Introducing your new favourite teacher - Teachoo Black, at only βΉ83 per month
Example 12 - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at April 19, 2021 by Teachoo
Examples
Example 1
Example 2
Example 3
Example 4 Important
Example 5
Example 6
Example 7
Example 8 Important
Example 9 Important
Example 10
Example 11 Important
Example 12 You are here
Example 13 Important
Example 14
Example 15
Example 16
Example 17 Important
Example 18
Example 19
Example 20
Example 21
Example 22
Example 23
Example 24
Example 25
Example 26
Example 27
Example 28 Important
Example 29
Example 30 Important
Example 31
Example 32 Important
Example 33 Important
Example 34
Example 35 Important
Example 36
Example 37 Important
Example 38 Important
Example 39
Example 40 Important
Example 41 Important
Example 42 Important
Example 43 Important
Example 44 Important
Example 45 Important
Example 46 Important
Example 47 Important
Example 48 Important
Example 49
Example 50 Important
Example 51
Chapter 6 Class 12 Application of Derivatives
Serial order wise
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]
