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JEE Advanced Mathematics Sequences and Series 2020 JEE Advanced 2020 (Paper 1)

JEE Advanced Mathematics Question (2020) — Solution

Question

Let m be the minimum possible value of log 3 3 y 1 + 3 y 2 + 3 y 3 , where y 1 ,   y 2 ,   y 3 are real numbers for which y 1 + y 2 + y 3 = 9 . Let M be the maximum possible value of log 3 x 1 + log 3   x 2 + log 3   x 3 , where x 1 ,   x 2 ,   x 3 are positive real numbers for which x 1 + x 2 + x 3 = 9 . Then the value of log 2   m 3 + log 3 M 2 is

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

Applying A.M-G.M inequality, 3 y 1 + 3 y 2 + 3 y 3 3 ≥ 3 y 1 ⋅ 3 y 2 ⋅ 3 y 3 1 3 = 3 y 1 + y 2 + y 3 1 3 ⇒ 3 y 1 + 3 y 2 + 3 y 3 ≥ 81 So,  m = log 3 81 = 4 Now,  log 3 x 1 + log 3 x 2 + log 3 x 3 = log 3 x 1 . x 2 .   x 3 ∵ x 1 + x 2 + x 3 3 ≥ x 1 . x 2 . x 3 1 3 ⇒ x 1 . x 2 . x 3 ≤ 27  (applying A.M-G.M inequality) So,  M = log 3 27 = 3 Thus, log 2 m 3 + log 3 M 2 = 8

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