Check sibling questions

The maximum value of (1/x) x   is:

(A) e                            (B) ee

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

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Transcript

Question 17 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.