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

Transcript

Ex 6.5, 29 The maximum value of ใ€–[๐‘ฅ(๐‘ฅโˆ’1)+1]ใ€—^(1/3) 0 โ‰ค x โ‰ค 1 is (A) (1/3)^(1/3) (B) 1/2 (C) 1 (D) 0 Let f(๐‘ฅ)=[๐‘ฅ(๐‘ฅโˆ’1)+1]^(1/3) Finding fโ€™(๐’™) ๐‘“(๐‘ฅ)=[๐‘ฅ[๐‘ฅโˆ’1]+1]^(1/3) ๐‘“(๐‘ฅ)=[๐‘ฅ^2โˆ’๐‘ฅ+1]^(1/3) ๐‘“^โ€ฒ (๐‘ฅ)=(๐‘‘(๐‘ฅ^2 โˆ’ ๐‘ฅ + 1)^(1/3))/๐‘‘๐‘ฅ ๐‘“^โ€ฒ (๐‘ฅ)=1/3 (๐‘ฅ^2โˆ’๐‘ฅ+1)^(1/3 โˆ’ 1) . ๐‘‘(๐‘ฅ^2 โˆ’ ๐‘ฅ + 1)/๐‘‘๐‘ฅ ๐‘“^โ€ฒ (๐‘ฅ)=1/3 (๐‘ฅ^2โˆ’๐‘ฅ+1)^((โˆ’2)/3) (2๐‘ฅโˆ’1) ๐‘“^โ€ฒ (๐‘ฅ)=1/(3(๐‘ฅ^2 โˆ’ ๐‘ฅ + 1)^(2/3) ) .(2๐‘ฅโˆ’1) ๐‘“^โ€ฒ (๐‘ฅ)=(2๐‘ฅ โˆ’ 1)/(3(๐‘ฅ^2 โˆ’ ๐‘ฅ + 1)^(2/3) ) Putting fโ€™(๐’™)=๐ŸŽ (2๐‘ฅโˆ’1)/(3(๐‘ฅ^2 โˆ’ ๐‘ฅ + 1)^(2/3) )=0 2๐‘ฅโˆ’1=0 2๐‘ฅ=1 ๐‘ฅ=1/2 Since, 0 โ‰ค x โ‰ค 1 Hence, critical points are ๐‘ฅ=0 ,1/2 , & 1 Hence, Maximum value is 1 at ๐‘ฅ=0 , 1 โˆด Hence, 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 9 years. He provides courses for Maths and Science at Teachoo.