Let $$f : R \rightarrow R$$ be a function such that $$f(2-x) = f(2+x)$$ and $$f(4-x) = f(4+x)$$, for all $$x \in R$$ and $$\int_0^2 f(x)dx = 5$$. Then the value of $$\int_{10}^{50} f(x)dx$$ is

The function $$f: \mathbb{R} \rightarrow \mathbb{R}$$ satisfies two symmetry conditions: $$f(2 - x) = f(2 + x)$$ and $$f(4 - x) = f(4 + x)$$ for all $$x \in \mathbb{R}$$. Additionally, it is given that $$\int_0^2 f(x) dx = 5$$. The goal is to find $$\int_{10}^{50} f(x) dx$$.

First, analyze the symmetry conditions. The condition $$f(2 - x) = f(2 + x)$$ indicates that the function is symmetric about the line $$x = 2$$. Similarly, $$f(4 - x) = f(4 + x)$$ indicates symmetry about the line $$x = 4$$. When a function has two different lines of symmetry, it is periodic. The distance between the lines $$x = 2$$ and $$x = 4$$ is $$4 - 2 = 2$$. The period $$T$$ of the function is twice this distance, so $$T = 2 \times |4 - 2| = 4$$. Therefore, $$f(x + 4) = f(x)$$ for all $$x \in \mathbb{R}$$.

To confirm periodicity, use the symmetries. From symmetry about $$x = 2$$, $$f(x) = f(4 - x)$$. From symmetry about $$x = 4$$, $$f(4 - x) = f(4 + x)$$. Now, consider $$f(x + 4)$$: $$f(x + 4) = f(4 + x) = f(4 - x) \quad \text{(by symmetry about } x = 4\text{)}.$$ But $$f(4 - x) = f(x)$$ (from symmetry about $$x = 2$$), so: $$f(x + 4) = f(x).$$ Thus, $$f$$ is periodic with period 4.

Next, evaluate the integral over one period. Given $$\int_0^2 f(x) dx = 5$$, compute $$\int_2^4 f(x) dx$$. Use the symmetry about $$x = 2$$. Substitute $$u = 4 - x$$ in the integral from 2 to 4: $$\int_2^4 f(x) dx = \int_2^0 f(4 - u) (-du) = \int_0^2 f(4 - u) du.$$ Since $$f(4 - u) = f(u)$$ (as established earlier), this becomes: $$\int_0^2 f(u) du = \int_0^2 f(x) dx = 5.$$ Therefore, $$\int_2^4 f(x) dx = 5$$. The integral over one full period from 0 to 4 is: $$\int_0^4 f(x) dx = \int_0^2 f(x) dx + \int_2^4 f(x) dx = 5 + 5 = 10.$$ Due to periodicity, the integral over any interval of length 4 is 10.

Now, compute $$\int_{10}^{50} f(x) dx$$. The length of the interval from 10 to 50 is $$50 - 10 = 40$$. Since the period is 4, the number of full periods in this interval is $$40 / 4 = 10$$. Therefore: $$\int_{10}^{50} f(x) dx = 10 \times \int_0^4 f(x) dx = 10 \times 10 = 100.$$ This can also be seen by dividing the interval into 10 subintervals of length 4: $$[10, 14]$$, $$[14, 18]$$, $$\ldots$$, $$[46, 50]$$. Each subinterval has the same integral value of 10, so the total is $$10 \times 10 = 100$$.

Hence, the correct answer is Option A.