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B
3π2
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C
π2
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D
π
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Solution
The correct option is Cπ2 Let f(x)=sin4x+cos4x =(sin2x+cos2x)2−2sin2xcos2x=1−4sin2xcos2x2=1−sin22x2=1−14(2sin22x)=1−(1−cos4x4)=34+14cos4x ∵ fundamental period of cosx=2π
Hence, fundamental period of function =2π4=π2