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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

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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.