Ex 6.2, 8 - Find x for which y = x(x - 2)2 is increasing

Ex 6.2,8 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.2,8 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.2,8 - Chapter 6 Class 12 Application of Derivatives - Part 4

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Ex 6.2, 8 Find the values of š‘„ for which y = [š‘„(š‘„ – 2)]2 is an increasing function š‘¦ = [š‘„(š‘„āˆ’2)]^2 Finding š’…š’š/š’…š’™ š‘¦ = [š‘„(š‘„āˆ’2)]^2 š‘¦ = [š‘„^2āˆ’2š‘„]^2 š‘¦ = (š‘„)^4+(2š‘„)^2āˆ’2(š‘„^2 )(2š‘„) š’š = š’™^šŸ’+šŸ’š’™^šŸāˆ’šŸ’š’™^šŸ‘ Differentiating w.r.t š‘„ š‘‘š‘¦/š‘‘š‘„=š‘‘(š‘„^4 + 4š‘„^2 āˆ’ 4š‘„^3 )/š‘‘š‘„ š‘‘š‘¦/š‘‘š‘„=4š‘„^3+8š‘„āˆ’12š‘„^2 š‘‘š‘¦/š‘‘š‘„=4š‘„(š‘„^2+2āˆ’3š‘„) š‘‘š‘¦/š‘‘š‘„=4š‘„(š‘„^2āˆ’3š‘„+2) š‘‘š‘¦/š‘‘š‘„=4š‘„(š‘„^2āˆ’2š‘„āˆ’š‘„+2) š‘‘š‘¦/š‘‘š‘„=4š‘„(š‘„(š‘„āˆ’2)āˆ’1(š‘„āˆ’2)) š‘‘š‘¦/š‘‘š‘„=4š‘„((š‘„āˆ’1)(š‘„āˆ’2)) š‘‘š‘¦/š‘‘š‘„=4š‘„(š‘„āˆ’1)(š‘„āˆ’2) Putting š’…š’š/š’…š’™=šŸŽ 4š‘„(š‘„āˆ’1)(š‘„āˆ’2)=0 So, š‘„=0 , š‘„=1 & š‘„=2 Plotting points on real line Thus, the function is strictly increasing for 0 <š’™<šŸ and š’™>šŸ

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