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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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