If $$f$$ & $$g$$ are differentiable functions in $$[0, 1]$$ satisfying $$f(0) = 2 = g(1)$$, $$g(0) = 0$$ & $$f(1) = 6$$, then for some $$c \in ]0, 1[$$:

We have two real-valued functions $$f$$ and $$g$$ which are differentiable on the open interval $$(0,1)$$ and continuous on the closed interval $$[0,1]$$. The given boundary values are

$$f(0)=2,\qquad f(1)=6, \qquad g(0)=0,\qquad g(1)=2.$$

Because both functions are continuous on $$[0,1]$$ and differentiable on $$(0,1)$$, we are allowed to apply the Cauchy Mean Value Theorem. First, let us recall the statement of that theorem.

Cauchy Mean Value Theorem. If functions $$p$$ and $$q$$ are continuous on $$[a,b]$$ and differentiable on $$(a,b)$$, then there exists at least one point $$c\in(a,b)$$ such that

$$\frac{p'(c)}{q'(c)}=\frac{p(b)-p(a)}{q(b)-q(a)}.$$

We now take $$p(x)=f(x)$$ and $$q(x)=g(x)$$, with $$a=0$$ and $$b=1$$. All the hypotheses are satisfied, so there exists some $$c\in(0,1)$$ for which

$$\frac{f'(c)}{g'(c)}=\frac{f(1)-f(0)}{g(1)-g(0)}.$$

Next, we substitute the known endpoint values.

$$f(1)-f(0)=6-2=4,$$

$$g(1)-g(0)=2-0=2.$$

Hence the right-hand side of the Cauchy Mean Value Theorem becomes

$$\frac{f(1)-f(0)}{g(1)-g(0)}=\frac{4}{2}=2.$$

So we have obtained

$$\frac{f'(c)}{g'(c)}=2.$$

Multiplying both sides of this equation by $$g'(c)$$ (which is permissible because differentiability implies continuity and we may assume $$g'(c)\neq 0$$ at the chosen point), we get

$$f'(c)=2\,g'(c).$$

This matches exactly the relation given in option B.

Hence, the correct answer is Option B.