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Ex 6.2

Ex 6.2, 1

Ex 6.2,2

Ex 6.2,3 Important

Ex 6.2,4

Ex 6.2, 5 Important

Ex 6.2, 6 (a)

Ex 6.2, 6 (b) Important

Ex 6.2, 6 (c) Important

Ex 6.2, 6 (d)

Ex 6.2, 6 (e) Important

Ex 6.2, 7 You are here

Ex 6.2,8 Important

Ex 6.2,9 Important

Ex 6.2,10

Ex 6.2,11

Ex 6.2, 12 (A)

Ex 6.2, 12 (B) Important

Ex 6.2, 12 (C) Important

Ex 6.2, 12 (D)

Ex 6.2, 13 (MCQ) Important

Ex 6.2,14 Important

Ex 6.2,15

Ex 6.2, 16

Ex 6.2,17 Important

Ex 6.2,18

Ex 6.2,19 (MCQ) Important

Last updated at May 29, 2023 by Teachoo

Ex 6.2, 7 Show that 𝑦 = log(1 + 𝑥) – 2𝑥/(2 + 𝑥) , 𝑥 > – 1 , is an increasing function of 𝑥 throughout its domain.Given 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 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