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Last updated at Jan. 7, 2020 by Teachoo

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Misc 14 Find the absolute maximum and minimum values of the function f given by f (π₯) = cos2 π₯ + sinβ‘π₯, π₯ β [0, π ] f(π₯)=cos^2 π₯+sin π₯ , π₯ β [0 , π] Finding fβ(π) f(π₯)=cos^2 π₯+sin π₯ fβ(π₯)= π(cos^2β‘γπ₯ + sinβ‘π₯ γ )/ππ₯ = 2cos π₯. π(cos π₯)/ππ₯ + cos π₯ = 2cos π₯(βsin π₯)+cosβ‘π₯ = cos π₯ (β2sin π₯+1) Putting fβ(π) = 0 cos π₯ (β2 sinβ‘γπ₯+1γ )=0 π₯ = π/6 & (π )/2 are Critical points. cos π₯ = 0 cos π₯ = 0 cos π₯ = cos π/2 π₯ = π/2 β 2 sin π₯ + 1 = 0 β 2 sin π₯ = β1 sin π₯ = (β1)/(β2) sin π₯ = 1/2 sin π₯ = sin π/6 π₯ = π/6 Since our interval is π₯ β [0, π ] Critical points are π₯=0, π/6 , π/2 , π Hence Absolute maximum value = π/π & Absolute minimum value = 1

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Chapter 6 Class 12 Application of Derivatives

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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 9 years. He provides courses for Maths and Science at Teachoo.