Ex 6.2, 7 - Show that y = log (1 + x) - 2x/2+x is increasing - Ex 6.2

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  1. Chapter 6 Class 12 Application of Derivatives
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Ex 6.2,7 Show that 𝑦 = log(1 + 𝑥) – ﷐2𝑥﷮2 + 𝑥﷯ , 𝑥 > – 1 , is an increasing function of 𝑥 throughout its domain. 𝑦 = log ﷐1+𝑥﷯ – ﷐2𝑥﷮2 + 𝑥﷯ , 𝑥 > –1 We need to show that y is strictly increasing function for 𝑥 > –1 i.e. we need to show that ﷐𝑑𝑦 ﷮𝑑𝑥﷯ > 0 for 𝑥 > –1 Step 1: Finding ﷐𝑑𝑦﷮𝑑𝑥﷯ 𝑦 = log ﷐1+𝑥﷯ – ﷐2𝑥 ﷮2 + 𝑥﷯ ﷐𝑑𝑦 ﷮𝑑𝑥﷯ = ﷐𝑑﷐﷐log﷮﷐1 + 𝑥﷯ − ﷐2𝑥﷮2 + 𝑥﷯﷯﷯﷮𝑑𝑥﷯ ﷐𝑑𝑦 ﷮𝑑𝑥﷯ = ﷐𝑑﷐𝑙𝑜𝑔﷐1+𝑥﷯﷯﷮𝑑𝑥﷯ – ﷐𝑑﷮𝑑𝑥﷯﷐﷐2𝑥﷮2+𝑥﷯﷯ ﷐𝑑𝑦 ﷮𝑑𝑥﷯ = ﷐1﷮1 + 𝑥﷯ . ﷐1+𝑥﷯’ – ﷐﷐﷐﷐2𝑥﷯﷮′﷯﷐2 + 𝑥﷯ −﷐ ﷐2 + 𝑥﷯﷮′﷯2𝑥﷮﷐2 + 𝑥﷯2﷯﷯ ﷐𝑑𝑦 ﷮𝑑𝑥﷯ = ﷐1﷮1 + 𝑥﷯ . ﷐0+1﷯ – ﷐﷐2﷐2 + 𝑥﷯ − ﷐0 + 1﷯2𝑥﷮﷐2 + 𝑥﷯2﷯﷯ ﷐𝑑𝑦 ﷮𝑑𝑥﷯ = ﷐1﷮1 + 𝑥﷯ –﷐﷐4 + 2𝑥 − 2𝑥﷮﷐2 + 𝑥﷯2﷯﷯ ﷐𝑑𝑦 ﷮𝑑𝑥﷯ = ﷐1﷮1 + 𝑥﷯ – ﷐﷐4﷮﷐2 + 𝑥﷯2﷯﷯ ﷐𝑑𝑦 ﷮𝑑𝑥﷯ = ﷐﷐2 + 𝑥﷯2 − 4﷐1 + 𝑥﷯﷮﷐1 + 𝑥﷯﷐2 + 𝑥﷯2﷯ ﷐𝑑𝑦 ﷮𝑑𝑥﷯ = ﷐﷐2﷯2 + ﷐𝑥﷯2 + 2﷐2﷯﷐𝑥﷯ − 4 − 4𝑥﷮﷐1 + 𝑥﷯﷐2 + 𝑥﷯2﷯ ﷐𝑑𝑦 ﷮𝑑𝑥﷯ = ﷐4 + 𝑥2 + 4𝑥 − 4 − 4𝑥﷮﷐1 + 𝑥﷯﷐2 + 𝑥﷯2﷯ ﷐𝑑𝑦 ﷮𝑑𝑥﷯ = ﷐𝑥2﷮﷐1 + 𝑥﷯﷐2 + 𝑥﷯2﷯ ﷐𝑑𝑦 ﷮𝑑𝑥﷯ = ﷐﷐﷐𝑥﷮2 + 𝑥﷯﷯﷮2﷯ ﷐1﷮1 + 𝑥﷯ Now, ﷐𝑑𝑦 ﷮𝑑𝑥﷯ = ﷐﷐﷐𝑥﷮2 + 𝑥﷯﷯﷮2﷯ ﷐1﷮1 + 𝑥﷯ Now, finding value where ﷐𝑑𝑦﷮𝑑𝑥﷯ > 0 ﷐𝑑𝑦﷮𝑑𝑥﷯ > 0 ﷐﷐﷐𝑥﷮2+𝑥﷯﷯﷮2﷯.﷐﷐1﷮1+𝑥﷯﷯ > 0 ﷐﷐1﷮1+𝑥 ﷯﷯>0 This is possible only when 1 + 𝑥 > 0 i.e. 𝑥 > –1 So, ﷐𝒅𝒚﷮𝒅𝒙﷯ > 0 for 𝒙 > –1 Hence proved

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