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Jee Main 2025 — Medium JEE math MCQ

Let $A = \left\{ x \in (0, \pi) - \left\{ \frac{\pi}{2} \right\} : \log_{2/\pi} |\sin x| + \log_{2/\pi} |\cos x| = 2 \right\}$ and $B = \{ x \geq 0 : \sqrt{\sqrt{x} - 4} - 3\sqrt{\sqrt{x} - 2} + 6 = 0 \}$. Then $n(A \cup B)$ is equal to
  1. A. 4
  2. B. 8
  3. C. 6
  4. D. 2

Solution

The correct option is **B**. (B. 8)

MATH

medium • PYQ Reworded
Question
Read carefully, then pick the best option.
Let A={x∈(0,π)−{π2}:log⁡2/π∣sin⁡x∣+log⁡2/π∣cos⁡x∣=2}A = \left\{ x \in (0, \pi) - \left\{ \frac{\pi}{2} \right\} : \log_{2/\pi} |\sin x| + \log_{2/\pi} |\cos x| = 2 \right\} and B={x≥0:x−4−3x−2+6=0}B = \{ x \geq 0 : \sqrt{\sqrt{x} - 4} - 3\sqrt{\sqrt{x} - 2} + 6 = 0 \}. Then n(A∪B)n(A \cup B) is equal to
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Jee Main 2025 — Medium JEE Mathematics MCQ | MyGoalPrep