Step-by-step Solution:

1. Analyze the given functions and relations. We are given $g(x) = \log f(x)$, where f(x) is a twice differentiable positive function on $(0, \infty)$. The function f(x) satisfies the functional equation f(x + 1) = xf(x).

2. Establish a relationship for g(x) using the functional equation. Take the natural logarithm of both sides of the functional equation for f(x): $\log(f(x + 1)) = \log(xf(x))$ Using the properties of logarithms, we can write: $\log(f(x + 1)) = \log(x) + \log(f(x))$ From the definition $g(x) = \log f(x)$, we have $g(x + 1) = \log f(x+1)$. Substituting this into the equation above gives: $g(x + 1) = \log(x) + g(x)$

3. Differentiate the relation for g(x) to find a relation for its derivatives. Differentiate the equation $g(x + 1) = \log(x) + g(x)$ with respect to x: ${d \over dx} g(x+1) = {d \over dx} (\log(x) + g(x))$ Applying the chain rule, we get: $g'(x + 1) \cdot 1 = {1 \over x} + g'(x)$ $g'(x + 1) = {1 \over x} + g'(x)$ Now, differentiate this equation again with respect to x: ${d \over dx} g'(x+1) = {d \over dx} \left({1 \over x} + g'(x)\right)$ $g''(x + 1) \cdot 1 = -{1 \over x^2} + g''(x)$ This gives us a recurrence relation for g''(x): $g''(x + 1) - g''(x) = -{1 \over x^2}$

4. Evaluate the expression g''(N + 1/2) - g''(1/2) using the recurrence relation. The expression can be written as a telescoping sum: $g''\left(N + {1 \over 2}\right) - g''\left({1 \over 2}\right) = \sum_{k=1}^{N} \left[ g''\left(k + {1 \over 2}\right) - g''\left(k - {1 \over 2}\right) \right]$ Let's verify this sum: For k=1: g''(3/2) - g''(1/2) For k=2: g''(5/2) - g''(3/2) ... For k=N: g''(N+1/2) - g''(N-1/2) Summing these terms, all intermediate terms cancel out, leaving g''(N+1/2) - g''(1/2).

5. Apply the recurrence relation to the terms in the sum. Our recurrence relation is $g''(x + 1) - g''(x) = -1/x^2$. For each term in the sum, let x = k - 1/2. Then x + 1 = k - 1/2 + 1 = k + 1/2. Substituting x = k - 1/2 into the recurrence relation gives: $g''\left(k + {1 \over 2}\right) - g''\left(k - {1 \over 2}\right) = -{1 \over (k - 1/2)^2} = -{1 \over (\frac{2k-1}{2})^2} = -{4 \over (2k-1)^2}$

6. Calculate the final sum. Now we can substitute this back into our telescoping sum expression: $g''\left(N + {1 \over 2}\right) - g''\left({1 \over 2}\right) = \sum_{k=1}^{N} \left( -{4 \over (2k-1)^2} \right)$ $g''\left(N + {1 \over 2}\right) - g''\left({1 \over 2}\right) = -4 \sum_{k=1}^{N} {1 \over (2k-1)^2}$ Expanding the sum: $-4 \left[ {1 \over (2(1)-1)^2} + {1 \over (2(2)-1)^2} + {1 \over (2(3)-1)^2} + \dots + {1 \over (2N-1)^2} \right]$ $-4 \left[ {1 \over 1^2} + {1 \over 3^2} + {1 \over 5^2} + \dots + {1 \over (2N-1)^2} \right]$ $-4 \left\{ 1 + {1 \over 9} + {1 \over 25} + \dots + {1 \over (2N-1)^2} \right\}$

7. Compare with the given options. The result matches option A.

Final Answer is option A.