Ex 6.3,11 - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Absolute minima/maxima
Last updated at April 16, 2024 by Teachoo
Ex 6.3, 11 It is given that at đĽ = 1, the function đĽ4 â 62đĽ2 + đđĽ+ 9 attains its maximum value, on the interval [0, 2]. Find the value of a.We have f(đĽ)=đĽ4 â 62đĽ2 + đđĽ+ 9 Finding fâ(đ) fâ(đĽ)=đ(đĽ^4â 62đĽ^2 + đđĽ + 9)/đđĽ = ă4đĽă^3â62 Ă2đĽ+đ = ă4đĽă^3â124đĽ+đ Given that at đĽ=1, f(đĽ)=đĽ^4â62đĽ^2+đđĽ+9 attain its Maximum Value i.e. f(đĽ) maximum at đĽ=1 â´ đâ(đĽ)=0 at đĽ=1 Now, fâ(1)=0 ă4đĽă^3â124đĽ+đ = 0 4(1)^3â124(1)+a=0 4 â 124 + a = 0 â120 + a = 0 a = 120 Hence, a = 120