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Last updated at April 19, 2021 by Teachoo

Transcript

Example 12 Find intervals in which the function given by f (x) = sin 3x, x, β [0, π/2] is (a) increasing (b) decreasing. f(π₯) = sin 3π₯ where π₯ β [0 ,π/2] Finding fβ(x) fβ(π₯) = π(sinβ‘3π₯ )/ππ₯ fβ(π₯) = cos 3π₯ Γ 3 fβ(π) = 3. cos 3π Putting fβ(π) = 0 3 cos 3π₯ = 0 cos 3π₯ = 0 We know that cos ΞΈ = 0 When ΞΈ = π/2 & 3π/2 So, for cos 3π = 0 3π₯ = π/2 & 3π₯ = 3π/2 π₯ = π/(2 Γ3) & π₯ = 3π/(2 Γ 3) π = π /π & π = π /π Since π₯ = π/6 β [π ,π /π] & π₯ = π/2 β [π,π /π] β΄ Both values of π₯ are valid Plotting points on number line So, point π₯ = π/6 divides the interval into two disjoint intervals [0 ,π/6) and (π/6, π/2] Checking sign of fβ(π) fβ(π₯) = 3. cos 3π₯ Case 1: For π β (π ,π /π) 0<π₯<π/6 3 Γ 0<3π₯<3π/6 π<ππ<π /π So when π₯ β (0 ,π/6), then 3π₯ β (0 , π/2) We know that cos π½>π for π½ β (π , π /π) cos 3x >0 for 3x β (0 , π/2) cos 3x >0 for x β (0 , π/6) 3 cos 3x >0 for x β (0 , π/6) πβ²(π)>π for x β (0 , π/6) Since fβ(0) = 3 and fβ(π /π) = 0 Therefore, fβ(x) β₯ 0 for π₯ β [0 , π/6] Thus, f(x) is increasing for π₯ β [0 , π/6] Case 2: For π β (π /π, π /π) π/6<π₯<π/2 3 Γ π/6<3π₯<3π/2 π /π<ππ<ππ /π So when π₯ β(π/6 , π/2), then 3π₯ β (π/2 , 3π/2) We know that, cos π<0 for π β (π/2 , 3π/2) cos 3π₯<0 for 3π₯ β (π/2 , 3π/2) cos 3π₯<0 for π₯ β (π/6 , π/2) 3 cos 3π₯<0 for π₯ β (π/6 , π/2) fβ(x) <π for π₯ β (π/6 , π/2) Since fβ(π /π) = 0 and fβ(π /π) = 0 Therefore fβ(x) β€ 0 for π₯ β [π/6,π/2] Thus, f(x) is decreasing for π₯ β [π/6,π/2] (As cos π is negative in 2nd and 3rd quadrant) Thus, f(x) is increasing for π β [π , π /π] & f(x) is decreasing for π β [π /π , π /π]

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Chapter 6 Class 12 Application of Derivatives (Term 1)

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About the Author

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.