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

Transcript

Example 13 Find the intervals in which the function f given by f (π‘₯)=sin⁑π‘₯+cos⁑π‘₯ , 0 ≀ π‘₯ ≀ 2πœ‹ is strictly increasing or strictly decreasing. f(π‘₯) = sin π‘₯ + cos π‘₯ 0 ≀ x ≀ 2Ο€ Finding f’(𝒙) f(π‘₯) = sin x + cos π‘₯ f’(π‘₯) = (𝑑 )/𝑑π‘₯ (sin π‘₯ + cos π‘₯) f’(π‘₯) = 𝑑(sin⁑π‘₯ )/𝑑π‘₯ + 𝑑(cos⁑π‘₯ )/𝑑π‘₯ f’(π‘₯) = "cos " π‘₯ + (βˆ’π‘ π‘–π‘›π‘₯) f’(π‘₯) = cos⁑π‘₯ – sin⁑π‘₯ Putting f’(𝒙) = 0 cos π‘₯ βˆ’ sin π‘₯ = 0 cos π‘₯ = sin π‘₯ ∴π‘₯=πœ‹/4 ,5πœ‹/4 π‘Žπ‘  0" ≀ " π‘₯ ≀ 2πœ‹ Plotting points Points π‘₯=πœ‹/4 ,5πœ‹/4 divides interval [0 , 2πœ‹] into 3 disjoint intervals [0 , πœ‹/4), (πœ‹/4,5πœ‹/4), (5πœ‹/4 , 2πœ‹] Checking sign of f’(𝒙) = cos 𝒙 – sin 𝒙 When 𝒙 ∈ [𝟎 , 𝝅/πŸ’) as 0 ≀ π‘₯ < πœ‹/4 At 𝒙 = 0 f’(π‘₯) = cos π‘₯ – sin π‘₯ f’(0) = cos 0 – sin 0 = 1 – 0 = 1 > 0 At 𝒙 = 𝝅/πŸ” ∈ (𝟎 , 𝝅/πŸ’) f’(π‘₯) = cos π‘₯ – sin⁑π‘₯ f’(πœ‹/6) = cos πœ‹/6 – sin πœ‹/6 = √3/2 – 1/2 = (√3 βˆ’ 1)/2 =(1.73 βˆ’ 1)/2=0.73/2 > 0 Thus, f’(π‘₯) > 0 for π‘₯ ∈ [0 , πœ‹/4) When 𝒙 ∈ (𝝅/πŸ’,πŸ“π…/πŸ’) As πœ‹/4 < x < 5πœ‹/4 Let us find value of f’(x) at any value of π‘₯ lies between πœ‹/4, 5πœ‹/4 Let 𝒙 = 𝝅/𝟐 ∈ (𝝅/πŸ’,πŸ“π…/πŸ’) f’ (π‘₯) = cos π‘₯ – sin π‘₯ f’(πœ‹/2) = cos πœ‹/2 – sin πœ‹/2 = 0 – 1 = – 1 < 0 f’(π‘₯) < 0 for π‘₯ ∈ (πœ‹/4,5πœ‹/4) When 𝒙 ∈ (πŸ“π…/πŸ’ , πŸπ…] as 5πœ‹/4 < π‘₯ ≀ 2πœ‹ At π‘₯ = 2Ο€ f’(π‘₯) = cos π‘₯ – sin π‘₯ f’(2πœ‹) = cos 2πœ‹ – sin 2πœ‹ = cos (πœ‹+πœ‹) – 0 = – cos Ο€ – 0 = βˆ’ (βˆ’1) βˆ’0 = 1 > 0 (As cos (πœ‹+πœƒ) = – cos ΞΈ) So, f’(π‘₯) > 0 at π‘₯ = 2Ο€ Hence f’(x) > 0 for π‘₯ ∈ (5πœ‹/4 , 2πœ‹] Thus, f is strictly increasing intervals [𝟎 , 𝝅/πŸ’)& (πŸ“π…/πŸ’ , πŸπ…] f is strictly increasing intervals (𝝅/πŸ’ , πŸ“π…/πŸ’)

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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.