1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
  3. Ex 6.3
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Ex 6.3,11 - Chapter 6 Class 12 Application of Derivatives

Last updated at Dec. 16, 2024 by Teachoo

Transcript

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
  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

    3. Ex 6.3

    Examples →
Examples →
Chapter 6 Class 12 Application of Derivatives
Serial order wise

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo