# Ex 6.2,8 - Chapter 6 Class 12 Application of Derivatives

Last updated at April 16, 2024 by Teachoo

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

Ex 6.2,8 Important You are here

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, 8 Find the values of 𝑥 for which y = [𝑥(𝑥 – 2)]2 is an increasing function 𝑦 = [𝑥(𝑥−2)]^2 Finding 𝒅𝒚/𝒅𝒙 𝑦 = [𝑥(𝑥−2)]^2 𝑦 = [𝑥^2−2𝑥]^2 𝑦 = (𝑥)^4+(2𝑥)^2−2(𝑥^2 )(2𝑥) 𝒚 = 𝒙^𝟒+𝟒𝒙^𝟐−𝟒𝒙^𝟑 Differentiating w.r.t 𝑥 𝑑𝑦/𝑑𝑥=𝑑(𝑥^4 + 4𝑥^2 − 4𝑥^3 )/𝑑𝑥 𝑑𝑦/𝑑𝑥=4𝑥^3+8𝑥−12𝑥^2 𝑑𝑦/𝑑𝑥=4𝑥(𝑥^2+2−3𝑥) 𝑑𝑦/𝑑𝑥=4𝑥(𝑥^2−3𝑥+2) 𝑑𝑦/𝑑𝑥=4𝑥(𝑥^2−2𝑥−𝑥+2) 𝑑𝑦/𝑑𝑥=4𝑥(𝑥(𝑥−2)−1(𝑥−2)) 𝑑𝑦/𝑑𝑥=4𝑥((𝑥−1)(𝑥−2)) 𝑑𝑦/𝑑𝑥=4𝑥(𝑥−1)(𝑥−2) Putting 𝒅𝒚/𝒅𝒙=𝟎 4𝑥(𝑥−1)(𝑥−2)=0 So, 𝑥=0 , 𝑥=1 & 𝑥=2 Plotting points on real line Thus, the function is strictly increasing for 0 <𝒙<𝟏 and 𝒙>𝟐