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

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 6.5,8 At what points in the interval [0, 2π ], does the function sin 2𝑥 attain its maximum value? Let f𝑥=sin2𝑥, 𝑥 ∈ 0 , 2𝜋 Step 1: Finding f’𝑥 f’𝑥=𝑑sin2𝑥𝑑𝑥 f’𝑥=2cos2𝑥 Step 2: Putting f’𝑥=0 2cos 2𝑥=0 cos 2𝑥=0 cos 2𝑥=cos𝜋2 2𝑥=2𝑛𝜋+𝜋2 𝑥=2𝑛+1𝜋4 Putting n = 0 𝑥=0+1𝜋4= 𝜋4 Putting n = 1 𝑥=2+1𝜋4= 3𝜋4 Putting n = 2 𝑥=22+1𝜋4= 5𝜋4 Putting n = 3 𝑥=23+1𝜋4= 7𝜋4 Putting n = 4 𝑥=24+1𝜋4= 9𝜋4 Putting n = 5 𝑥=25+1𝜋4 =11𝜋4>2𝜋 Since 𝑥 ∈ 0 , 2𝜋 So, Critical Point are 𝑥 = 0 , 𝜋4 , 3𝜋4 , 5𝜋4 , 7𝜋4 & 2π Step 3: Finding value of f(x) at critical points i.e. 𝑥 = 0 , 𝜋4 , 3𝜋4 , 5𝜋4 , 7𝜋4 & 2π

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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 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.