Mean value theorem $f(b) -f(a) = (b -a) f '(x_1);$ from $a < x_1 < b,$ if $f(x) = 1/x$ then $x_1 = ?$

If the equation

${a_n}{x^{n - 1}} + \,{a_{n - 1}}{x^{n - 1}} + \,......\, + \,{a_1}x = 0,\,{a_1} \ne 0,n\, \geqslant \,2,$

has a positive root $x= \alpha ,$ then the equation 

$n{a_n}{x^{n - 1}} + \,(n - 1){a_{n - 1}}{x^{n - 1}} + \,......\, + \,{a_1} = 0$

has a positive root which is

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

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 $x \in[1,2]$

If the function $f(x) = a{x^3} + b{x^2} + 11x - 6$ satisfies the conditions of Rolle's theorem for the interval $[1, 3$] and $f'\left( {2 + \frac{1}{{\sqrt 3 }}} \right) = 0$, then the values of $a$ and $b$ are respectively

Let $f :[2,4] \rightarrow R$ be a differentiable function such that $\left(x \log _e x\right) f^{\prime}(x)+\left(\log _e x\right) f(x)+f(x) \geq 1$, $x \in[2,4]$ with $f(2)=\frac{1}{2}$ and $f(4)=\frac{1}{4}$.

Consider the following two statements:

$(A): f(x) \leq 1$, for all $x \in[2,4]$

$(B)$ : $f(x) \geq \frac{1}{8}$, for all $x \in[2,4]$

Then,