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

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

Transcript

Ex 6.2, 15 Let I be any interval disjoint from [โ€“1, 1]. Prove that the function f given by ๐‘“(๐‘ฅ) = ๐‘ฅ + 1/๐‘ฅ is strictly increasing on I. I is any interval disjoint from [โ€“1, 1] Hence I = (โˆ’โˆž, โˆ’1)โˆช(1, โˆž) f(๐‘ฅ) = ๐‘ฅ + 1/๐‘ฅ We need to show f(๐‘ฅ) is strictly increasing on I i.e. we need to show fโ€™(๐‘ฅ) > 0 for ๐‘ฅ โˆˆ(โˆ’โˆž, โˆ’1)โˆช(1, โˆž) Finding fโ€™(๐’™) f(๐‘ฅ) = ๐‘ฅ + 1/๐‘ฅ fโ€™(๐‘ฅ) = 1 โ€“ 1/๐‘ฅ2 fโ€™(๐‘ฅ) = (๐‘ฅ2 โˆ’ 1)/๐‘ฅ2 Putting fโ€™(๐’™) = 0 (๐‘ฅ2 โˆ’ 1)/๐‘ฅ2 = 0 ๐‘ฅ2โˆ’1 = 0 (๐‘ฅโˆ’1)(๐‘ฅ+1)=0 So, ๐‘ฅ=1 & ๐‘ฅ=โˆ’1 Plotting point on real line The point ๐‘ฅ = โ€“1 , 1 into three disjoint intervals i.e. (โˆ’โˆž , โˆ’1) (โˆ’1 , 1) & (1 , โˆž) โˆด f(x) is strictly increasing on (โˆ’โˆž , โˆ’๐Ÿ) & (๐Ÿ , โˆž) Hence proved

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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.