If $f(x) = \sqrt{x}$ and $g(x) = \frac{1}{\sqrt{x}}$ for $x \in [3, 12]$,then the value of $c \in (3, 12)$ for which $\frac{f^{\prime}(c)}{g^{\prime}(c)} = \frac{f(12) - f(3)}{g(12) - g(3)}$ holds,is

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

Given $f(x) = 4 - (\frac{1}{2} - x)^{2/3}$,$g(x) = \begin{cases} \frac{\tan([x])}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases}$,$h(x) = \{x\}$,and $k(x) = 5^{\log_2(x + 3)}$. Then,in the interval $[0, 1]$,Lagrange's Mean Value Theorem is $NOT$ applicable to:

If the function $f(x)=x^3+ax^2+bx+40$ satisfies the conditions of Rolle's theorem on the interval $[-5,4]$ and $-5,4$ are two roots of the equation $f(x)=0$, then one of the values of $c$ as stated in that theorem is

Let $f$ be a polynomial function defined on $[2,7]$. If $f(2)=3$ and $f^{\prime}(x) \leq 5$ for all $x$ in $(2,7)$,then the maximum possible value attained by $f$ at $x=7$ is

Rolle's theorem holds for a monic quadratic polynomial $f(x)$ on the interval $[\alpha, \alpha + 3]$ where $f(\alpha) = 0$. Similarly, $g(x) = f(x) + 2$ also follows Rolle's theorem on the interval $[\beta, 3]$ where $g(3) = 0$, such that the value of $c$ (where $f'(c) = g'(c) = 0$) is the same for both $f(x)$ and $g(x)$. Then the value of $(f \circ g)(\alpha)$ is...