Let $f^{\prime}(0)=-3$ and $f^{\prime}(x) \leq 5$ for all real values of $x$. The $f(2)$ can have possible maximum value as

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

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

Consider a 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 $2:$ The Rolle's theorem is applicable to function $g(x)$ on the interval $[0, 1].$

Verify Rolle's theorem for the function $y=x^{2}+2$ on the interval $[-2, 2]$.

Let $f(x)$ be a differentiable function for all $x$ such that $f'(x) \le 5$ and $f(1) = 4$. The maximum value of $f(5)$ is...