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Ex 6.2,7 - Ex 6.2

Ex 6.2,7 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.2,7 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.2,7 - Chapter 6 Class 12 Application of Derivatives - Part 4

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Ex 6.2, 7 Show that ๐‘ฆ = log(1 + ๐‘ฅ) โ€“ 2๐‘ฅ/(2 + ๐‘ฅ) , ๐‘ฅ > โ€“ 1 , is an increasing function of ๐‘ฅ throughout its domain.Given ๐‘ฆ = log (1+๐‘ฅ) โ€“ 2๐‘ฅ/(2 + ๐‘ฅ) , ๐‘ฅ > โ€“1 We need to show that y is strictly increasing function for ๐‘ฅ > โ€“1 i.e. we need to show that (๐’…๐’š )/๐’…๐’™ > 0 for ๐’™ > โ€“1 Finding ๐’…๐’š/๐’…๐’™ ๐‘ฆ = log (1+๐‘ฅ) โ€“ (2๐‘ฅ )/(2 + ๐‘ฅ) (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ = ๐‘‘(logโกใ€–(1 + ๐‘ฅ) โˆ’ 2๐‘ฅ/(2 + ๐‘ฅ)ใ€— )/๐‘‘๐‘ฅ (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ = ๐‘‘(๐‘™๐‘œ๐‘”(1 + ๐‘ฅ))/๐‘‘๐‘ฅ โ€“ ๐‘‘/๐‘‘๐‘ฅ (๐Ÿ๐’™/(๐Ÿ+๐’™)) (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ = 1/(1 + ๐‘ฅ) . (1+๐‘ฅ)โ€™ โ€“ [((๐Ÿ๐’™)^โ€ฒ (๐Ÿ + ๐’™) โˆ’ใ€– (๐Ÿ + ๐’™)ใ€—^โ€ฒ ๐Ÿ๐’™)/(๐Ÿ + ๐’™)๐Ÿ] (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ = 1/(1 + ๐‘ฅ) . (0+1) โ€“ [(2(2 + ๐‘ฅ) โˆ’ (0 + 1)2๐‘ฅ)/(2 + ๐‘ฅ)2] (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ = 1/(1 + ๐‘ฅ) โ€“[(4 + 2๐‘ฅ โˆ’ 2๐‘ฅ)/(2 + ๐‘ฅ)2] (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ = 1/(1 + ๐‘ฅ) โ€“ [4/(2 + ๐‘ฅ)2] (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ = ((2 + ๐‘ฅ)2 โˆ’ 4(1 + ๐‘ฅ))/(1 + ๐‘ฅ)(2 + ๐‘ฅ)2 (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ = ((2)2 + (๐‘ฅ)2 + 2(2)(๐‘ฅ) โˆ’ 4 โˆ’ 4๐‘ฅ)/(1 + ๐‘ฅ)(2 + ๐‘ฅ)2 (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ = (4 + ๐‘ฅ2 + 4๐‘ฅ โˆ’ 4 โˆ’ 4๐‘ฅ)/(1 + ๐‘ฅ)(2 + ๐‘ฅ)2 (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ = ๐‘ฅ2/(1 + ๐‘ฅ)(2 + ๐‘ฅ)2 (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ = (๐‘ฅ/(2 + ๐‘ฅ))^2 1/(1 + ๐‘ฅ) Now, (๐‘‘๐‘ฆ )/๐‘‘๐‘ฅ = (๐’™/(๐Ÿ + ๐’™))^๐Ÿ 1/(1 + ๐‘ฅ) Now, finding value where ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ > 0 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ > 0 (๐‘ฅ/(2 + ๐‘ฅ))^2.(1/(1 + ๐‘ฅ)) > 0 (๐Ÿ/(๐Ÿ + ๐’™ ))>๐ŸŽ This is possible only when 1 + ๐‘ฅ > 0 i.e. ๐’™ > โ€“1 So, ๐’…๐’š/๐’…๐’™ > 0 for ๐’™ > โ€“1 Hence proved

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.