Ex 6.2, 7 - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Ex 6.2
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 April 16, 2024 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