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B
2eπ/2
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C
eπ/4
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D
2eπ/4
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Solution
The correct option is D2eπ/4 For positive numbers we have, AM≥GM esin−1x+ecos−1x2≥√e(sin−1x+cos−1x) esin−1x+ecos−1x≥2eπ/4[sin−1x+cos−1x=π2] Hence, option 'D' is correct.