GTA 2026 Part 7 | 2 Questions That Will Surprise You 🤯 #jeeadvanced

In this video, I solve 2 excellent questions from Narayana GTA (2026) based on student doubts. At first glance, both problems seem simple — but they require sharp thinking and the right perspective to solve. This session focuses on developing intuition, pattern recognition, and smart approaches that can make a huge difference in competitive exams. Try them yourself first… and see where you stand! 🔥 0:00 Question 1 Let \(f(x), f'(x), f''(x)\) are continuous on \([0,\ln 2]\) and \[ f(0)=0,\; f'(0)=3,\; f'(\ln 2)=4,\; f(\ln 2)=6 \] and \[ \int_{0}^{\ln 2} e^{-2x} f(x)\,dx = 3. \] Also \(g(x)\) be a continuous function satisfying \[ \int_{0}^{1} g(x)\bigl(4x^2 - g(x)\bigr)\,dx = \frac{4}{5}. \] Now answer the following: \begin{enumerate} \item The value of \(\displaystyle \int_{0}^{\ln 2} e^{-2x} f''(x)\,dx\) equals \blank. \item The value of \(m+n\), if the arithmetic mean of \[ g\!\left(\frac{1}{2}\right), g\!\left(\frac{1}{2^2}\right), \ldots, g\!\left(\frac{1}{2^{10}}\right) \quad \text{is} \quad \frac{1}{m}\left(1 - \frac{1}{2^n}\right), \quad (m,n \in \mathbb{N}), \] is equal to \blank. \end{enumerate} 20:00 Question 2 Let \(Z_1, Z_2, Z_3\) be complex numbers such that \[ Z_1^2 + Z_2^2 + Z_3^2 = Z_1Z_2 + Z_2Z_3 + Z_3Z_1 \] and \[ |Z_1 + Z_2 + Z_3| = 21. \] Given that \(|Z_1 - Z_2| = 2\sqrt{3}\), \(|Z_1| = 3\sqrt{3}\), then answer the following: \begin{enumerate} \item The value of \(|Z_2|^2 + |Z_3|^2\) is equal to \blank. \item The value of \( Z_1 \conj{Z_2}+ \conj{Z_2} Z_1+ Z_2 \conj{Z_3}+ \conj{Z_3} Z_2 + Z_1 \conj{Z_3}+ \conj{Z_3} Z_1\) is equal to \blank. \end{enumerate}