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 $a =$ ..............

Let $g: R \rightarrow R$ be a non constant twice differentiable such that $g^{\prime}\left(\frac{1}{2}\right)=g^{\prime}\left(\frac{3}{2}\right)$. If a real valued function $f$ is defined as $\mathrm{f}(\mathrm{x})=\frac{1}{2}[\mathrm{~g}(\mathrm{x})+\mathrm{g}(2-\mathrm{x})]$, then

If $g(x) = 2f (2x^3 - 3x^2) + f(6x^2 - 4x^3 - 3)$, $\forall  x \in R$ and $f"(x) > 0, \forall  x \in R$ , then $g'(x) > 0$ for $x$ belonging to

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 function $f(x) = x(x + 3) e^{-x/2} ;$ satisfies the rolle's theorem in the interval $[-3, 0],$ then find $C$

The function $f(x) = {x^3} - 6{x^2} + ax + b$ satisfy the conditions of Rolle's theorem in $[1, 3]. $ The values of  $a $ and $ b $ are