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Ex 6.5, 2 (iv) - 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 You are here
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
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, 2 Find the maximum and minimum values, if any, of the following functions given by (iv) π (π₯)=|sinβ‘4π₯+3|π (π₯)=| sinβ‘4π₯+3| We know that β1 β€ sin ΞΈ β€ 1 So, β1 β€ sin 4π₯ β€ 1 Adding 3 both sides β1 + 3 β€ sin 4π₯ + 3 β€ 1 + 3 2 β€ sin 4π₯ +3 β€ 4 Taking modulus |2| β€ | sinβ‘4π₯+3| β€ |4| 2 β€ | sinβ‘4π₯+3| β€ |4| 2 β€ f(π₯)β€4 Hence Maximum value of f(π) is 4 & Minimum value of f(π) is 2
