Ex 6.2

Chapter 6 Class 12 Application of Derivatives
Serial order wise

### Transcript

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 )/๐๐ > 0 for ๐ > โ1 Finding ๐๐/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) โ 2๐ฅ/(2 + ๐ฅ)ใ )/๐๐ฅ (๐๐ฆ )/๐๐ฅ = ๐(๐๐๐(1 + ๐ฅ))/๐๐ฅ โ ๐/๐๐ฅ (๐๐/(๐+๐)) (๐๐ฆ )/๐๐ฅ = 1/(1 + ๐ฅ) . (1+๐ฅ)โ โ [((๐๐)^โฒ (๐ + ๐) โใ (๐ + ๐)ใ^โฒ ๐๐)/(๐ + ๐)๐] (๐๐ฆ )/๐๐ฅ = 1/(1 + ๐ฅ) . (0+1) โ [(2(2 + ๐ฅ) โ (0 + 1)2๐ฅ)/(2 + ๐ฅ)2] (๐๐ฆ )/๐๐ฅ = 1/(1 + ๐ฅ) โ[(4 + 2๐ฅ โ 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 โ 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) โ 4 โ 4๐ฅ)/(1 + ๐ฅ)(2 + ๐ฅ)2 (๐๐ฆ )/๐๐ฅ = (4 + ๐ฅ2 + 4๐ฅ โ 4 โ 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 + ๐))^๐ 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/(๐ + ๐ ))>๐ This is possible only when 1 + ๐ฅ > 0 i.e. ๐ > โ1 So, ๐๐/๐๐ > 0 for ๐ > โ1 Hence proved