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Question

Let n be a positive integer such that sinπ2n+cosπ2n=n2. Then


A

6n8

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B

4 < n8

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C

4 n < 8

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D

4 < n < 8

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Solution

The correct option is B

4 < n8


sinπ2n+cosπ2n=n22(sinπ2n.cosπ4+cosπ2n.sinπ4)=n22 sin(π4+π2n)=n2

Since sin(π4+π2n)1n22n22n8.Again n2=2 sin(π4+π2n)>2.12=1(sin (π4+π2n)> sinπ4)

n > 4, Hence, 4 < n 8


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