# Ex 6.5,8 - Chapter 6 Class 12 Application of Derivatives

Last updated at Jan. 7, 2020 by Teachoo

Last updated at Jan. 7, 2020 by Teachoo

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Ex 6.5, 8 At what points in the interval [0, 2Ο ], does the function sin 2π₯ attain its maximum value? Let f(π₯)=sinβ‘2π₯, π₯ β [0 , 2π] Finding fβ(π) fβ(π₯)=π(sinβ‘2π₯ )/ππ₯ fβ(π₯)=2 cosβ‘2π₯ Putting fβ(π)=π 2cos 2π₯=0 cos 2π₯=0 cos 2π₯=cosβ‘γπ/2γ If cos π₯=cosβ‘π¦ Then general solution of π₯=2ππΒ±π¦ 2π₯=2ππ+π/2 π₯=(2π+1) π/4 Putting n = 0 π₯=(0+1) π/4= π/4 Putting n = 1 π₯=(2+1) π/4= 3π/4 Putting n = 2 π₯=(2(2)+1) π/4= 5π/4 Putting n = 3 π₯=(2(3)+1) π/4= 7π/4 Putting n = 4 π₯=(2(4)+1) π/4= 9π/4 Putting n = 5 π₯=(2(5)+1) π/4 =11π/4>2π Since π₯ β [0 , 2π] So, Critical Point are π₯ = 0 , π/4 , 3π/4 , 5π/4 , 7π/4 & 2Ο Finding value of f(x) at critical points i.e. π₯ = 0 , π/4 , 3π/4 , 5π/4 , 7π/4 & 2Ο π₯ = 5π/4 f(5π/4)= sin 2(5π/4) = sin(5π/2) = sin(2π+ π/2) = sin (π/2) = 1 (sin (2Ο + ΞΈ) = sin ΞΈ) f(7π/4)= sin 2(7π/4) = sin(7π/2) = sin(3π+ π/2) = β sin (π/2) = β 1 (sin (Ο + ΞΈ) = β sin ΞΈ) f(2π" " )=sin 2(2π" " ) = sin 4Ο = 0 Hence Maximum Value of f(π)=π at π = π /π , ππ /π

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Chapter 6 Class 12 Application of Derivatives

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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.