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Question

If alpha2sin1x+cos1xβ, then


A

α=π2β=π2

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B

α=π2β=3π2

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C

α=0β=π

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D

α=0β=2π

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Solution

The correct option is C

α=0β=π


αsin1x+π2βsin1xϵ[π2,π2]α=0,β=π


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