# If $f$ and $g$ are differentiable function in $[0, 1]$ satisfying $f(0)=2 = g(1),g(0)= 0$ and $f(1)=6$, then for some $c\in [0,1]$

$\begin{array}{1 1}(A)\;2f'(c)=g'(c)\\(B)\;2f'(c)=3g'(c)\\(C)\;f'(c)=g'(c)\\(D)\;f'(c)=2g'(c)\end{array}$

By MTV : $f'(c) = \frac{f(b) - f(a)}{b-a}$
$f'(c) = \frac{f(1)-f(0)}{1-0} = \frac{6-2}{1} = 4$
$g'(c) = \frac{g(1)- g(0)}{1-0} = \frac{2-0}{1} = 2$
$f'(c) = 2.g'(c)$
edited Nov 7, 2017