Ex 6.3, 5 (ii) - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Ex 6.3
Ex 6.3, 1 (i)
Important
Ex 6.3, 1 (ii)
Ex 6.3, 1 (iii) Important
Ex 6.3, 1 (iv)
Ex 6.3, 2 (i)
Ex 6.3, 2 (ii) Important
Ex 6.3, 2 (iii)
Ex 6.3, 2 (iv) Important
Ex 6.3, 2 (v) Important
Ex 6.3, 3 (i)
Ex 6.3, 3 (ii)
Ex 6.3, 3 (iii)
Ex 6.3, 3 (iv) Important
Ex 6.3, 3 (v)
Ex 6.3, 3 (vi)
Ex 6.3, 3 (vii) Important
Ex 6.3, 3 (viii)
Ex 6.3, 4 (i)
Ex 6.3, 4 (ii) Important
Ex 6.3, 4 (iii)
Ex 6.3, 5 (i)
Ex 6.3, 5 (ii) You are here
Ex 6.3, 5 (iii) Important
Ex 6.3, 5 (iv)
Ex 6.3,6
Ex 6.3,7 Important
Ex 6.3,8
Ex 6.3,9 Important
Ex 6.3,10
Ex 6.3,11 Important
Ex 6.3,12 Important
Ex 6.3,13
Ex 6.3,14 Important
Ex 6.3,15 Important
Ex 6.3,16
Ex 6.3,17
Ex 6.3,18 Important
Ex 6.3,19 Important
Ex 6.3, 20 Important
Ex 6.3,21
Ex 6.3,22 Important
Ex 6.3,23 Important
Ex 6.3,24 Important
Ex 6.3,25 Important
Ex 6.3, 26 Important
Ex 6.3, 27 (MCQ)
Ex 6.3,28 (MCQ) Important
Ex 6.3,29 (MCQ)
Transcript
Ex 6.3, 5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (ii) f (π₯) = sinβ‘π₯ + cosβ‘π₯ , π₯ β [0, π ] Finding fβ(π) fβ(π₯)=π(π πππ₯ + πππ π₯)/ππ₯ fβ(π₯)=cosβ‘γπ₯ βsinβ‘π₯ γ Putting fβ(π) cosβ‘γπ₯ βsinβ‘π₯ γ= 0 cosβ‘γπ₯=sinβ‘π₯ γ 1 = sinβ‘π₯/(cosβ‘ π₯) 1 = tan π₯ tan π₯ = 1 We know that know tan ΞΈ = 1 at ΞΈ = π/4 β΄ π₯ = π/4 Since given interval π₯ β [0 , π] Hence calculating f(π₯) at π₯=0 , π/4 ,π Absolute Maximum value of f(π₯) is βπ at π = π /π & Absolute Minimum value of f(π₯) is β1 at π = Ο
