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Question

Let n be a positive integer such that
sin(π2n)+cos(π2n)=n2

A
<4
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B
n>8
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C
4n<8
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D
6n8
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Solution

The correct option is C 4n<8

(sin(π2n)+cos(π2n))2=n2

1+2sinπ2ncosπ2n=n4......[cos2θ+sin2θ=1]

1+sinπ2n1=n4


sin(π2n1)=n44
AS n=+ve.1 and sinθ1
0<n441
4<n8


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