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

  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

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 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 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.