A Beautiful Functional Equation Question from JEE Mains 2025

A Beautiful Functional Equation Question from JEE Mains 2025 | Gems of JEE Mains In this video, I discuss a beautiful functional equation problem that looks technical but collapses elegantly with one clean idea (telescoping via f(2x) − f(x) = x). Question (JEE style): Let f: R → R be continuous with f(0) = 1 and f(2x) − f(x) = x for all real x. If G(x) = lim (n → ∞) { f(x) − f(x/2ⁿ) }, then find Σ (r = 1 to 10) G(r²). Idea: From f(2z) − f(z) = z, put z = x/2^{k+1} to get f(x/2^{k}) − f(x/2^{k+1}) = x/2^{k+1}. Telescoping gives f(x) − f(x/2ⁿ) = x(1 − 2^{-n}) ⇒ G(x) = x. Hence Σ G(r²) = Σ r² (r = 1..10) = 385. Perfect for building FE intuition: rewriting, telescoping, and using limits cleanly. 📌 Homework (similar style): Let f: R → R be continuous with f(0) = 0 and f(3x) − f(x) = 2x. Define H(x) = lim (n → ∞) { f(x) − f(x/3ⁿ) }. Compute Σ (k = 1 to 6) H(k). Telegram (PDF + solutions): https://t.me/academyfactorial #GemsOfJEEMains #JEEMains #FunctionalEquations #JEE2026 #JEEMaths2026 #factorialacademy