If $f$ is defined in $[1,3]$ by $f(x)=x^3+b x^2+a x$,such that $f(1)-f(3)=0$ and $f^{\prime}(c)=0$,where $c=2+\frac{1}{\sqrt{3}}$,then $(a, b)$ is equal to

If Rolle's theorem holds for the function $f(x) = 2x^3 + bx^2 + cx$ on the interval $x \in [-1, 1]$ at the point $x = \frac{1}{2}$,then the value of $2b + c$ is:

If the equation $a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x = 0$,where $a_1 \neq 0$ and $n \ge 2$,has a positive root $x = \alpha$,then the equation $n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \dots + a_1 = 0$ has a positive root,which is:

If $f(x)$ is a differentiable function,$f^{\prime}(x) \geq 5$ for all $x \in [2, 6]$,$f(2) = 4$ and $f(3) = 15$,then a possible value of $f(6)$ is:

Let $a, b, c$ be real numbers such that $2a + 3b + 6c = 0$ and $g(x) = ax^2 + bx + c = 0$ has at least one root in the interval $(1, 2)$. If a function $f: [1, 2] \rightarrow \mathbb{R}$ for which Rolle's Theorem holds is such that $f(x)$ is a primitive of $g(x)$,then $f(x) = $

The value of $c$ satisfied by Rolle's theorem for the function $f(x) = x^2(1 - x)^2$ on the interval $x \in [0, 1]$ is...