let f(θ)=sin2θcos2θ1+4sin4θsin2θ1+cos2θ4sin4θ1+sin2θcos2θ4sin4θ then f is
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a
a non periodic function
b
periodic with period π
c
periodic with period π2
d
odd function
answer is C.
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Detailed Solution
f(θ)=−101−1101+sin2θcos2θ4sin2θ (R1→R1-R3 R2 →R2-R3)=001−1101+sin2θcos2θ4sin4θ+4sin4θ (C1 →C1 + C3)=−cos2θ+1+sin2θ+4sin4θ⇒-2(1+2sin 4 θ)which is periodic function with period 2π4=π2.