For which interval, the function ${{{x^2} - 3x} \over {x - 1}}$ satisfies all the conditions of Rolle's theorem

Consider a quadratic equation $ax^2 + bx + c = 0,$ where $2a + 3b + 6c = 0$ and let $g(x) = a\frac{{{x^3}}}{3} + b\frac{{{x^2}}}{2} + cx.$

Statement $1:$ The quadratic equation has at least one root in the interval $(0, 1).$

Statement $2:$ The Rolle's theorem is applicable to function $g(x)$ on the interval $[0, 1 ].$

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 $(0,2)$

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 $f:[-5,5] \rightarrow \mathrm{R}$ is a differentiable function and if $f^{\prime}(x)$ does not vanish anywhere, then prove that $f(-5) \neq f(5).$

If the function $f(x) = {x^3} - 6{x^2} + ax + b$ satisfies Rolle’s theorem in the interval $[1,\,3]$ and $f'\left( {{{2\sqrt 3 + 1} \over {\sqrt 3 }}} \right) = 0$, then $a =$ ..............