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Question

Let n be a positive integer such that sinπ2n+cosπ2n=n2 , then

A
5n<8
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B
5<n<9
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C
4n<8
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D
4<n8
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Solution

The correct option is D 4<n8

sin(π2n)+cos(π2n)=n2, n0


Squaring on both sides,

sin2(π2n)+cos2(π2n)+sin(2π2n)=n4


1+sin(π2n1)=n4


sin(π2n1)=n44
0<n441


0<n44


4<n8


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