Let f : S → S where S = (0, ∞) be a twice differentiable function such that f(x + 1) = xf(x). If g : S → R be defined as g(x) = logₑ f(x), then the value of |g''(5) - g''(1)| is equal to :

Solution

Maths sir · 4 steps

Take log first: the product xf(x) becomes a sum, and differentiating twice leaves only -1/x².

1. Take the log. Since f maps into (0, ∞), ln f is defined. Taking logₑ of f(x+1) = x f(x) gives g(x+1) = ln x + g(x) for x > 0.
2. Differentiate twice. First derivative: g'(x+1) = 1/x + g'(x). Second derivative: g''(x+1) = -1/x² + g''(x), so g''(x+1) - g''(x) = -1/x².
3. Telescope. Put x = 1, 2, 3, 4 and add the four equations. The middle terms g''(2), g''(3), g''(4) cancel, leaving g''(5) - g''(1) on the left.

   g''(5) - g''(1) = -(1 + 1/4 + 1/9 + 1/16) = -(144 + 36 + 16 + 9)/144 = -205/144

4. So. |g''(5) - g''(1)| = 205/144, which is option (d).

The idea to keep. A functional equation like f(x+1) = xf(x) turns into a sum once you take log. Differentiate that, then add the steps from 1 to n-1 so the middle terms cancel.

Answer from the NTA final answer key. Solution by PrepWiser.