The value of $\left[ {\frac{{\log \left( {\frac{x}{e}} \right)}}{{x - \,e}}} \right]\,\forall x\, > \,e$ is equal to (where [.] denotes greatest integer function)

In the mean value theorem, $f(b) - f(a) = (b - a)f'(c) $ if $a = 4$, $b = 9$ and $f(x) = \sqrt x $ then the value of $c$  is

Verify Rolle's theorem for the function $y=x^{2}+2, a=-2$ and $b=2$

Consider the function $f (x) = 8x^2 - 7x + 5$ on the interval $[-6, 6]$. The value of $c$ that satisfies the conclusion of the mean value theorem, is

For the function $f(x) = {e^x},a = 0,b = 1$, the value of $ c$ in mean value theorem will be

If $f$ and $g$ are differentiable functions in $[0, 1]$ satisfying $f\left( 0 \right) = 2 = g\left( 1 \right)\;,\;\;g\left( 0 \right) = 0,$ and $f\left( 1 \right) = 6,$ then for some $c \in \left] {0,1} \right[$  . .