Misc 6 - Find intervals in which f(x) = 4 sin x - 2x - x cos x

Misc 6 - Chapter 6 Class 12 Application of Derivatives - Part 2
Misc 6 - Chapter 6 Class 12 Application of Derivatives - Part 3 Misc 6 - Chapter 6 Class 12 Application of Derivatives - Part 4 Misc 6 - Chapter 6 Class 12 Application of Derivatives - Part 5 Misc 6 - Chapter 6 Class 12 Application of Derivatives - Part 6 Misc 6 - Chapter 6 Class 12 Application of Derivatives - Part 7

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

Transcript

Misc 6 Find the intervals in which the function f given by f(π‘₯) = (4 sin⁑〖π‘₯ βˆ’ 2π‘₯ βˆ’ π‘₯π‘π‘œπ‘  π‘₯γ€—)/(2 + cos⁑π‘₯ ) is (i) increasing (ii) decreasing.f(π‘₯) = (4 sin⁑〖π‘₯ βˆ’ 2π‘₯ βˆ’ π‘₯π‘π‘œπ‘  π‘₯γ€—)/(2 + cos⁑π‘₯ ) Let’s consider the interval [𝟎 , πŸπ…] Finding f’(𝒙) f(π‘₯) = (4 sin⁑〖π‘₯ βˆ’ 2π‘₯ βˆ’ π‘₯ π‘π‘œπ‘  π‘₯γ€—)/(2 + cos⁑π‘₯ ) f(π‘₯) = (4 sin⁑〖π‘₯ βˆ’ π‘₯(2 + cos⁑π‘₯ )γ€—)/(2 + cos⁑π‘₯ ) f(π‘₯) = (4 sin⁑π‘₯)/(2 + π‘π‘œπ‘ ) – π‘₯(2 + cos⁑π‘₯ )/(2 + cos⁑π‘₯ ) f(𝒙) = (πŸ’ π’”π’Šπ’β‘π’™)/(𝟐 + 𝒄𝒐𝒔⁑𝒙 )βˆ’π’™ Differentiating w.r.t π‘₯ f’(π‘₯) = 𝑑/𝑑π‘₯ ((4 sin⁑π‘₯)/(2 + π‘π‘œπ‘  π‘₯) βˆ’ π‘₯) = 𝑑/𝑑π‘₯ ((4 sin⁑π‘₯)/(2 + cos⁑π‘₯ )) – 𝑑(π‘₯)/𝑑π‘₯ = 𝑑/𝑑π‘₯ ((4 sin⁑π‘₯)/(2 + cos⁑π‘₯ )) – 1 Using quotient rule as (𝑒/𝑣)^β€² = (𝑒^β€² 𝑣 βˆ’π‘£^β€² 𝑒)/𝑣^2 Using quotient rule as (𝑒/𝑣)^β€² = (𝑒^β€² 𝑣 βˆ’π‘£^β€² 𝑒)/𝑣^2 = (8 π‘π‘œπ‘  π‘₯ βˆ’ cos^2⁑π‘₯ βˆ’ 4 cos⁑π‘₯ )/(2 +γ€– cos〗⁑π‘₯ )^2 = (4 cos⁑〖π‘₯ βˆ’γ€– cosγ€—^2⁑π‘₯ γ€—)/(2 + cos⁑π‘₯ )^2 ∴ f’(𝒙) = 𝒄𝒐𝒔⁑〖𝒙 (πŸ’ βˆ’ 𝒄𝒐𝒔⁑𝒙 )γ€—/(𝟐 + 𝒄𝒐𝒔⁑𝒙 )^𝟐 Putting f’(𝒙) = 0 cos⁑π‘₯(4 βˆ’ cos⁑π‘₯ )/(2 + π‘π‘œπ‘ β‘π‘₯ )^2 = 0 Thus, Numerator is 0 cos 𝒙 (πŸ’βˆ’π’„π’π’”β‘π’™ ) = 0 Thus, cos 𝒙 = 0 Since 0 ≀ π‘₯ ≀ 2πœ‹ So, values of 𝒙 are 𝝅/𝟐 & πŸ‘π…/𝟐 4 – cos π‘₯ = 0 cos π‘₯ = 4 But βˆ’1" ≀" cos⁑π‘₯ ≀ 1 So cos π‘₯ = 4 is not possible 4 – cos π‘₯ = 0 cos π‘₯ = 4 But βˆ’1" ≀" cos⁑π‘₯ ≀ 1 So cos π‘₯ = 4 is not possible Plotting value of 𝒙 on number line So, f(π‘₯) is strictly increasing on (0 , πœ‹/2) & (3πœ‹/2 , 2πœ‹) f(π‘₯) is strictly decreasing on (πœ‹/2 ,3πœ‹/2) But we need to find Increasing & Decreasing f’(π‘₯) = π‘π‘œπ‘ β‘γ€–π‘₯ (4 βˆ’ π‘π‘œπ‘ β‘π‘₯ )γ€—/(2 + π‘π‘œπ‘ β‘π‘₯ )^2 Thus, f(π‘₯) is increasing on [𝟎 , 𝝅/𝟐] & [πŸ‘π…/𝟐 , πŸπ…] f(π‘₯) is decreasing on [𝝅/𝟐 ,πŸ‘π…/𝟐] For x = 0 f’(0) = 1/3 For x = πœ‹/2 f’(πœ‹/2) = 0 For x = 3πœ‹/2 f’(3πœ‹/2) = 0 For x = 2πœ‹ f’(2πœ‹) = 1/3

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.