# Ex 6.2,8 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 14, 2021 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 14, 2021 by Teachoo

Hello! Teachoo has made these pages with hours (even days!) of effort and love. Your Board exams are coming, if Teachoo has been of any help in the whole year of studying... please consider making a donation to support us.

Hello! Teachoo has made these pages with hours (even days!) of effort and love. Your Board exams are coming, if Teachoo has been of any help in the whole year of studying... please consider making a donation to support us.

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