Suppose $f(x)$ is twice differentiable in the interval $[1, 3]$ and $f(1)=f(3)$. If $|f^{\prime \prime}(x)| \leq 2$,then for all $x$ in $[1, 3]$,which one of the following is true?

If $c = \frac{1}{2}$ and $f(x) = 2x - x^2$,then the interval $(a, b)$ of $x$ in which the Lagrange's Mean Value Theorem $(LMVT)$ is applicable for the function $f(x)$ is:

Let $f(x)$ be continuous on $[0,4]$, differentiable on $(0,4)$, $f(0)=4$ and $f(4)=-2$. If $g(x)=\frac{f(x)}{x+2}$, then the value of $g^{\prime}(c)$ for some Lagrange's constant $c \in (0,4)$ is

$A$ value of $c$ according to the Lagrange's mean value theorem for $f(x)=(x-1)(x-2)(x-3)$ in $[0,4]$ 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...

Let $f(x) = \log(1 + x^2)$ and $A$ be a constant such that $\frac{|f(x) - f(y)|}{|x - y|} \leq A$ for all real $x, y$ where $x \neq y$. Then,the least possible value of $A$ is