Ex 6.3,11 - Chapter 6 Class 12 Application of Derivatives

Last updated at April 16, 2024 by

Slide34.JPG

Slide35.JPG

Go Ad-free

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

Davneet Singh's photo - Co-founder, Teachoo

Made by

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, Social Science, Physics, Chemistry, Computer Science at Teachoo.