JEE Main 2025January, Unspecified (session-level dataset)MathematicsMediumTrigonometry118× since 2002Let A={x∈(0,π)−{π2}:log(2/π)∣sinx∣+log(2/π)∣cosx∣=2}\mathbf{A} = \{x \in (0, \pi) - \{\frac{\pi}{2}\} : \log_{(2 / \pi)} |\sin x| + \log_{(2 / \pi)} |\cos x| = 2\}A={x∈(0,π)−{2π}:log(2/π)∣sinx∣+log(2/π)∣cosx∣=2} and B={x⩾0:x(x−4)−3∣x−2∣+6=0}\mathbf{B} = \{x \geqslant 0 : \sqrt{x} (\sqrt{x} - 4) - 3 |\sqrt{x} - 2| + 6 = 0\}B={x⩾0:x(x−4)−3∣x−2∣+6=0}. Then n(A∪B)n(\mathbf{A} \cup \mathbf{B})n(A∪B) is equal to:AOption A. 4BOption B. 8COption C. 6DOption D. 2CheckSkip · show solution
