In the Mean Value Theorem,$f(b) - f(a) = (b - a)f'(c)$. If $a = 4$,$b = 9$ and $f(x) = \sqrt{x}$,then the value of $c$ is:

$A$ value of $c$ for which the conclusion of the Mean Value Theorem holds for the function $f(x) = \log_{e}x$ on the interval $[1, 3]$ is

If $f(x) = \begin{cases} x, & 0 \leq x \leq 1 \\ 2-x, & 1 < x \leq 2 \end{cases}$, then Rolle's theorem is not applicable to $f(x)$ because

Let $f(x)=2+\cos x$ for all real $x$.
$STATEMENT-1$: For each real $t$,there exists a point $c$ in $[t, t+\pi]$ such that $f^{\prime}(c)=0$. because
$STATEMENT-2$: $f(t)=f(t+2\pi)$ for each real $t$.

If the $L.M.V.T.$ holds for the function $f(x) = x + \frac{1}{x}$ on the interval $x \in [1, 3]$,then $c$ is:

Let $f(x) = \begin{cases} \frac{x^p}{(\sin x)^q} & \text{if } 0 < x \leq \frac{\pi}{2} \\ 0 & \text{if } x = 0 \end{cases}$ where $p, q \in \mathbb{R}$. Then, Lagrange's Mean Value Theorem is applicable to $f(x)$ in the closed interval $[0, \frac{\pi}{2}]$ if: