The maximum value of (1/x) x   is:

(A) e                            (B) ee

(C) e (1/e)                       (D) 1/e (1/e)



  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise


Question 30 The maximum value of (1/๐‘ฅ)^๐‘ฅ is: (A) e (B) ee (C) ๐‘’^(1/๐‘’) (D) ใ€–1/๐‘’ใ€—^(1/๐‘’) Let f (๐‘ฅ) = (1/๐‘ฅ)^๐‘ฅ To find maximum value, we need to differentiate f(x) For differentiating f (๐‘ฅ), we use logarithmic differentiation f (๐‘ฅ) = (1/๐‘ฅ)^๐‘ฅ log (f(x)) = ๐’™ log (๐Ÿ/๐’™) Differentiating wrt ๐‘ฅ ๐Ÿ/(๐’‡(๐’™)) fโ€™(x) = 1โˆ™log (๐Ÿ/๐’™) + ๐’™ ร— (๐Ÿ/(๐Ÿ/๐’™)) ร— ((โˆ’๐Ÿ)/๐’™^๐Ÿ ) 1/(๐‘“(๐‘ฅ)) fโ€™(x) = log (1/๐‘ฅ) + ๐‘ฅ ร— (๐‘ฅ) ร— ((โˆ’1)/๐‘ฅ^2 ) 1/(๐‘“(๐‘ฅ)) fโ€™(x) = log (1/๐‘ฅ) + ๐‘ฅ^2 ร— ((โˆ’1)/๐‘ฅ^2 ) 1/(๐‘“(๐‘ฅ)) fโ€™(x) = log (1/๐‘ฅ) โˆ’ 1 fโ€™(x) = f(x) [logโกใ€–(1/๐‘ฅ)โˆ’1ใ€— ] Putting f (๐‘ฅ) =(1/๐‘ฅ)^๐‘ฅ fโ€™(x) = (๐Ÿ/๐’™)^๐’™ (๐ฅ๐จ๐ โกใ€–(๐Ÿ/๐’™)โˆ’๐Ÿใ€— ) Putting fโ€™(x) = 0 (1/๐‘ฅ)^๐‘ฅ (logโกใ€–(1/๐‘ฅ)โˆ’1ใ€— ) = 0 Since, there is only one critical point, so it will be point of maxima Either (๐Ÿ/๐’™)^๐’™ = 0 Since, it is an exponential function It can never be zero. Or (๐’๐’๐’ˆโกใ€–(๐Ÿ/๐’™)โˆ’๐Ÿใ€— ) = 0 log 1/๐‘ฅ = 1 Taking exponential on both sides e^logโกใ€–1/xใ€— = ๐‘’^1 1/๐‘ฅ = e ๐’™ = ๐’†^(โˆ’๐Ÿ) Putting ๐‘ฅ = 1/๐‘’ in f (x) f (๐Ÿ/๐’†) = (1/(1/๐‘’))^(1/๐‘’) f (๐Ÿ/๐’†) = ๐’†^(๐Ÿ/๐’†) So, the correct answer is (C)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.