Let $f(x)$ be a function continuous on $[1,2]$ and differentiable on $(1,2)$ satisfying
$f(1) = 2, f(2) = 3$ and $f'(x) \geq 1 \forall x \in (1,2)$.Define $g(x)=\int\limits_1^x {f(t)\,dt\,\forall \,x\, \in [1,2]} $ then the greatest value of $g(x)$ on $[1,2]$ is-

If $f$ and $g$ are differentiable functions in $[0, 1]$ satisfying $f\left( 0 \right) = 2 = g\left( 1 \right)\;,\;\;g\left( 0 \right) = 0,$ and $f\left( 1 \right) = 6,$ then for some $c \in \left] {0,1} \right[$  . .

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:

In each of the intervals $(-1,0)$ and $(0,2)$ the function $(f-3 g)^{\prime \prime}$ never vanishes. Then the correct statement(s) is(are)

$(A)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly three solutions in $(-1,0) \cup(0,2)$

$(B)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(-1,0)$

$(C)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly one solution in $(0,2)$

$(D)$ $f^{\prime}(x)-3 g^{\prime}(x)=0$ has exactly two solutions in $(-1,0)$ and exactly two solutions in $(0,2)$

Examine if Rolle's Theorem is applicable to any of the following functions. Can you say some thing about the converse of Roller's Theorem from these examples?

$f(x)=x^{2}-1$ for $x \in[1,2]$

For which interval, the function ${{{x^2} – 3x} \over {x – 1}}$ satisfies all the conditions of Rolle's theorem