Ex 6.5, 5 (iii) - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at Aug. 19, 2021 by Teachoo
Ex 6.5
Ex 6.5, 1 (i)
Important
Ex 6.5, 1 (ii)
Ex 6.5, 1 (iii) Important
Ex 6.5, 1 (iv)
Ex 6.5, 2 (i)
Ex 6.5, 2 (ii) Important
Ex 6.5, 2 (iii)
Ex 6.5, 2 (iv) Important
Ex 6.5, 2 (v) Important
Ex 6.5, 3 (i)
Ex 6.5, 3 (ii)
Ex 6.5, 3 (iii)
Ex 6.5, 3 (iv) Important
Ex 6.5, 3 (v)
Ex 6.5, 3 (vi)
Ex 6.5, 3 (vii) Important
Ex 6.5, 3 (viii)
Ex 6.5, 4 (i)
Ex 6.5, 4 (ii) Important
Ex 6.5, 4 (iii)
Ex 6.5, 5 (i)
Ex 6.5, 5 (ii)
Ex 6.5, 5 (iii) Important You are here
Ex 6.5, 5 (iv)
Ex 6.5,6
Ex 6.5,7 Important
Ex 6.5,8
Ex 6.5,9 Important
Ex 6.5,10
Ex 6.5,11 Important
Ex 6.5,12 Important
Ex 6.5,13
Ex 6.5,14 Important
Ex 6.5,15 Important
Ex 6.5,16
Ex 6.5,17
Ex 6.5,18 Important
Ex 6.5,19 Important
Ex 6.5, 20 Important
Ex 6.5,21
Ex 6.5,22 Important
Ex 6.5,23 Important
Ex 6.5,24 Important
Ex 6.5,25 Important
Ex 6.5,26 Important
Ex 6.5, 27 (MCQ)
Ex 6.5,28 (MCQ) Important
Ex 6.5,29 (MCQ)
Transcript
Ex 6.5, 5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (iii) f(π₯) = 4π₯ β 1/2 π₯2 , π₯ β [β2, 9/2] f(π₯) = 4π₯ β 1/2 π₯2 Finding fβ(π) fβ(π₯)=π(4π₯ β 1/2 π₯^2 )/ππ₯ = 4 β 1/2 Γ2π₯ = 4 β π₯ Putting fβ(π)=π 4 β π₯=0 π₯=4 β΄ π₯=4 is only critical point Since given interval π₯ β [β2 , 9/2] Hence , calculating f(π₯) at π₯ = β 2 , 4 , 9/2 Absolute Maximum value of f(x) is 8 at π = 4 & Absolute Minimum value of f(x) is β10 at π = β2
