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

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-

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 $f (0)$

For the function$x + {1 \over x},x \in [1,\,3]$, the value of $ c$  for the mean value theorem is

Let $f(x) = (x-4)(x-5)(x-6)(x-7)$ then -

Let $f (1) = - 2$ and $f ' (x) \ge 4.2$ for $1 \le x \le 6$. The smallest possible value of $f (6)$, is