# Ex 6.5,12 - Chapter 6 Class 12 Application of Derivatives

Last updated at Jan. 7, 2020 by Teachoo

Last updated at Jan. 7, 2020 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 " " If cos π₯=cosβ‘π¦ Then general solution of π₯=2ππΒ±π¦ 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(0)=0+sinβ‘γ2(0)γ = 0 + sin 0 = 0 π(π/3)=π/3+sinβ‘ 2(π/3) = π/3+sin(πβπ/3) = π/3+ sin π/3 = π/3+β3/2 π₯ = 2π/3 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 β("sin " (π+ π/3)=βπ ππ π/3 @" " ππππ ππ "3rd quadrant " ) π₯ = 4π/3 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 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₯ = 2Ο 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

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Chapter 6 Class 12 Application of Derivatives

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.