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

Transcript

Ex 6.5, 28 For all real values of x, the minimum value of (1 โˆ’ ๐‘ฅ + ๐‘ฅ2)/(1 + ๐‘ฅ + ๐‘ฅ2) is (A) 0 (B) 1 (C) 3 (D) 1/3Ex 6.5, 28 For all real values of x, the minimum value of (1 โˆ’ ๐‘ฅ + ๐‘ฅ2)/(1 + ๐‘ฅ + ๐‘ฅ2) is (A) 0 (B) 1 (C) 3 (D) 1/3 Let ๐‘“(๐‘ฅ)=(1 โˆ’ ๐‘ฅ + ๐‘ฅ2)/(1 + ๐‘ฅ + ๐‘ฅ2) Finding ๐’‡โ€ฒ(๐’™) ๐‘“(๐‘ฅ)=(1 โˆ’ ๐‘ฅ + ๐‘ฅ2)/(1 + ๐‘ฅ + ๐‘ฅ2) ๐‘“^โ€ฒ(๐‘ฅ) =((1 โˆ’ ๐‘ฅ + ๐‘ฅ^2 )^โ€ฒ (1 + ๐‘ฅ + ๐‘ฅ^2 ) โˆ’ (1 โˆ’ ๐‘ฅ + ๐‘ฅ^2 ) (1 + ๐‘ฅ + ๐‘ฅ^2 )^โ€ฒ)/(1 + ๐‘ฅ + ๐‘ฅ^2 )^2 Using quotient Rule As (๐‘ข/๐‘ฃ)^โ€ฒ=(๐‘ข^โ€ฒ ๐‘ฃโˆ’๐‘ฃ^โ€ฒ ๐‘ข)/๐‘ฃ^2 ๐‘“โ€ฒ(๐‘ฅ)=(โˆ’1 โˆ’ ๐‘ฅ โˆ’ ๐‘ฅ^2 + 2๐‘ฅ + 2๐‘ฅ^2+ 2๐‘ฅ^3 โˆ’ (1 โˆ’ ๐‘ฅ + ๐‘ฅ^2+ 2๐‘ฅ โˆ’ 2๐‘ฅ^2 + 2๐‘ฅ^3 ))/(1 + ๐‘ฅ + ๐‘ฅ^2 )^2 ๐‘“(๐‘ฅ)=(โˆ’1 + ๐‘ฅ + ๐‘ฅ^2 + 2๐‘ฅ^3 โˆ’ (1 + ๐‘ฅ โˆ’ ๐‘ฅ^2 + 2๐‘ฅ^3 ))/(1 + ๐‘ฅ + ๐‘ฅ^2 )^2 ๐‘“โ€ฒ(๐‘ฅ)=(โˆ’2 + 2๐‘ฅ^2)/(1 + ๐‘ฅ + ๐‘ฅ^2 )^2 Putting ๐’‡^โ€ฒ (๐’™)=๐ŸŽ (โˆ’2 + 2๐‘ฅ^2)/(1 + ๐‘ฅ + ๐‘ฅ^2 )^2 =0 2๐‘ฅ^2โˆ’2=0 2๐‘ฅ^2=2 ๐‘ฅ^2=1 ๐‘ฅ=ยฑ1 Hence, x = 1 or x = โ€“1 are the critical points Finding value of ๐’‡(๐’™) at critical points Hence, minimum value of f(x) is 1/3. So, (D) is the correct answer

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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.