If $f(x) = ax^3 + bx^2 + 11x - 6$ for $x \in [1, 3]$ satisfies the conditions of Rolle's theorem and $f'\left( 2 + \frac{1}{\sqrt{3}} \right) = 0$,find $a$ and $b$.

If the equation $a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x = 0$,where $a_1 \neq 0$ and $n \geq 2$,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:

Consider the 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-$2$: Rolle's theorem can be applied to the function $g(x)$ in the interval $[0, 1]$.

If the function $f(x) = ax^3 + bx^2 + 11x - 6$, defined on $[1, 3]$, satisfies all the conditions of Rolle's theorem for $c = 2 + \frac{1}{\sqrt{3}}$, then

Examine if Rolle's Theorem is applicable to the following function. Can you say something about the converse of Rolle's Theorem from this example?
$f(x) = [x]$ for $x \in [-2, 2]$

Consider $f(x) = \int\limits_0^x {\left( {t + \frac{1}{t}} \right)\,dt}$ and $g(x) = f'(x)$ for $x \in \left[ {\frac{1}{2}, 3} \right]$. If $P$ is a point on the curve $y = g(x)$ such that the tangent to this curve at $P$ is parallel to a chord joining the points $\left( {\frac{1}{2}, g\left( {\frac{1}{2}} \right)} \right)$ and $(3, g(3))$ of the curve,then the coordinates of the point $P$ are: