Example 13 - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at May 6, 2021 by Teachoo
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Chapter 6 Class 12 Application of Derivatives
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Transcript
Example 13 Find the intervals in which the function f given by f (π₯)=sinβ‘π₯+cosβ‘π₯ , 0 β€ π₯ β€ 2π is strictly increasing or strictly decreasing.f(π₯) = sin π₯ + cos π₯ Finding fβ(π) fβ(π₯) = (π )/ππ₯ (sin π₯ + cos π₯) fβ(π₯) = π(sinβ‘π₯ )/ππ₯ + π(cosβ‘π₯ )/ππ₯ fβ(π₯) = "cos " π₯ + (βπ πππ₯) fβ(π) = πππβ‘π β πππβ‘π Putting fβ(π) = 0 cos π₯ β sin π₯ = 0 cos π = sin π β΄π₯=π /π ,ππ /π ππ 0" β€ " π₯ β€ 2π Plotting points So, points π₯=π/4 ,5π/4 divides interval into 3 disjoint intervals [0 , π/4), (π/4,5π/4), (5π/4 , 2π] Checking sign of π^β² (π) π^β² (π₯)" "=" cos " π₯" β sin " π₯ When π β [π , π /π) Let us find value of fβ(x) at any value of π₯ lies between 0, π/4 Thus, fβ(π) > 0 for π₯ β [0 , π/4) At π = 0 fβ(0) = cos 0 β sin 0 = 1 β 0 = 1 > 0 At π = π /π β (π , π /π) fβ(π/6) = cos π/6 β sin π/6 = β3/2 β 1/2 = (β3 β 1)/2 =(1.73 β 1)/2=0.73/2 > 0 When π β (π /π,ππ /π) As π/4 < x < 5π/4 Let us find value of fβ(x) at any value of π₯ lies between π/4, 5π/4 Thus, fβ(π) < 0 for π₯ β (π/4,5π/4) Let π = π /π β (π /π,ππ /π) fβ (π₯) = cos π₯ β sin π₯ fβ(π/2) = cos π/2 β sin π/2 = 0 β 1 = β 1 < 0 When π β (ππ /π , ππ ] As 5π/4 < π₯ β€ 2π Let us find value of fβ(x) at any value of π₯ lies between 5π/4, 2π At π = 2Ο fβ(π₯) = cos π₯ β sin π₯ fβ(2π) = cos 2π β sin 2π = 1β0 = 1 > 0 Hence, fβ(x) > 0 for π₯ β (5π/4 , 2π] When π β (ππ /π , ππ ] As 5π/4 < π₯ β€ 2π Let us find value of fβ(x) at any value of π₯ lies between 5π/4, 2π At π = 2Ο fβ(π₯) = cos π₯ β sin π₯ fβ(2π) = cos 2π β sin 2π = 1β0 = 1 > 0 Hence, fβ(x) > 0 for π₯ β (5π/4 , 2π] Thus, f is strictly increasing in intervals [π , π /π)& (ππ /π , ππ ] f is strictly increasing in intervals (π /π , ππ /π)
