JEE MAIN - Mathematics Hindi (2023 - 13th April Morning Shift - No. 14)
$$\mathop {\max }\limits_{0 \le x \le \pi } \left\{ {x - 2\sin x\cos x + {1 \over 3}\sin 3x} \right\} = $$
$${{5\pi + 2 + 3\sqrt 3 } \over 6}$$
0
$${{\pi + 2 - 3\sqrt 3 } \over 6}$$
$$\pi$$
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