# Ex 6.5,11 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 15, 2021 by Teachoo

Last updated at April 15, 2021 by Teachoo

Transcript

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

Ex 6.5

Ex 6.5, 1 (i)
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Ex 6.5, 1 (ii)

Ex 6.5, 1 (iii) Important

Ex 6.5, 1 (iv)

Ex 6.5, 2 (i)

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Ex 6.5, 2 (iv) Important

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Ex 6.5, 3 (iv) Important

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Ex 6.5, 4 (iii)

Ex 6.5, 5 (i)

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Ex 6.5, 5 (iv)

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Ex 6.5,11 Important You are here

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Ex 6.5, 27 (MCQ)

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Ex 6.5,29 (MCQ)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.