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Question

limxπ4sec2x2f(t)dtx2π216equals

A
8πf(2)
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B
2πf(2)
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C
2πf(12)
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D
4f (2)
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Solution

The correct option is A 8πf(2)
Required limits is of the form 00, so it is equal to
limxπ42 secx secx tan xf(sec2x)2x=limxπ4sec2x tanx f(sec2x)x=8πf(2)

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