Ex 6.3, 11 It is given that at 𝑥 = 1, the function 𝑥4 – 62𝑥2 + 𝑎𝑥+ 9 attains its maximum value, on the interval [0, 2]. Find the value of a.We have f(𝑥)=𝑥4 – 62𝑥2 + 𝑎𝑥+ 9 Finding f’(𝒙) f’(𝑥)=𝑑(𝑥^4− 62𝑥^2 + 𝑎𝑥 + 9)/𝑑𝑥 = 〖4𝑥〗^3−62 ×2𝑥+𝑎 = 〖4𝑥〗^3−124𝑥+𝑎 Given that at 𝑥=1, f(𝑥)=𝑥^4−62𝑥^2+𝑎𝑥+9 attain its Maximum Value i.e. f(𝑥) maximum at 𝑥=1 ∴ 𝑓’(𝑥)=0 at 𝑥=1 Now, f’(1)=0 〖4𝑥〗^3−124𝑥+𝑎 = 0 4(1)^3−124(1)+a=0 4 – 124 + a = 0 –120 + a = 0 a = 120 Hence, a = 120

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Davneet Singh

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.

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