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

If $f$ is a differentiable function such that $f(2x + 1) = f(1 -2x)$ $\forall \,\,x \in R$ then minimum number of roots of the equation $f'(x) = 0$ in $x \in \left( { - 5,10} \right)$ ,given that $f(2) = f(5) = f(10)$ , is

Let  $f(x)$  satisfy the requirement of lagranges mean value theorem in $[0,2]$ . If $f(x)=0$ ; $\left| {f'\left( x \right)} \right| \leqslant \frac{1}{2}$ for all $x \in \left[ {0,2} \right]$, then-

If from mean value theorem, $f'({x_1}) = {{f(b) - f(a)} \over {b - a}}$, then

In  $[0, 1]$ Lagrange's mean value theorem is $ NOT$  applicable to

Let $y = f (x)$ and $y = g (x)$ be two differentiable function in $[0,2]$ such that  $f(0) = 3,$ $f(2) = 5$ , $g (0) = 1$ and $g(2) = 2$. If there exist atlellst one $c \in \left( {0,2} \right)$ such that $f'(c)=kg'(c)$,then $k$ must be