Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12




Last updated at Jan. 7, 2020 by Teachoo
Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12
Transcript
Misc 2 Show that the function given by f(x) = logโก๐ฅ/๐ฅ has maximum at x = e. Let f(๐ฅ) = logโก๐ฅ/๐ฅ Finding fโ(๐) fโ(๐ฅ) = ๐/๐๐ฅ (๐๐๐โก๐ฅ/๐ฅ) fโ(๐ฅ) = (๐(๐๐๐โก๐ฅ )/๐๐ฅ " " . ๐ฅ โ๐(๐ฅ)/๐๐ฅ " . " ๐๐๐โก๐ฅ)/๐ฅ2 fโ(๐ฅ) = (1/๐ฅ ร ๐ฅ โ ๐๐๐โก๐ฅ)/๐ฅ2 fโ(๐ฅ) = (1 โ ๐๐๐โก๐ฅ)/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 ๐ฅ = 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 ๐ฅ=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[1 + 2 โ 2 logโก๐ฅ ])/๐ฅ^4 = (โ๐ฅ[3 โ 2 logโก๐ฅ ])/๐ฅ4 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 (๐๐ logโกใ๐=1ใ ) (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.
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