Let f(x) be a real differentiable function such that f(0)=1 and f(x+y)=f(x) f′(y)+f′(x) f(y) for all x, y ∈ R. Then Σn=1¹⁰⁰ log ₑ f(n) is equal to :

Solution

Maths sir · 4 steps

Put y = 0 and the functional equation turns into a simple differential equation for f.

1. Find f'(0). Put x = y = 0: f(0) = 2f(0)f'(0). Since f(0) = 1, this gives f'(0) = 1/2.
2. Put y = 0. Now f(x) = f(x)f'(0) + f'(x)f(0) = 1/2f(x) + f'(x). So f'(x) = 1/2f(x) for every x.

   f'(x)/f(x) = 1/2 ⇒ f(x) = e^(x/2) (using f(0) = 1)

3. Check it works. With f(x) = e^(x/2), the right side is e^(x/2)·1/2e^(y/2) + 1/2e^(x/2)· e^(y/2) = e^((x+y)/2) = f(x+y). The equation holds.
4. So. logₑ f(n) = n/2, so the sum is 1/2(1 + 2 + … + 100) = 1/2× 5050 = 2525, option (c).

The idea to keep. In a functional equation with derivatives, substitute 0 first. It usually gives f'(0) and then a first-order differential equation you can solve directly.

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