JEE Advanced 2025 | Limit PYQ | Maths PYQ | Full Concepts Explain скачать в хорошем качестве

JEE Advanced 2025 | Limit PYQ | Maths PYQ | Full Concepts Explain

Let x0 be the real number such that e^(x_0 )+x_0=0.For a given real number α,define g(x)=(3xe^x+3x-αe^x-αx)/(3(e^x+1)) for all real numbers x. Then which one of the following statements is TRUE? (A)For α=2, lim┬(x→x_0 )⁡|(g(x)+e^(x_0 ))/(x-x_0 )| = 0 (B)For α=2, lim┬(x→x_0 )⁡|(g(x)+e^(x_0 ))/(x-x_0 )| = 1 (C)For α=3, lim┬(x→x_0 )⁡|(g(x)+e^(x_0 ))/(x-x_0 )| = 0 (D)For α=3, lim┬(x→x_0 )⁡|(g(x)+e^(x_0 ))/(x-x_0 )| = 2/3 JEE Advanced 2025 One of the most important and concept-based limit questions expected in JEE Advanced 2025. In this video, we solve a high-level limit problem involving 👉 exponential functions 👉 root-based limit 👉 advanced substitution trick 👉 concept of special point x₀ ⸻ 🧠 Question: Let x_0 be the real number such that e^{x_0} + x_0 = 0. For a real number α, define g(x)=\frac{3xe^x + 3x − αe^x − αx}{3(e^x+1)} Then find the correct statement \lim_{x→x_0}\left|\frac{g(x)+e^{x_0}}{x-x_0}\right| (A) For α=2 → value 0 (B) For α=2 → value 1 (C) For α=3 → value 0 (D) For α=3 → value 2/3 ⸻ 🎯 Why this question is important • Expected in JEE Advanced 2025 • Strong concept of limits + expansion • Repeating pattern in IIT papers • Rank booster question ⸻ 📚 Topics covered • Limit using special root • Exponential approximation • JEE Advanced level algebra • PYQ pattern analysis ⸻ 🔥 For JEE 2025/2026 aspirants If you are preparing for: • IIT JEE Advanced • JEE Main • Olympiad level maths This question is a must-do. ⸻ 👍 Support If this helped you: LIKE 👍 SUBSCRIBE 📌 Share with JEE friends More JEE Advanced PYQ coming daily.