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

Given, f (0) = 2 = g (1), g (0) = 0   and  f (1) = 6
f and g are differentiable in (0,1).
Let    h(x) = f (x) – 2g(x)                                         …(i)
        h (0) = f (0) -2g (0) = 2 - 0 = 2 and  h (1) = f (1) -2g (1) = 6 – 2(2) = 2, h (0) = h (1) = 2
Hence, using Rolle’s theorem, h’(c) = 0, such that Differentiating  Eq.(i) at c, we get 

$\Rightarrow f\text{'}\left(c\right)-2g\text{'}\left(c\right)=0\Rightarrow f\text{'}\left(c\right)=2g\text{'}\left(c\right)$