Mean Value Theorem states that for a function $f:[a, b] \rightarrow R,$ if

a) $f$ is continuous on $[a, b]$

b) $f$ is continuous on $(a, b)$

Then, there exists some $c \in(a, b)$ such that $f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$

Therefore, Mean Value Theorem is not applicable to those functions that do not satisfy any of the two conditions of the hypothesis.

$(i)$ $f(x)=[x]$ for $x \in[5,9]$

It is evident that the given function $f(x)$ is not continuous at every integral point.

In particular, $f(x)$ is not continuous at $x=5$ and $x=9$

is not continuous in $[5,9]$

The differentiability of $f$ in $(5,9)$ is checked as follows,

Let $\mathrm{n}$ be an integer such that $n \in(5,9)$

The left hand limit of $f$ at $x=n$ is.

$\mathop {\lim }\limits_{h \to {0^ – }} \frac{{f(n + h) – f(n)}}{h} = \mathop {\lim }\limits_{h \to {0^ – }} \frac{{[n + h] – [n]}}{h} = \mathop {\lim }\limits_{h \to {0^ – }} \frac{{n – 1 – n}}{h} = \mathop {\lim }\limits_{h \to {0^ – }} \frac{{ – 1}}{h} = \infty $

The right hand limit of $f$ at $x=n$ is.

$\mathop {\lim }\limits_{h \to {0^ + }} \frac{{f(n + h) – f(n)}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{[n + h] – [n]}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} \frac{{n – n}}{h} = \mathop {\lim }\limits_{h \to {0^ + }} 0 = 0$

Since the left and right hand limits of $f$ at $x=n$ are not equal, $f$ is not differentiable at $x=n$

$\therefore f$ is not differentiable in $(5,9).$

It is observed that $f$ does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is not applicable for $f(x)=[x]$ for $x \in[5,9]$

$(ii)$ $f(x)=[x]$ for $x \in[-2,2]$

It is evident that the given function $f(x)$ is not continuous at every integral point.

In particular, $f(x)$ is not continuous at $x=-2$ and $x=2$

$\Rightarrow f(x)$ is not continuous in $[-2,2].$

The differentiability of $f$ in $(-2,2)$ is checked as follows.

Let $n$ be an integer such that $n \in(-2,2)$

The left hand limit of $f$ at $x=n$ is.

$\mathop {\lim }\limits_{h \to {0^\prime }} \frac{{f(n + h) – f(n)}}{h} = \mathop {\lim }\limits_{h \to {0^\prime }} \frac{{[n + h] – [n]}}{h} = \mathop {\lim }\limits_{h \to {0^\prime }} \frac{{n – 1 – n}}{h} = \mathop {\lim }\limits_{h \to {0^\prime }} \frac{{ – 1}}{h} = \infty $

The right hand limit of $f$ at $x=n$ is.

$\mathop {\lim }\limits_{h \to {0^\prime }} \frac{{f(n + h) – f(n)}}{h} = \mathop {\lim }\limits_{h \to {0^\prime }} \frac{{[n + h] – [n]}}{h} = \mathop {\lim }\limits_{h \to {0^\prime }} \frac{{n – n}}{h} = \mathop {\lim }\limits_{h \to {0^\prime }} 0 = 0$

Since the left and right hand limits of $f$ at $x=\mathrm{n}$ are not equal, $f$ is not differentiable at $x=n$

$\therefore f$ is not differentiable in $(-2,2)$

It is observed that $f$ does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is not applicable for $f(x)=[x]$ for $x \in[-2,2].$

$(iii)$ $f(x)=x^{2}-1$ for $x \in[1,2]$

It is evident that $f$, being a polynomial function, is a continuous in $[1,2]$ and is differentiable in $(1,2)$

It is observed that $f$ satisfies all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is applicable for $f(x)=x^{2}-1$ for $x \in[1,2]$

It can be proved as follows.

$f(1)=1^{2}-1=0, f(2)=2^{2}-1=3$

$\therefore \frac{f(b)-f(a)}{b-a}=\frac{f(2)-f(1)}{2-1}=\frac{3-0}{1}=3$

$f^{\prime}(x)=2 x$

$\therefore f^{\prime}(c)=3$

$\Rightarrow 2 c=3$

$\Rightarrow c=\frac{3}{2}=1.5,$ where $1.5 \in[1,2]$