Let $$f(x) = x^5 + 2x^3 + 3x + 1, x \in \mathbb{R}$$, and $$g(x)$$ be a function such that $$g(f(x)) = x$$ for all $$x \in \mathbb{R}$$. Then $$\frac{g(7)}{g'(7)}$$ is equal to :

Given $$f(x) = x^5 + 2x^3 + 3x + 1$$ and $$g(f(x)) = x$$, so $$g = f^{-1}$$. First, $$f(1) = 1 + 2 + 3 + 1 = 7$$, which implies $$g(7) = 1$$.

By the inverse function theorem, $$g'(f(x)) = \frac{1}{f'(x)}$$. Since $$f'(x) = 5x^4 + 6x^2 + 3$$, at $$x = 1$$ one gets $$f'(1) = 5 + 6 + 3 = 14$$ and hence $$g'(7) = g'(f(1)) = \frac{1}{f'(1)} = \frac{1}{14}$$.

$$\frac{g(7)}{g'(7)} = \frac{1}{1/14} = 14$$

The correct answer is Option (1): 14.