Let $f$ be a twice differentiable function on $(1,6)$. If $f(2)=8$,$f'(2)=5$,$f'(x) \geq 1$ and $f''(x) \geq 4$ for all $x \in (1,6)$,then:

Given that $f(x)$ is continuously differentiable on $a \le x \le b$ where $a < b, f(a) < 0$ and $f(b) > 0$,which of the following are always true?
$(i)$ $f(x)$ is bounded on $a \le x \le b$.
$(ii)$ The equation $f(x) = 0$ has at least one solution in $a < x < b$.
$(iii)$ The maximum and minimum values of $f(x)$ on $a \le x \le b$ occur at points where $f'(c) = 0$.
$(iv)$ There is at least one point $c$ with $a < c < b$ where $f'(c) > 0$.
$(v)$ There is at least one point $d$ with $a < d < b$ where $f'(d) < 0$.

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 $b$.

The constant $c$ of Lagrange's mean value theorem for $f(x)=\cos x-\sin 2x$ in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ is

If from the Mean Value Theorem,$f'({x_1}) = \frac{f(b) - f(a)}{b - a}$,then

The function $f(x) = x(x + 3)e^{-(1/2)x}$ satisfies all the conditions of Rolle's theorem in $[-3, 0]$. The value of $c$ is