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Question

The value of limxπ422(cosx+sinx)31sin2x is

A
32
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B
23
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C
12
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D
2
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Solution

The correct option is A 32
limxπ422(cosx+sinx)31sin2x (00orm)
=limxπ43(cosx+sinx)2(sinx+cosx)2cos2x
=limxπ43(cosx+sinx)(cos2xsin2x)2cos2x
=limxπ43(cosx+sinx)cos2x2cos2x
=limxπ43(cosx+sinx)2=32.(12+12)=32

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