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)

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 $c \in \left] {0,1} \right[$  . .

If $f(x) = \cos x,0 \le x \le {\pi \over 2}$, then the real number $ ‘c’ $ of the mean value theorem is

If the function $f(x) = {x^3} - 6a{x^2} + 5x$ satisfies the conditions of Lagrange's mean value theorem for the interval $[1, 2] $ and the tangent to the curve $y = f(x) $ at $x = {7 \over 4}$ is parallel to the chord that joins the points of intersection of the curve with the ordinates $x = 1$ and $x = 2$. Then the value of $a$ is

Let $f (1) = - 2$ and $f ' (x) \ge 4.2$ for $1 \le x \le 6$. The smallest possible value of $f (6)$, is

If the function  $f(x) =  - 4{e^{\left( {\frac{{1 - x}}{2}} \right)}} + 1 + x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3}$ and $g(x)=f^{-1}(x) \,;$ then the value of $g'(-\frac{7}{6})$ equals