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

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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 Diff. 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 The points 𝑥 = 0 , 1 and 2 divide the real line into 4 disjoint intervals i.e. (−∞,0) , (0 , 1) , (1 , 2) & (2 ,∞) Thus the function y = [𝑥(𝑥−1)^2 ] is strictly increasing for 0 <𝑥<1 and 𝑥>2