If $f(x)$ satisfies the conditions of Rolle’s theorem in $[1,\,2]$ and $f(x)$ is continuous in $[1,\,2]$ then $\int_1^2 {f'(x)dx} $ is equal to

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

In the mean value theorem, $f(b) - f(a) = (b - a)f'(c)$if $a = 4$, $b = 9$ and $f(x) = \sqrt x $ then the value of  $c$  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 $f (0)$

Consider a quadratic equation $ax^2 + bx + c = 0,$ where $2a + 3b + 6c = 0$ and let $g(x) = a\frac{{{x^3}}}{3} + b\frac{{{x^2}}}{2} + cx.$

Statement $1:$ The quadratic equation has at least one root in the interval $(0, 1).$

Statement $2:$ The Rolle's theorem is applicable to function $g(x)$ on the interval $[0, 1 ].$

If $f(x) = \cos x,0 \le x \le {\pi \over 2}$, then the real number $ ‘c’ $ of the mean value theorem is