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  1. Chapter 6 Class 12 Application of Derivatives
  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€ π‘₯ = 5πœ‹/4 f(5πœ‹/4)= sin 2(5πœ‹/4) = sin(5πœ‹/2) = sin(2πœ‹+ πœ‹/2) = sin (πœ‹/2) = 1 (sin (2Ο€ + ΞΈ) = sin ΞΈ) f(7πœ‹/4)= sin 2(7πœ‹/4) = sin(7πœ‹/2) = sin(3πœ‹+ πœ‹/2) = – sin (πœ‹/2) = – 1 (sin (Ο€ + ΞΈ) = – sin ΞΈ) f(2πœ‹" " )=sin 2(2πœ‹" " ) = sin 4Ο€ = 0 Hence Maximum Value of f(𝒙)=𝟏 at 𝒙 = 𝝅/πŸ’ , πŸ“π…/πŸ’

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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.