Miscellaneous

Chapter 6 Class 12 Application of Derivatives
Serial order wise

This video is only available for Teachoo black users

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Transcript

Misc 11 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) fβ(π₯)= π(cos^2β‘γπ₯ + sinβ‘π₯ γ )/ππ₯ = 2cos π₯. π(cos π₯)/ππ₯ + cos π₯ = 2cos π₯(βsin π₯)+cosβ‘π₯ = cos π (βππ¬π’π§ π+π) Putting fβ(π) = 0 cos π₯ (β2 sinβ‘γπ₯+1γ )=0 π₯ = π/6 , 5π/6 & π/2 are Critical points. cos π = 0 cos π₯ = 0 cos π₯ = cos π/2 π = π/π β 2 sin π + 1 = 0 β 2 sin π₯ = β1 sin π₯ = (β1)/(β2) sin π₯ = 1/2 sin π₯ = sin π/6 π = π/π Also, π = π βπ/6=ππ/π Since our interval is π β [0, π] Critical points are π₯=π, π/6 , π/2 ,5π/6,π Hence Absolute maximum value = π/π & Absolute minimum value = 1