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Jee Main 2025 — Hard JEE math MCQ

Let \( y = y(x) \) be the solution of the differential equation \[ \cos x \left( \log_e (\cos x) \right)^2 dy + (\sin x - 3y \sin x \log_e (\cos x)) dx = 0, \quad x \in \left( 0, \frac{\pi}{2} \right).\] If \( y \left( \frac{\pi}{6} \right) = \frac{-1}{\log_2 2} \), then \( y \left( \frac{\pi}{8} \right) \) is equal to:
  1. A. \frac{1}{\log_2 (5) - \log_2 (4)} \\
  2. B. \frac{2}{\log_2
  3. C. - \log_2
  4. D. } \\ (3) \frac{1}{\log_2 (4) - \log_2 (3)} \\ (4) \frac{1}{\log_2 (4)}

Solution

The correct option is **A**. (A. \frac{1}{\log_2 (5) - \log_2 (4)} \\)

MATH

hard • PYQ Reworded
Question
Read carefully, then pick the best option.
Let y=y(x) y = y(x) be the solution of the differential equation cos⁡x(log⁡e(cos⁡x))2dy+(sin⁡x−3ysin⁡xlog⁡e(cos⁡x))dx=0,x∈(0,π2). \cos x \left( \log_e (\cos x) \right)^2 dy + (\sin x - 3y \sin x \log_e (\cos x)) dx = 0, \quad x \in \left( 0, \frac{\pi}{2} \right). If y(π6)=−1log⁡22 y \left( \frac{\pi}{6} \right) = \frac{-1}{\log_2 2} , then y(π8) y \left( \frac{\pi}{8} \right) is equal to:
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Jee Main 2025 — Hard JEE Mathematics MCQ | MyGoalPrep