Let $f(x)$ be continuous on $[0,6]$ and differentiable on $(0,6)$. Let $f(0)=12$ and $f(6)=-4$. If $g(x)=\frac{f(x)}{x+1}$, then for some Lagrange's constant $c \in(0,6)$, $g^{\prime}(c)=$

Given that $f(x)$ is continuously differentiable on $a \le x \le b$ where $a < b, f(a) < 0$ and $f(b) > 0$,which of the following are always true?
$(i)$ $f(x)$ is bounded on $a \le x \le b$.
$(ii)$ The equation $f(x) = 0$ has at least one solution in $a < x < b$.
$(iii)$ The maximum and minimum values of $f(x)$ on $a \le x \le b$ occur at points where $f'(c) = 0$.
$(iv)$ There is at least one point $c$ with $a < c < b$ where $f'(c) > 0$.
$(v)$ There is at least one point $d$ with $a < d < b$ where $f'(d) < 0$.

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:

The value $C$ of the Lagrange's mean value theorem for the function $f(x)=x(x-1)(x-2)$ in the interval $[0, 1/2]$ is

Values of $c$ as per Rolle's theorem for $f(x)=\sin x+\cos x+6$ on $[0, 2\pi]$ are

Let $f$ be any continuous function on $[0,2]$ and twice differentiable on $(0,2)$. If $f(0)=0, f(1)=1$ and $f(2)=2$,then