If the function $f(x) = ax^3 + bx^2 + 26x - 24$ satisfies the conditions of Rolle's theorem in $[2, 4]$ and $f^{\prime}\left(3 + \frac{1}{\sqrt{3}}\right) = 0$,then the value of $ab$ is equal to

Suppose $f$ is differentiable for all $x$. If $f(1) = -2$ and $f'(x) \geq 2$ for all $x \in [1, 6]$,then:

Let $f: R \rightarrow R$ 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 $(3 f^{\prime} f^{\prime \prime} + f f^{\prime \prime \prime})(x)$ is....................

Find the value of $p$ and $q$ if the function $f(t) = t^3 - 6t^2 + pt + q$ defined on $[1, 3]$ satisfies Rolle's theorem for $c = \frac{2\sqrt{3} + 1}{\sqrt{3}}$.

If $2a + 3b + 6c = 0$,then at least one root of the equation $ax^2 + bx + c = 0$ lies in the interval:

If $c$ is a point at which Rolle's theorem holds for the function $f(x) = \log_{e}\left(\frac{x^{2}+\alpha}{7x}\right)$ in the interval $[3, 4]$,where $\alpha \in R$,then $f''(c)$ is equal to