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  1. Class 12
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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‘₯ ) Since we have to find increasing and decreasing, We consider the open interval (0 , 2πœ‹) Step 1: Finding f’(π‘₯) f(π‘₯) = (4 sin⁑〖π‘₯ βˆ’ 2π‘₯ βˆ’ π‘₯π‘π‘œπ‘  π‘₯γ€—)/(2 + cos⁑π‘₯ ) = (4 sin⁑〖π‘₯ βˆ’ π‘₯(2 +cos⁑π‘₯ )γ€—)/(2 + cos⁑π‘₯ ) = (4 sin⁑π‘₯)/(2 + π‘π‘œπ‘ ) – π‘₯(2 + cos⁑π‘₯ )/(2 + cos⁑π‘₯ ) = (4 sin⁑π‘₯)/(2 + cos⁑π‘₯ )βˆ’π‘₯ Therefore, f(π‘₯)= (4 sin⁑π‘₯)/(2 + cos⁑π‘₯ )βˆ’π‘₯ Diff 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^2⁑π‘₯ +γ€– sin^2〗⁑π‘₯ ))/(2 + cos⁑π‘₯ )^2 βˆ’1 = (8 cos⁑π‘₯ + 4)/(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’(π‘₯) = cos⁑〖π‘₯ (4 βˆ’ cos⁑π‘₯ )γ€—/(2 + cos⁑π‘₯ )^2 Step 2:Putting f’(π‘₯) = 0 cos⁑π‘₯(4 βˆ’ cos⁑π‘₯ )/(2 + cos⁑π‘₯ )^2 = 0 ∴ cos π‘₯ (4βˆ’cos⁑π‘₯ ) = 0 ∴ cos π‘₯ = 0 π‘₯ = (2𝑛+1) πœ‹/2 , n ∈ Z Putting n = 0 π‘₯ = (2(0)+1) πœ‹/2 = πœ‹/2 Putting n = 1 π‘₯ = (2(1)+1) πœ‹/2 = 3πœ‹/2 cos π‘₯ = 0 4 – cos π‘₯ = 0 cos π‘₯ = 4 But βˆ’1" ≀" cos⁑π‘₯ ≀ 1 So cos π‘₯ = 4 is not possible ∴ cos π‘₯ = 0 We know that cos ΞΈ = 0 at ΞΈ = (2𝑛+1) Ο€ /2 , n ∈ Z π‘₯ = (2𝑛+1) πœ‹/2 , n ∈ Z Putting n = 0 π‘₯ = (2(0)+1) πœ‹/2 = πœ‹/2 Putting n = 1 π‘₯ = (2(1)+1) πœ‹/2 = 3πœ‹/2 Putting n = 2 π‘₯ = (2(2)+1) πœ‹/2 = 5πœ‹/2 Since π‘₯ ∈ (0, 2πœ‹) So value of π‘₯ are πœ‹/2 & 3πœ‹/2 Step 3: Plotting value of π‘₯ Thus, we divide the interval (0 , 2πœ‹) into three disjoint intervals (0 ,πœ‹/2), (πœ‹/2 ,3πœ‹/2) & (3πœ‹/2,2"Ο€" ) f(π‘₯) 𝑖𝑠 strictly increasing on (𝟎 , 𝝅/𝟐) & (πŸ‘π…/𝟐 , πŸπ…) & f(π‘₯) is strictly decreasing on (𝝅/𝟐 ,πŸ‘π…/𝟐)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.