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Question

Let n be a positive integer such that sin π2n + cos π2n = n2. Then

A
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B
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C
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Solution

The correct option is B

(b) sin π2n + cos π2n = n2
2 (sinπ2n.cosπ4+cosπ2n.sinπ4) = n2
2 sin(π4+π2n) = n2
Since sin(π4+π2n) 1
n2 2 2 22 n 8 .
Again n2 = 2 sin(π4+π2n) > 2.12 = 1
(sin(π4+π2n)sinπ4)
n > 4, Hence, 4 < n 8.


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