note: The students must remember that “Every differentiable function is continuous” is just a necessary condition, not sufficient. Hence, converse need not be true. So, do not use it conversely.
Do you wonder that when we talk about continuity, we take closed brackets but when we talk about differentiability we take open brackets. Why is that?
It is because we cannot talk about differentiability of a function without approaching it from both the sides (left and right). If we take a closed bracket in left, we cannot approach from left at point a and same for point b if that is closed.