Ex 6.3

Chapter 6 Class 12 Application of Derivatives
Serial order wise

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### Transcript

Ex 6.3, 12 Find the maximum and minimum values of π₯ + sin 2π₯ on [0, 2Ο ] Let f(π₯)=π₯ + sin 2π₯ Finding fβ(π) πβ(π₯)=π(π₯ + sin 2π₯)/ππ₯ =1+2 cosβ‘2π₯ Putting fβ(π)=π 1 + 2 cos 2π₯=0 2 cos 2π₯=β1 cos 2π₯=(β1)/2 cos 2π₯=cosβ‘γ2π/3γ General solution for cos 2π₯ is 2π₯=2ππΒ±2π/3 π₯=(2ππ Β± 2π/3)/2 π₯= nΟ Β± π/3 Putting π=0 π₯=0(π)Β±π/3 =Β±π/3 " " cos 2π₯=cosβ‘γ2π/3γ General solution for cos 2π₯ is 2π₯=2ππΒ±2π/3 π₯=(2ππ Β± 2π/3)/2 π₯= nΟ Β± π/3 Putting π=0 π₯=0(π)Β±π/3 =Β±π/3 " " We know that πππ  60Β°=1/2 And cos is negative in 2nd & 3rd quadrant π = 180 β 60 = 120 = 120 Γ π/180 = 2π/3 So, π₯=(βπ)/3,π/3 Since Given π₯ β[0 , 2π] β΄ π₯=π/3 only Putting π=1 π₯=(1)πΒ±π/3 =πΒ±π/3 =(3π + π)/3 , (3π β π)/3 =4π/3 , 2π/3 Putting π=2 π₯=2(π)Β±π/3 π₯=2πβπ/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‘γ4π/3 γ = 2π/3+ sin(π+π/3) = 2π/3β sin π/3 = 2π/3ββ3/2 π(4π/3)=4π/3+sinβ‘2 (4π/3) = 4π/3+sinβ‘γ8π/3 γ = 4π/3+ sin(3πβπ/3) = 4π/3+ sin π/3 = 4π/3+β3/2 π(4π/3)=4π/3+sinβ‘2 (4π/3) = 4π/3+sinβ‘γ8π/3 γ = 4π/3+ sin(3πβπ/3) = 4π/3+ sin π/3 = 4π/3+β3/2 Since 3Ο β π/3 lies in 2nd quadrant, sin is positive π(5π/3)=5π/3+sinβ‘ 2(5π/3) = 5π/3+sinβ‘γ10π/3 γ = 5π/3+sin(3π+π/3) = 5π/3β sin π/3 = 5π/3ββ3/2 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)=ππ & f(π₯) is Minimum at π₯=0 Minimum value of f(π)=π