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Ex 6.2, 8 Find the values of ๐‘ฅ for which y = [๐‘ฅ(๐‘ฅ โ€“ 2)]2 is an increasing function ๐‘ฆ = [๐‘ฅ(๐‘ฅโˆ’2)]^2 Finding ๐’…๐’š/๐’…๐’™ ๐‘ฆ = [๐‘ฅ(๐‘ฅโˆ’2)]^2 ๐‘ฆ = [๐‘ฅ^2โˆ’2๐‘ฅ]^2 ๐‘ฆ = (๐‘ฅ)^4+(2๐‘ฅ)^2โˆ’2(๐‘ฅ^2 )(2๐‘ฅ) ๐’š = ๐’™^๐Ÿ’+๐Ÿ’๐’™^๐Ÿโˆ’๐Ÿ’๐’™^๐Ÿ‘ Differentiating w.r.t ๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=๐‘‘(๐‘ฅ^4 + 4๐‘ฅ^2 โˆ’ 4๐‘ฅ^3 )/๐‘‘๐‘ฅ ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=4๐‘ฅ^3+8๐‘ฅโˆ’12๐‘ฅ^2 ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=4๐‘ฅ(๐‘ฅ^2+2โˆ’3๐‘ฅ) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=4๐‘ฅ(๐‘ฅ^2โˆ’3๐‘ฅ+2) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=4๐‘ฅ(๐‘ฅ^2โˆ’2๐‘ฅโˆ’๐‘ฅ+2) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=4๐‘ฅ(๐‘ฅ(๐‘ฅโˆ’2)โˆ’1(๐‘ฅโˆ’2)) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=4๐‘ฅ((๐‘ฅโˆ’1)(๐‘ฅโˆ’2)) ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ=4๐‘ฅ(๐‘ฅโˆ’1)(๐‘ฅโˆ’2) Putting ๐’…๐’š/๐’…๐’™=๐ŸŽ 4๐‘ฅ(๐‘ฅโˆ’1)(๐‘ฅโˆ’2)=0 So, ๐‘ฅ=0 , ๐‘ฅ=1 & ๐‘ฅ=2 Plotting points on real line Thus, the function is strictly increasing for 0 <๐’™<๐Ÿ and ๐’™>๐Ÿ

  1. Chapter 6 Class 12 Application of Derivatives
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About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo