Consider the function $f(x) = e^{-2x} \sin 2x$ over the interval $(0, \pi/2)$. $A$ real number $c \in (0, \pi/2)$,as guaranteed by Rolle's theorem,such that $f'(c) = 0$ is

Let $S$ be the set of all functions $f:[0,1] \rightarrow \mathbb{R}$ which are continuous on $[0,1]$ and differentiable on $(0,1)$. Then for every $f \in S$,there exists a $c \in (0,1)$,depending on $f$,such that:

If Rolle's theorem is applicable for the function $f(x) = \log \left( \frac{x^2 + a}{x} \right)$ on $[3, 4]$ with $c \in (3, 4)$ such that $f'(c) = 0$, then the value of $f''(c)$ is

The abscissa of the points of the curve $y = x^3$ in the interval $[-2, 2]$,where the slope of the tangents is equal to the slope of the secant line as per the Mean Value Theorem for the interval $[-2, 2]$,are:

If $f:[-5,5] \rightarrow R$ is a differentiable function and if $f^{\prime}(x)$ does not vanish anywhere,then prove that $f(-5) \neq f(5).$

$f:[1,3] \rightarrow R$ is a function defined as $f(x)=x^3+a x^2+b x$. If $f(1)-f(3)=0$ and $f^{\prime}\left(\frac{2 \sqrt{3}+1}{\sqrt{3}}\right)=0$, then $a-b$ is equal to