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

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Ex 6.5, 8 At what points in the interval [0, 2π ], does the function sin 2𝑥 attain its maximum value? Let f(𝑥)=sin⁡2𝑥, 𝑥 ∈ [0 , 2𝜋] Finding f’(𝒙) f’(𝑥)=𝑑(sin⁡2𝑥 )/𝑑𝑥 f’(𝑥)=2 cos⁡2𝑥 Putting f’(𝒙)=𝟎 2cos 2𝑥=0 cos 2𝑥=0 cos 2𝑥=cos⁡〖𝜋/2〗 If cos 𝑥=cos⁡𝑦 Then general solution of 𝑥=2𝑛𝜋±𝑦 2𝑥=2𝑛𝜋+𝜋/2 𝑥=(2𝑛+1) 𝜋/4 Putting n = 0 𝑥=(0+1) 𝜋/4= 𝜋/4 Putting n = 1 𝑥=(2+1) 𝜋/4= 3𝜋/4 Putting n = 2 𝑥=(2(2)+1) 𝜋/4= 5𝜋/4 Putting n = 3 𝑥=(2(3)+1) 𝜋/4= 7𝜋/4 Putting n = 4 𝑥=(2(4)+1) 𝜋/4= 9𝜋/4 Putting n = 5 𝑥=(2(5)+1) 𝜋/4 =11𝜋/4>2𝜋 Since 𝑥 ∈ [0 , 2𝜋] So, Critical Point are 𝑥 = 0 , 𝜋/4 , 3𝜋/4 , 5𝜋/4 , 7𝜋/4 & 2π Finding value of f(x) at critical points i.e. 𝑥 = 0 , 𝜋/4 , 3𝜋/4 , 5𝜋/4 , 7𝜋/4 & 2π 𝑥 = 5𝜋/4 f(5𝜋/4)= sin 2(5𝜋/4) = sin(5𝜋/2) = sin(2𝜋+ 𝜋/2) = sin (𝜋/2) = 1 (sin (2π + θ) = sin θ) f(7𝜋/4)= sin 2(7𝜋/4) = sin(7𝜋/2) = sin(3𝜋+ 𝜋/2) = – sin (𝜋/2) = – 1 (sin (π + θ) = – sin θ) f(2𝜋" " )=sin 2(2𝜋" " ) = sin 4π = 0 Hence Maximum Value of f(𝒙)=𝟏 at 𝒙 = 𝝅/𝟒 , 𝟓𝝅/𝟒