Let $$f$$ and $$g$$ be two differentiable functions on R such that $$f'(x) > 0$$ and $$g'(x) < 0$$ for all $$x \in R$$. Then for all x:

We are given that $$f$$ and $$g$$ are differentiable functions on $$\mathbb{R}$$ with $$f'(x) > 0$$ and $$g'(x) < 0$$ for all $$x \in \mathbb{R}$$. This means $$f$$ is strictly increasing and $$g$$ is strictly decreasing. A strictly increasing function preserves inequalities: if $$a > b$$, then $$f(a) > f(b)$$. A strictly decreasing function reverses inequalities: if $$a > b$$, then $$g(a) < g(b)$$.

Now, let's evaluate each option step by step.

Starting with Option A: $$f(g(x)) > f(g(x-1))$$.

Compare $$x$$ and $$x-1$$. Since $$x > x-1$$ (as $$x - (x-1) = 1 > 0$$), and $$g$$ is strictly decreasing, we have $$g(x) < g(x-1)$$. Now, because $$f$$ is strictly increasing, $$g(x) < g(x-1)$$ implies $$f(g(x)) < f(g(x-1))$$. This contradicts the option's claim $$f(g(x)) > f(g(x-1))$$. Thus, Option A is false.

Next, Option B: $$f(g(x)) > f(g(x+1))$$.

Compare $$x$$ and $$x+1$$. Since $$x < x+1$$ (as $$x+1 - x = 1 > 0$$), and $$g$$ is strictly decreasing, we have $$g(x) > g(x+1)$$. Now, because $$f$$ is strictly increasing, $$g(x) > g(x+1)$$ implies $$f(g(x)) > f(g(x+1))$$. This matches the option's statement. Thus, Option B appears true, but we'll verify the others.

Option C: $$g(f(x)) > g(f(x-1))$$.

Compare $$x$$ and $$x-1$$. Since $$x > x-1$$, and $$f$$ is strictly increasing, we have $$f(x) > f(x-1)$$. Now, because $$g$$ is strictly decreasing, $$f(x) > f(x-1)$$ implies $$g(f(x)) < g(f(x-1))$$. This contradicts the option's claim $$g(f(x)) > g(f(x-1))$$. Thus, Option C is false.

Option D: $$g(f(x)) < g(f(x+1))$$.

Compare $$x$$ and $$x+1$$. Since $$x < x+1$$, and $$f$$ is strictly increasing, we have $$f(x) < f(x+1)$$. Now, because $$g$$ is strictly decreasing, $$f(x) < f(x+1)$$ implies $$g(f(x)) > g(f(x+1))$$. This contradicts the option's claim $$g(f(x)) < g(f(x+1))$$. Thus, Option D is false.

Therefore, only Option B is correct. Hence, the correct answer is Option B.