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Last updated at April 16, 2024 by Teachoo

Misc 3 Find the intervals in which the function f given by f(π₯) = (4 sinβ‘γπ₯ β 2π₯ β π₯πππ π₯γ)/(2 + cosβ‘π₯ ) is (i) increasing (ii) decreasing.f(π₯) = (4 sinβ‘γπ₯ β 2π₯ β π₯πππ π₯γ)/(2 + cosβ‘π₯ ) Letβs consider the interval [π , ππ ] Finding fβ(π) f(π₯) = (4 sinβ‘γπ₯ β 2π₯ β π₯ πππ π₯γ)/(2 + cosβ‘π₯ ) f(π₯) = (4 sinβ‘γπ₯ β π₯(2 + cosβ‘π₯ )γ)/(2 + cosβ‘π₯ ) f(π₯) = (4 sinβ‘π₯)/(2 + πππ ) β π₯(2 + cosβ‘π₯ )/(2 + cosβ‘π₯ ) f(π) = (π πππβ‘π)/(π + πππβ‘π )βπ Differentiating w.r.t π₯ fβ(π₯) = π/ππ₯ ((4 sinβ‘π₯)/(2 + πππ π₯) β π₯) = π/ππ₯ ((4 sinβ‘π₯)/(2 + cosβ‘π₯ )) β π(π₯)/ππ₯ = π/ππ₯ ((4 sinβ‘π₯)/(2 + cosβ‘π₯ )) β 1 = [((4 sinβ‘π₯ )^β² (2 + cosβ‘π₯ )β(2 + cosβ‘π₯ )^β² (4 sinβ‘π₯ ))/(2+cosβ‘π₯ )^2 ] β1 = [(4 cosβ‘π₯ (2 + cosβ‘π₯ ) β (βsinβ‘π₯ )(4 sinβ‘π₯ ))/(2 + cosβ‘π₯ )^2 ] β1 = [(8 cosβ‘π₯ + 4 cos^2β‘π₯ + 4 sin^2β‘π₯)/(2 + cosβ‘π₯ )^2 ] β1 = (8 cosβ‘π₯ + 4(γπππγ^πβ‘π +γ γπππγ^πγβ‘π ))/(2 + cosβ‘π₯ )^2 β1 = (8 cosβ‘π₯ + 4(π))/(2 + cosβ‘π₯ )^2 β1 = (8 cosβ‘π₯ + 4 β(2 + cosβ‘π₯ )^2)/(2+ cosβ‘π₯ )^2 = (8 cosβ‘γπ₯ + 4 β (4 + cos^2β‘γπ₯ + 4 cosβ‘π₯ γ )γ)/((2 + γcosβ‘π₯γ^2 ) ) = (8 πππ π₯ β cos^2β‘π₯ β 4 cosβ‘π₯ )/(2 +γ cosγβ‘π₯ )^2 = (4 cosβ‘γπ₯ βγ cosγ^2β‘π₯ γ)/(2 + cosβ‘π₯ )^2 β΄ fβ(π) = πππβ‘γπ (π β πππβ‘π )γ/(π + πππβ‘π )^π Putting fβ(π) = 0 cosβ‘π₯(4 β cosβ‘π₯ )/(2 + πππ β‘π₯ )^2 = 0 Thus, Numerator is 0 cos π (πβπππβ‘π ) = 0 Thus, cos π = 0 Since 0 β€ π₯ β€ 2π So, values of π are π /π & ππ /π 4 β cos π₯ = 0 cos π₯ = 4 But β1" β€" cosβ‘π₯ β€ 1 So cos π₯ = 4 is not possible Plotting value of π on number line So, f(π₯) is strictly increasing on (0 , π/2) & (3π/2 , 2π) f(π₯) is strictly decreasing on (π/2 ,3π/2) But we need to find Increasing & Decreasing fβ(π₯) = πππ β‘γπ₯ (4 β πππ β‘π₯ )γ/(2 + πππ β‘π₯ )^2 Thus, f(π₯) is increasing on [π , π /π] & [ππ /π , ππ ] f(π₯) is decreasing on [π /π ,ππ /π] For x = 0 fβ(0) = 1/3 For x = π/2 fβ(π/2) = 0 For x = 3π/2 fβ(3π/2) = 0 For x = 2π fβ(2π) = 1/3