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


Transcript

Misc 3 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 = [((4 sin⁑π‘₯ )^β€² (2 + cos⁑π‘₯ )βˆ’(2 + cos⁑π‘₯ )^β€² (4 sin⁑π‘₯ ))/(2+cos⁑π‘₯ )^2 ] βˆ’1 = [(4 cos⁑π‘₯ (2 + cos⁑π‘₯ ) βˆ’ (βˆ’sin⁑π‘₯ )(4 sin⁑π‘₯ ))/(2 + cos⁑π‘₯ )^2 ] βˆ’1 = [(8 cos⁑π‘₯ + 4 cos^2⁑π‘₯ + 4 sin^2⁑π‘₯)/(2 + cos⁑π‘₯ )^2 ] βˆ’1 = (8 cos⁑π‘₯ + 4(〖𝒄𝒐𝒔〗^πŸβ‘π’™ +γ€– γ€–π’”π’Šπ’γ€—^πŸγ€—β‘π’™ ))/(2 + cos⁑π‘₯ )^2 βˆ’1 = (8 cos⁑π‘₯ + 4(𝟏))/(2 + cos⁑π‘₯ )^2 βˆ’1 = (8 cos⁑π‘₯ + 4 βˆ’(2 + cos⁑π‘₯ )^2)/(2+ cos⁑π‘₯ )^2 = (8 cos⁑〖π‘₯ + 4 βˆ’ (4 + cos^2⁑〖π‘₯ + 4 cos⁑π‘₯ γ€— )γ€—)/((2 + γ€–cos⁑π‘₯γ€—^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 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.