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Last updated at Jan. 7, 2020 by Teachoo

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Ex 6.5, 11 It is given that at ๐ฅ = 1, the function ๐ฅ4 โ 62๐ฅ2 + ๐๐ฅ+ 9 attains its maximum value, on the interval [0, 2]. Find the value of a. We have f(๐ฅ)=๐ฅ4 โ 62๐ฅ2 + ๐๐ฅ+ 9 Finding fโ(๐) fโ(๐ฅ)=๐(๐ฅ^4โ 62๐ฅ^2 + ๐๐ฅ + 9)/๐๐ฅ = ใ4๐ฅใ^3โ62 ร2๐ฅ+๐ = ใ4๐ฅใ^3โ124๐ฅ+๐ Given that at ๐ฅ=1, f(๐ฅ)=๐ฅ^4โ62๐ฅ^2+๐๐ฅ+9 attain its Maximum Value i.e. f(๐ฅ) maximum at ๐ฅ=1 โด ๐โ(๐ฅ)=0 at ๐ฅ=1 Now, fโ(1)=0 ใ4๐ฅใ^3โ124๐ฅ+๐ = 0 4(1)^3โ124(1)+a=0 4 โ 124 + a = 0 โ120 + a = 0 a = 120 Hence, a = 120

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Chapter 6 Class 12 Application of Derivatives

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.