The value of $c$ in the Lagrange's mean value theorem for $f(x)=\sqrt{x-2}$ in the interval $[2,6]$ is

Suppose that $f$ is differentiable for all $x$ and that $f'(x) \le 2$ for all $x$. If $f(1) = 2$ and $f(4) = 8$,then $f(2)$ has the value equal to

If Rolle's theorem holds for the function $f(x) = 2x^3 + ax^2 + bx$ in the interval $[-1, 1]$ for the point $c = \frac{1}{2}$,then the value of $2a + b$ is

The number of admissible values of $C$ obtained when the Lagrange's mean value theorem is applied for the function $f(x)=x$ on $[2,5]$ is

Let $f, g:[-1,2] \rightarrow R$ be continuous functions which are twice differentiable on the interval $(-1,2)$. Let the values of $f$ and $g$ at the points $-1, 0$ and $2$ be as given in the following table:

$x$	$x=-1, 0, 2$
$f(x)$	$3, 6, 0$
$g(x)$	$0, 1, -1$

In each of the intervals $(-1,0)$ and $(0,2)$ the function $(f-3g)^{\prime \prime}$ never vanishes. Then the correct statement$(s)$ is(are):
$(A)$ $f^{\prime}(x)-3g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup (0,2)$
$(B)$ $f^{\prime}(x)-3g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$
$(C)$ $f^{\prime}(x)-3g^{\prime}(x)=0$ has exactly one solution in $(0,2)$
$(D)$ $f^{\prime}(x)-3g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two solutions in $(0,2)$

Examine the applicability of the Mean Value Theorem for the following functions:
$(i)$ $f(x) = [x]$ for $x \in [5, 9]$
$(ii)$ $f(x) = [x]$ for $x \in [-2, 2]$
$(iii)$ $f(x) = x^{2} - 1$ for $x \in [1, 2]$