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Ex 6.5, 12 - Find max and min of x + sin 2x on [0, 2pi] - Ex 6.5

Ex 6.5,12 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.5,12 - Chapter 6 Class 12 Application of Derivatives - Part 3 Ex 6.5,12 - Chapter 6 Class 12 Application of Derivatives - Part 4 Ex 6.5,12 - Chapter 6 Class 12 Application of Derivatives - Part 5 Ex 6.5,12 - Chapter 6 Class 12 Application of Derivatives - Part 6 Ex 6.5,12 - Chapter 6 Class 12 Application of Derivatives - Part 7 Ex 6.5,12 - Chapter 6 Class 12 Application of Derivatives - Part 8 Ex 6.5,12 - Chapter 6 Class 12 Application of Derivatives - Part 9

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Ex 6.5, 12 Find the maximum and minimum values of π‘₯ + sin 2π‘₯ on [0, 2Ο€ ] Let f(π‘₯)=π‘₯ + sin 2π‘₯ Finding f’(𝒙) 𝑓’(π‘₯)=𝑑(π‘₯ + sin 2π‘₯)/𝑑π‘₯ =1+2 cos⁑2π‘₯ Putting f’(𝒙)=𝟎 1 + 2 cos 2π‘₯=0 2 cos 2π‘₯=βˆ’1 cos 2π‘₯=(βˆ’1)/2 cos 2π‘₯=cos⁑〖2πœ‹/3γ€— General solution for cos 2π‘₯ is 2π‘₯=2π‘›πœ‹Β±2πœ‹/3 π‘₯=(2π‘›πœ‹ Β± 2πœ‹/3)/2 π‘₯= nΟ€ Β± πœ‹/3 Putting 𝑛=0 π‘₯=0(πœ‹)Β±πœ‹/3 =Β±πœ‹/3 " " cos 2π‘₯=cos⁑〖2πœ‹/3γ€— General solution for cos 2π‘₯ is 2π‘₯=2π‘›πœ‹Β±2πœ‹/3 π‘₯=(2π‘›πœ‹ Β± 2πœ‹/3)/2 π‘₯= nΟ€ Β± πœ‹/3 Putting 𝑛=0 π‘₯=0(πœ‹)Β±πœ‹/3 =Β±πœ‹/3 " " We know that π‘π‘œπ‘  60Β°=1/2 And cos is negative in 2nd & 3rd quadrant πœƒ = 180 βˆ’ 60 = 120 = 120 Γ— πœ‹/180 = 2πœ‹/3 So, π‘₯=(βˆ’πœ‹)/3,πœ‹/3 Since Given π‘₯ ∈[0 , 2πœ‹] ∴ π‘₯=πœ‹/3 only Putting 𝑛=1 π‘₯=(1)πœ‹Β±πœ‹/3 =πœ‹Β±πœ‹/3 =(3πœ‹ + πœ‹)/3 , (3πœ‹ βˆ’ πœ‹)/3 =4πœ‹/3 , 2πœ‹/3 Putting 𝑛=2 π‘₯=2(πœ‹)Β±πœ‹/3 π‘₯=2πœ‹βˆ’πœ‹/3 & 2πœ‹+πœ‹/3 π‘₯=(6πœ‹ βˆ’ πœ‹)/3 & (6πœ‹ + πœ‹)/3 π‘₯=5πœ‹/3 & 7πœ‹/3 So, π‘₯=5πœ‹/3 only Also, We are given interval π‘₯ ∈[0 , 2πœ‹] Hence , calculating f(π‘₯) at π‘₯=0 , πœ‹/3 , 2πœ‹/3 , 4πœ‹/3 , 5πœ‹/3 , 2πœ‹ Since given π‘₯ ∈[0 , 2πœ‹] & 7πœ‹/3>2πœ‹ 𝑓(2πœ‹/3)=πœ‹/3+sin⁑ 2(2πœ‹/3) = 2πœ‹/3+sin⁑〖4πœ‹/3 γ€— = 2πœ‹/3+ sin(πœ‹+πœ‹/3) = 2πœ‹/3βˆ’ sin πœ‹/3 = 2πœ‹/3βˆ’βˆš3/2 𝑓(4πœ‹/3)=4πœ‹/3+sin⁑2 (4πœ‹/3) = 4πœ‹/3+sin⁑〖8πœ‹/3 γ€— = 4πœ‹/3+ sin(3πœ‹βˆ’πœ‹/3) = 4πœ‹/3+ sin πœ‹/3 = 4πœ‹/3+√3/2 𝑓(4πœ‹/3)=4πœ‹/3+sin⁑2 (4πœ‹/3) = 4πœ‹/3+sin⁑〖8πœ‹/3 γ€— = 4πœ‹/3+ sin(3πœ‹βˆ’πœ‹/3) = 4πœ‹/3+ sin πœ‹/3 = 4πœ‹/3+√3/2 Since 3Ο€ – πœ‹/3 lies in 2nd quadrant, sin is positive 𝑓(5πœ‹/3)=5πœ‹/3+sin⁑ 2(5πœ‹/3) = 5πœ‹/3+sin⁑〖10πœ‹/3 γ€— = 5πœ‹/3+sin(3πœ‹+πœ‹/3) = 5πœ‹/3βˆ’ sin πœ‹/3 = 5πœ‹/3βˆ’βˆš3/2 Since 3Ο€ – πœ‹/3 lies in 2nd quadrant, sin is positive 𝑓(2πœ‹)=2πœ‹+sin⁑ 2πœ‹ =2πœ‹+0 =2πœ‹ Hence, f(π‘₯) is Maximum at π‘₯=2πœ‹ Maximum value of f(𝒙)=πŸπ… & f(π‘₯) is Minimum at π‘₯=0 Minimum value of f(𝒙)=𝟎

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.