Ex 6.5, 8 - At what points in interval [0, 2pi], does sin 2x

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

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

Transcript

Ex 6.5, 8 At what points in the interval [0, 2Ο€ ], does the function sin 2π‘₯ attain its maximum value? Let f(π‘₯)=sin⁑2π‘₯, π‘₯ ∈ [0 , 2πœ‹] Finding f’(𝒙) f’(π‘₯)=𝑑(sin⁑2π‘₯ )/𝑑π‘₯ f’(π‘₯)=2 cos⁑2π‘₯ Putting f’(𝒙)=𝟎 2cos 2π‘₯=0 cos 2π‘₯=0 cos 2π‘₯=cosβ‘γ€–πœ‹/2γ€— If cos π‘₯=cos⁑𝑦 Then general solution of π‘₯=2π‘›πœ‹Β±π‘¦ 2π‘₯=2π‘›πœ‹+πœ‹/2 π‘₯=(2𝑛+1) πœ‹/4 Putting n = 0 π‘₯=(0+1) πœ‹/4= πœ‹/4 Putting n = 1 π‘₯=(2+1) πœ‹/4= 3πœ‹/4 Putting n = 2 π‘₯=(2(2)+1) πœ‹/4= 5πœ‹/4 Putting n = 3 π‘₯=(2(3)+1) πœ‹/4= 7πœ‹/4 Putting n = 4 π‘₯=(2(4)+1) πœ‹/4= 9πœ‹/4 Putting n = 5 π‘₯=(2(5)+1) πœ‹/4 =11πœ‹/4>2πœ‹ Since π‘₯ ∈ [0 , 2πœ‹] So, Critical Point are π‘₯ = 0 , πœ‹/4 , 3πœ‹/4 , 5πœ‹/4 , 7πœ‹/4 & 2Ο€ Finding value of f(x) at critical points i.e. π‘₯ = 0 , πœ‹/4 , 3πœ‹/4 , 5πœ‹/4 , 7πœ‹/4 & 2Ο€

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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.