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Ex 6.5, 11 - At x = 1, function x4 - 62x2 + ax + 9 attains max

Ex 6.5,11 - Chapter 6 Class 12 Application of Derivatives - Part 2

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Ex 6.5, 11 It is given that at π‘₯ = 1, the function π‘₯4 – 62π‘₯2 + π‘Žπ‘₯+ 9 attains its maximum value, on the interval [0, 2]. Find the value of a.We have f(π‘₯)=π‘₯4 – 62π‘₯2 + π‘Žπ‘₯+ 9 Finding f’(𝒙) f’(π‘₯)=𝑑(π‘₯^4βˆ’ 62π‘₯^2 + π‘Žπ‘₯ + 9)/𝑑π‘₯ = γ€–4π‘₯γ€—^3βˆ’62 Γ—2π‘₯+π‘Ž = γ€–4π‘₯γ€—^3βˆ’124π‘₯+π‘Ž Given that at π‘₯=1, f(π‘₯)=π‘₯^4βˆ’62π‘₯^2+π‘Žπ‘₯+9 attain its Maximum Value i.e. f(π‘₯) maximum at π‘₯=1 ∴ 𝑓’(π‘₯)=0 at π‘₯=1 Now, f’(1)=0 γ€–4π‘₯γ€—^3βˆ’124π‘₯+π‘Ž = 0 4(1)^3βˆ’124(1)+a=0 4 – 124 + a = 0 –120 + a = 0 a = 120 Hence, a = 120

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.