The value of $c$ for which Rolle's theorem holds for the function $f(x)=x^3-3x^2+2x$ in the interval $[0,2]$ is:

Let the function $f:[-7,0] \rightarrow R$ be continuous on $[-7,0]$ and differentiable on $(-7,0)$. If $f(-7)=-3$ and $f'(x) \leq 2$ for all $x \in (-7,0)$,then for all such functions $f$,$f(-1)+f(0)$ lies in the interval:

If $f(x)=x^3+p x^2+q x$ is defined on $[0,2]$ such that $f(0)=f(2)$ and $f^{\prime}\left(1+\frac{1}{\sqrt{3}}\right)=0$,then $p^2+q^2=$

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be such that $f(0)=0$ and $|f^{\prime}(x)| \leq 5$ for all $x$. Then $f(1)$ is in

Let $g: R \rightarrow R$ be a non-constant twice differentiable function such that $g^{\prime}\left(\frac{1}{2}\right)=g^{\prime}\left(\frac{3}{2}\right)$. If a real-valued function $f$ is defined as $f(x)=\frac{1}{2}[g(x)+g(2-x)]$,then:

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 $b$.