Скачать или смотреть 📘 JEE Mains 2025 (4th April Shift 1) | Math Solution | Maths With Vishal

📘 JEE Main 2025 (4th April Shift 1) Math Solution
In this video, we solve a challenging Functional Equation and Limits problem from JEE Main 2025 (4 April Shift 1).
The question involves:
✅ A continuous function f:R→Rf: \mathbb{R} \to \mathbb{R}f:R→R with conditions f(0)=1f(0) = 1f(0)=1 and f(2x)−f(x)=xf(2x) - f(x) = xf(2x)−f(x)=x.
✅ Evaluating the limit:
lim⁡n→∞(f(x)−f ⁣(x2n))=G(x)\lim_{n \to \infty} \Big(f(x) - f\!\left(\tfrac{x}{2^n}\right)\Big) = G(x)n→∞lim(f(x)−f(2nx))=G(x)
✅ Finally computing the sum:
∑r=110G(r2)\sum_{r=1}^{10} G(r^2)r=1∑10G(r2)
This problem beautifully tests functional equations, limits, and series concepts, making it a must-watch for JEE aspirants.
✨ Perfect for JEE Main and Advanced preparation!
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