Last updated at April 19, 2021 by Teachoo

Transcript

Misc 2 Show that the function given by f(x) = logโก๐ฅ/๐ฅ is maximum at x = e.Let f(๐ฅ) = logโก๐ฅ/๐ฅ Finding fโ(๐) fโ(๐ฅ) = ๐/๐๐ฅ (logโก๐ฅ/๐ฅ) fโ(๐ฅ) = (๐(logโก๐ฅ )/๐๐ฅ " " . ๐ฅ โ ๐(๐ฅ)/๐๐ฅ " . log " ๐ฅ)/๐ฅ2 fโ(๐ฅ) = (1/๐ฅ ร ๐ฅ โ logโก๐ฅ)/๐ฅ2 fโ(๐ฅ) = (1 โ logโก๐ฅ)/๐ฅ2 Using quotient rule as (๐ข/๐ฃ)^โฒ = (๐ข^โฒ ๐ฃโ๐ฃ^โฒ ๐ข)/๐ฃ^2 Putting fโ(๐) = 0 (1 โ logโก๐ฅ)/๐ฅ2=0 1 โ log ๐ฅ = 0 log ๐ฅ = 1 ๐ = e Finding fโโ(๐) fโ(๐ฅ) = (1 โ logโก๐ฅ)/๐ฅ2 Diff w.r.t. ๐ฅ fโโ(๐ฅ) = ๐/๐๐ฅ ((1 โ logโก๐ฅ)/๐ฅ2) log ๐ฅ=1 is only possible when ๐=๐ Using quotient rule as (๐ข/๐ฃ)^โฒ=(๐ข^โฒ ๐ฃ โ ๐ฃ^โฒ ๐ข)/(๐ฃ2 .) fโโ(๐ฅ) = (๐(1 โ logโก๐ฅ )/๐๐ฅ . ๐ฅ2โ ๐(๐ฅ2)/๐๐ฅ . (1 โ logโก๐ฅ ))/(๐ฅ^2 )^2 = ((0 โ 1/๐ฅ) . ๐ฅ2 โ 2๐ฅ(1 โ logโก๐ฅ ))/๐ฅ4 = ((โ1)/๐ฅ ร ๐ฅ2 โ 2๐ฅ(1 โ logโก๐ฅ ))/๐ฅ^4 = (โ๐ฅ โ 2๐ฅ(1 โ logโก๐ฅ ))/๐ฅ^4 = (โ๐ฅ[1 + 2(1 โ logโก๐ฅ )])/๐ฅ4 = (โ๐ฅ[3 โ 2 logโก๐ฅ ])/๐ฅ4 โด fโโ(๐ฅ) = (โ(3 โ 2 logโก๐ฅ ))/๐ฅ3 Putting ๐ = e fโโ(๐) = (โ(3 โ 2 logโก๐ ))/๐3 = (โ(3 โ 2))/๐3 = (โ1)/๐3 = โ(1/๐3) < 0 (โต1/๐๐ฅ>0 ๐๐๐ ๐๐๐ ๐ฃ๐๐๐ข๐ ๐๐ ๐ฅ โ๐น) Since fโโ(๐ฅ) < 0 at ๐ฅ = e . โด ๐ฅ = e is point of maxima Hence, f(๐ฅ) is maximum at ๐ = e.

Miscellaneous

Misc 1
Important
Deleted for CBSE Board 2022 Exams

Misc 2 Important You are here

Misc 3 Important

Misc 4

Misc 5 Important

Misc 6 Important

Misc 7

Misc 8 Important

Misc 9 Important

Misc 10

Misc 11 Important

Misc 12 Important

Misc 13 Important

Misc 14 Important

Misc 15 Important

Misc 16 Important

Misc 17 Important

Misc 18 Important

Misc. 19 Deleted for CBSE Board 2022 Exams

Misc 20 Important

Misc 21 Important

Misc 22

Misc. 23 Important

Misc 24 Important

Chapter 6 Class 12 Application of Derivatives (Term 1)

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About the Author

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.