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Question: Answered & Verified by Expert
Let fix=sin2pix for i=1,2,3 & piN. It is given that the fundamental periods of f1x+f2x+f3x, f1x+f2x and f1x+f3x are π,π2 and π3 respectively, then the minimum value of p1+p2+p3 is
MathematicsFunctionsJEE Main
Solution:
2214 Upvotes Verified Answer
The correct answer is: 11
f1x=sin2p1x
f2x=sin2p2x
f3x=sin2p3x
Now, π=LCM2π2p1,2π2p2,2π2p3HCFp1,p2,p3=1
π2=LCM2π2p1,2π2p2HCFp1,p2=2p1=2ap2=2b
π3=LCM2π2p1,2π2p3HCFp1,p3=3p1=3xp3=3y
Hence, p1=6, p2=2, p3=3 (for minimum value)

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