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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

Transcript

Misc 14 Find the absolute maximum and minimum values of the function f given by f (π‘₯) = cos2 π‘₯ + sin⁑π‘₯, π‘₯ ∈ [0, πœ‹ ] f(π‘₯)=cos^2 π‘₯+sin π‘₯ , π‘₯ ∈ [0 , πœ‹] Finding f’(𝒙) f(π‘₯)=cos^2 π‘₯+sin π‘₯ f’(π‘₯)= 𝑑(cos^2⁑〖π‘₯ + sin⁑π‘₯ γ€— )/𝑑π‘₯ = 2cos π‘₯. 𝑑(cos π‘₯)/𝑑π‘₯ + cos π‘₯ = 2cos π‘₯(βˆ’sin π‘₯)+cos⁑π‘₯ = cos π‘₯ (βˆ’2sin π‘₯+1) Putting f’(𝒙) = 0 cos π‘₯ (βˆ’2 sin⁑〖π‘₯+1γ€— )=0 π‘₯ = πœ‹/6 & (πœ‹ )/2 are Critical points. cos π‘₯ = 0 cos π‘₯ = 0 cos π‘₯ = cos πœ‹/2 π‘₯ = πœ‹/2 – 2 sin π‘₯ + 1 = 0 – 2 sin π‘₯ = –1 sin π‘₯ = (βˆ’1)/(βˆ’2) sin π‘₯ = 1/2 sin π‘₯ = sin πœ‹/6 π‘₯ = πœ‹/6 Since our interval is π‘₯ ∈ [0, πœ‹ ] Critical points are π‘₯=0, πœ‹/6 , πœ‹/2 , πœ‹ Hence Absolute maximum value = πŸ“/πŸ’ & Absolute minimum value = 1

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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.