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

Let $a > 0$ and $f$ be continuous in $[-a, a]$. Suppose that $f'(x)$ exists and $f'(x) \le 1$ for all $x \in (-a, a)$. If $f(a) = a$ and $f(-a) = -a$,then $f(0)$ is:

The position of a moving car at time $t$ is given by $f(t) = at^{2} + bt + c, t > 0,$ where $a, b,$ and $c$ are real numbers greater than $1.$ Then the average speed of the car over the time interval $[t_{1}, t_{2}]$ is attained at the point

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:

Let $y = f(x)$ and $y = g(x)$ be two differentiable functions in $[0, 2]$ such that $f(0) = 3$,$f(2) = 5$,$g(0) = 1$,and $g(2) = 2$. If there exists at least one $c \in (0, 2)$ such that $f'(c) = k g'(c)$,then $k$ must be:

Let $f(x)$ be a non-constant twice differentiable function defined on $(-\infty, \infty)$ such that $f(x)=f(1-x)$ and $f^{\prime}\left(\frac{1}{4}\right)=0$. Then
$(A)$ $f^{\prime \prime}(x)$ vanishes at least twice on $[0,1]$
$(B)$ $f^{\prime}\left(\frac{1}{2}\right)=0$
$(C)$ $\int_{-1 / 2}^{1 / 2} f\left(x+\frac{1}{2}\right) \sin x d x=0$
$(D)$ $\int_0^{1 / 2} f(t) e^{\sin \pi t} d t=\int_{1 / 2}^1 f(1-t) e^{\sin \pi t} d t$