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Question

The least value of esin1x+ecos1x is

A
1+eπ/2
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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 D 2eπ/4
For positive numbers we have,
AMGM
esin1x+ecos1x2e(sin1x+cos1x)
esin1x+ecos1x2eπ/4 [sin1x+cos1x=π2]
Hence, option 'D' is correct.

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