Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 28

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Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 29

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Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 30 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 31 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 32 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 33 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 34 Ex 6.5,3 - Chapter 6 Class 12 Application of Derivatives - Part 35

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

Transcript

Ex 6.5, 3 Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be: (viii) f(๐‘ฅ) = ๐‘ฅโˆš(1โˆ’๐‘ฅ), ๐‘ฅ > 0Finding fโ€™(๐’™) fโ€™(๐‘ฅ)=๐‘‘(๐‘ฅโˆš(1 โˆ’ ๐‘ฅ))/๐‘‘๐‘ฅ fโ€™(๐‘ฅ)=๐‘‘(๐‘ฅ)/๐‘‘๐‘ฅ . โˆš(1โˆ’๐‘ฅ) + ๐‘‘(โˆš(1 โˆ’ ๐‘ฅ))/๐‘‘๐‘ฅ . ๐‘ฅ = 1 . โˆš(1โˆ’๐‘ฅ) + 1/(2โˆš(1 โˆ’ ๐‘ฅ)) . ๐‘‘(1 โˆ’ ๐‘ฅ)/๐‘‘๐‘ฅ . ๐‘ฅ = โˆš(1โˆ’๐‘ฅ) + 1/(2โˆš(1 โˆ’ ๐‘ฅ)) (0 โˆ’1) . ๐‘ฅ = โˆš(1โˆ’๐‘ฅ) โ€“ ๐‘ฅ/(2โˆš(1 โˆ’ ๐‘ฅ)) = (2(โˆš(1 โˆ’ ๐‘ฅ) )^2โˆ’ ๐‘ฅ)/(2โˆš(1 โˆ’ ๐‘ฅ)) = (2(1 โˆ’ ๐‘ฅ) โˆ’ ๐‘ฅ)/(2โˆš(1 โˆ’ ๐‘ฅ)) = (2 โˆ’ 2๐‘ฅ โˆ’ ๐‘ฅ)/(2โˆš(1 โˆ’ ๐‘ฅ)) = (2 โˆ’ 3๐‘ฅ)/(2โˆš(1 โˆ’ ๐‘ฅ)) Putting fโ€™(๐’™)=๐ŸŽ (2 โˆ’ 3๐‘ฅ)/(2โˆš(1 โˆ’ ๐‘ฅ))=0 2 โ€“ 3๐‘ฅ = 0 ร— 2โˆš(1โˆ’๐‘ฅ) 2 โ€“ 3๐‘ฅ=0 โ€“ 3๐‘ฅ=โˆ’2 ๐‘ฅ =2/3 Finding fโ€™โ€™(๐’™) fโ€™(๐‘ฅ)=(2 โˆ’ 3๐‘ฅ)/(2โˆš(1 โˆ’ ๐‘ฅ)) fโ€™โ€™(๐‘ฅ)=๐‘‘/๐‘‘๐‘ฅ ((2 โˆ’ 3๐‘ฅ)/(2โˆš(1 โˆ’ ๐‘ฅ))) = 1/2 [(๐‘‘(2 โˆ’ 3๐‘ฅ)/๐‘‘๐‘ฅ . โˆš(1 โˆ’ ๐‘ฅ) โˆ’ ๐‘‘(โˆš(1 โˆ’ ๐‘ฅ))/๐‘‘๐‘ฅ . (2 โˆ’ 3๐‘ฅ))/(โˆš(1 โˆ’ ๐‘ฅ))^2 ] = 1/2 [((0 โˆ’ 3) โˆš(1 โˆ’ ๐‘ฅ) โˆ’ 1/(2โˆš(1 โˆ’ ๐‘ฅ)) . ๐‘‘(1 โˆ’ ๐‘ฅ)/๐‘‘๐‘ฅ . (2 โˆ’ 3๐‘ฅ))/((1 โˆ’ ๐‘ฅ) )] Using quotient rule as (๐‘ข/๐‘ฃ)^โ€ฒ=(๐‘ข^โ€ฒ ๐‘ฃ โˆ’ ๐‘ฃ^โ€ฒ ๐‘ข)/๐‘ฃ^2 = 1/2 [(โˆ’3โˆš(1 โˆ’ ๐‘ฅ) โˆ’ 1/(2โˆš(1 โˆ’ ๐‘ฅ)) (0 โˆ’ 1) . (2 โˆ’ 3๐‘ฅ))/((1 โˆ’ ๐‘ฅ) )] = 1/2 [(โˆ’3โˆš(1 โˆ’ ๐‘ฅ) + (2 โˆ’ 3๐‘ฅ)/(2โˆš(1 โˆ’ ๐‘ฅ)) )/(1 โˆ’ ๐‘ฅ)] = 1/2 [((โˆ’3โˆš(1 โˆ’ ๐‘ฅ)) (2โˆš(1 โˆ’ ๐‘ฅ)) + 2 โˆ’ 3๐‘ฅ )/(2(1 โˆ’ ๐‘ฅ) โˆš(1 โˆ’ ๐‘ฅ))] = 1/2 [(โˆ’6(1 โˆ’ ๐‘ฅ) + 2 โˆ’ 3๐‘ฅ )/(2(1 โˆ’ ๐‘ฅ) โˆš(1 โˆ’ ๐‘ฅ))] = 1/2 [(โˆ’6 + 6๐‘ฅ + 2 โˆ’ 3๐‘ฅ )/(2(1 โˆ’ ๐‘ฅ) โˆš(1 โˆ’ ๐‘ฅ))] = 1/4 [(โˆ’4 + 3๐‘ฅ )/(1 + ๐‘ฅ)^(3/2) ] Hence, fโ€™โ€™(๐‘ฅ)=1/4 [(โˆ’4 + 3๐‘ฅ )/(1 + ๐‘ฅ)^(3/2) ] Putting ๐‘ฅ=2/3 fโ€™โ€™(2/3)=1/4 [(โˆ’4 + 3(2/3))/(1 + 2/3)^(3/2) ] =1/4 [(โˆ’4 + 2)/(5/3)^(3/2) ] =1/4 [(โˆ’2)/(5/3)^(3/2) ] = (โˆ’ 1)/2 (3/5)^(3/2) < 0 Since fโ€™โ€™(๐‘ฅ)<0 when ๐‘ฅ = 2/3 Hence, ๐‘ฅ=2/3 is the maxima Finding Maximum value of f(๐’™)=๐’™โˆš(๐Ÿโˆ’๐’™) Putting ๐‘ฅ=2/3 f(2/3)=2/3 โˆš(1โˆ’2/3) =2/3 โˆš((3 โˆ’ 2)/3) Finding Maximum value of f(๐’™)=๐’™โˆš(๐Ÿโˆ’๐’™) Putting ๐‘ฅ=2/3 f(๐‘ฅ) = ๐‘ฅโˆš(1โˆ’๐‘ฅ) f(2/3)=2/3 โˆš(1โˆ’2/3) =2/3 โˆš((3 โˆ’ 2)/3) =2/3 โˆš(1/3) =2/(3โˆš3) =2/(3โˆš3) ร— โˆš3/โˆš3 =(2โˆš3)/9 Maximum value of f(๐’™) is (๐Ÿโˆš๐Ÿ‘)/๐Ÿ— at x = ๐Ÿ/๐Ÿ‘

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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.