# Ex 6.5,12 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at April 15, 2021 by Teachoo

Last updated at April 15, 2021 by Teachoo

Transcript

Ex 6.5, 12 Find the maximum and minimum values of π₯ + sin 2π₯ on [0, 2Ο ] Let f(π₯)=π₯ + sin 2π₯ Finding fβ(π) πβ(π₯)=π(π₯ + sin 2π₯)/ππ₯ =1+2 cosβ‘2π₯ Putting fβ(π)=π 1 + 2 cos 2π₯=0 2 cos 2π₯=β1 cos 2π₯=(β1)/2 cos 2π₯=cosβ‘γ2π/3γ General solution for cos 2π₯ is 2π₯=2ππΒ±2π/3 π₯=(2ππ Β± 2π/3)/2 π₯= nΟ Β± π/3 Putting π=0 π₯=0(π)Β±π/3 =Β±π/3 " " cos 2π₯=cosβ‘γ2π/3γ General solution for cos 2π₯ is 2π₯=2ππΒ±2π/3 π₯=(2ππ Β± 2π/3)/2 π₯= nΟ Β± π/3 Putting π=0 π₯=0(π)Β±π/3 =Β±π/3 " " We know that πππ 60Β°=1/2 And cos is negative in 2nd & 3rd quadrant π = 180 β 60 = 120 = 120 Γ π/180 = 2π/3 So, π₯=(βπ)/3,π/3 Since Given π₯ β[0 , 2π] β΄ π₯=π/3 only Putting π=1 π₯=(1)πΒ±π/3 =πΒ±π/3 =(3π + π)/3 , (3π β π)/3 =4π/3 , 2π/3 Putting π=2 π₯=2(π)Β±π/3 π₯=2πβπ/3 & 2π+π/3 π₯=(6π β π)/3 & (6π + π)/3 π₯=5π/3 & 7π/3 So, π₯=5π/3 only Also, We are given interval π₯ β[0 , 2π] Hence , calculating f(π₯) at π₯=0 , π/3 , 2π/3 , 4π/3 , 5π/3 , 2π Since given π₯ β[0 , 2π] & 7π/3>2π π(2π/3)=π/3+sinβ‘ 2(2π/3) = 2π/3+sinβ‘γ4π/3 γ = 2π/3+ sin(π+π/3) = 2π/3β sin π/3 = 2π/3ββ3/2 π(4π/3)=4π/3+sinβ‘2 (4π/3) = 4π/3+sinβ‘γ8π/3 γ = 4π/3+ sin(3πβπ/3) = 4π/3+ sin π/3 = 4π/3+β3/2 π(4π/3)=4π/3+sinβ‘2 (4π/3) = 4π/3+sinβ‘γ8π/3 γ = 4π/3+ sin(3πβπ/3) = 4π/3+ sin π/3 = 4π/3+β3/2 Since 3Ο β π/3 lies in 2nd quadrant, sin is positive π(5π/3)=5π/3+sinβ‘ 2(5π/3) = 5π/3+sinβ‘γ10π/3 γ = 5π/3+sin(3π+π/3) = 5π/3β sin π/3 = 5π/3ββ3/2 Since 3Ο β π/3 lies in 2nd quadrant, sin is positive π(2π)=2π+sinβ‘ 2π =2π+0 =2π Hence, f(π₯) is Maximum at π₯=2π Maximum value of f(π)=ππ & f(π₯) is Minimum at π₯=0 Minimum value of f(π)=π

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

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 You are here

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)

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.