Suppose f: ℝ →(0, ∞) be a differentiable function such that 5 f(x+y)=f(x) · f(y), ∀ x, y ∈ ℝ. If f(3)=320, then Σn=0⁵ f(n) is equal to :

Solution

Maths sir · 4 steps

Divide f by 5 and the messy equation becomes the clean exponential rule g(x + y) = g(x) g(y).

1. Find f(0). Put x = y = 0: 5f(0) = f(0)². Since f(0) > 0, we get f(0) = 5.
2. Remove the 5. Let g(x) = f(x)/5. Dividing the equation by 25 gives g(x+y) = g(x) g(y). A positive, differentiable g with this property must be g(x) = aˣ, so f(x) = 5aˣ.
3. Use f(3) = 320. 5a³ = 320, so a³ = 64 and a = 4. Hence f(n) = 5 × 4ⁿ.

   Σₙ₌₀⁵ f(n) = 5(1 + 4 + 16 + 64 + 256 + 1024) = 5 × 1365

4. So. The sum is 5 × 1365 = 6825, which is option (d).

The idea to keep. When you see k f(x+y) = f(x)f(y), divide f by k. The new function satisfies g(x+y) = g(x)g(y), so f(x) = k aˣ.

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