From mean value theorem $f(b) - f(a) = $ $(b - a)f'({x_1});$   $a < {x_1} < b$ if $f(x) = {1 \over x}$, then ${x_1} = $

Verify Mean Value Theorem, if $f(x)=x^{2}-4 x-3$ in the interval $[a, b],$ where $a=1$ and $b=4$

Let  $f(x)$  satisfy the requirement of lagranges mean value theorem in $[0,2]$ . If $f(x)=0$ ; $\left| {f'\left( x \right)} \right| \leqslant \frac{1}{2}$ for all $x \in \left[ {0,2} \right]$, then-

Let $f, g:[-1,2] \rightarrow R$ be continuous functions which are twice differentiable on the interval $(-1,2)$. Let the values of $f$ and $g$ at the points $-1.0$ and $2$ be as given in the following table:

In each of the intervals $(-1,0)$ and $(0,2)$ the function $(f-3 g)^{\prime \prime}$ never vanishes. Then the correct statement(s) is(are)

$(A)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup(0,2)$

$(B)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$

$(C)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(0,2)$

$(D)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two solutions in $(0,2)$

Examine the applicability of Mean Value Theorem:

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

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

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

Let $f(x) = 8x^3 - 6x^2 - 2x + 1,$ then