# Misc 14 - Chapter 6 Class 12 Application of Derivatives

Last updated at Jan. 7, 2020 by Teachoo

Last updated at Jan. 7, 2020 by Teachoo

Transcript

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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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.