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 $f (x)$ and $g (x)$ be two continuous functions defined from $R \rightarrow R$, such that $f (x_1) > f (x_2)$ and $g (x_1) < g (x_2), \forall x_1 > x_2$ , then solution set of $f\,\left( {\,g({\alpha ^2} – 2\alpha )\,} \right) >f\,\left( {\,g(3\alpha – 4)\,} \right)$ is

Let $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ be a thrice differentiable function such that $f(0)=0, f(1)=1, f(2)=-1, f(3)=2$ and $f(4)=-2$. Then, the minimum number of zeros of $\left(3 f^{\prime} f^{\prime \prime}+f f^{\prime \prime \prime}\right)(x)$ is………………..

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

If from mean value theorem, $f'({x_1}) = {{f(b) – f(a)} \over {b – a}}$, then