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MathematicsSequence & SeriesMeans / inequalityMedium2 minPYQ_2020
MathematicsMediumnumerical

Letmbe the minimum possible value oflog33y1+3y2+3y3, wherey1, y2, y3are real numbers for whichy1+y2+y3=9. LetMbe the maximum possible value oflog3x1+log3 x2+log3 x3, wherex1, x2, x3are positive real numbers for whichx1+x2+x3=9. Then the value oflog2 m3+log3M2is

Answer:
8.00
Solution:

Applying A.M-G.M inequality,

3y1+3y2+3y333y13y23y313=3y1+y2+y313

3y1+3y2+3y381

So, m=log381=4

Now, log3x1+log3x2+log3x3=log3x1.x2. x3

x1+x2+x33x1.x2.x313x1.x2.x327 (applying A.M-G.M inequality)

So, M=log327=3

Thus, log2m3+log3M2=8

Stream:JEE_ADVSubject:MathematicsTopic:Sequence & SeriesSubtopic:Means / inequality
2mℹ️ Source: PYQ_2020

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