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

Transcript

Misc 6 Find the intervals in which the function f given by f(π₯) = (4 sinβ‘γπ₯ β 2π₯ β π₯πππ  π₯γ)/(2 + cosβ‘π₯ ) is (i) increasing (ii) decreasing. f(π₯) = (4 sinβ‘γπ₯ β 2π₯ β π₯πππ  π₯γ)/(2 + cosβ‘π₯ ) Since we have to find increasing and decreasing, We consider the open interval (0 , 2π) Step 1: Finding fβ(π₯) f(π₯) = (4 sinβ‘γπ₯ β 2π₯ β π₯πππ  π₯γ)/(2 + cosβ‘π₯ ) = (4 sinβ‘γπ₯ β π₯(2 +cosβ‘π₯ )γ)/(2 + cosβ‘π₯ ) = (4 sinβ‘π₯)/(2 + πππ ) β π₯(2 + cosβ‘π₯ )/(2 + cosβ‘π₯ ) = (4 sinβ‘π₯)/(2 + cosβ‘π₯ )βπ₯ Therefore, f(π₯)= (4 sinβ‘π₯)/(2 + cosβ‘π₯ )βπ₯ Diff w.r.t π₯ fβ(π₯) = π/ππ₯ ((4 sinβ‘π₯)/(2 + πππ  π₯) β π₯) = π/ππ₯ ((4 sinβ‘π₯)/(2 + cosβ‘π₯ )) β π(π₯)/ππ₯ = π/ππ₯ ((4 sinβ‘π₯)/(2 + cosβ‘π₯ )) β 1 = [((4 sinβ‘π₯ )^β² (2+cosβ‘π₯ )β(2+cosβ‘π₯ )^β² (4 sinβ‘π₯ ))/(2+cosβ‘π₯ )^2 ] β1 = [(4 cosβ‘π₯ (2 + cosβ‘π₯ ) β (βsinβ‘π₯ )(4 sinβ‘π₯ ))/(2 + cosβ‘π₯ )^2 ] β1 = [(8 cosβ‘π₯ + 4 cos^2β‘π₯ + 4 sin^2β‘π₯)/(2 + cosβ‘π₯ )^2 ] β1 = (8 cosβ‘π₯ + 4(cos^2β‘π₯ +γ sin^2γβ‘π₯ ))/(2 + cosβ‘π₯ )^2 β1 = (8 cosβ‘π₯ + 4)/(2 + cosβ‘π₯ )^2 β1 = (8 cosβ‘π₯ + 4 β(2 + cosβ‘π₯ )^2)/(2+ cosβ‘π₯ )^2 = (8 cosβ‘γπ₯ + 4 β (4 + cos^2β‘γπ₯ + 4 cosβ‘π₯ γ )γ)/((2 + γcosβ‘π₯γ^2 ) ) = (8 πππ  π₯ β cos^2β‘π₯ β 4 cosβ‘π₯ )/(2 +γ cosγβ‘π₯ )^2 = (4 cosβ‘γπ₯ βγ cosγ^2β‘π₯ γ)/(2 + cosβ‘π₯ )^2 fβ(π₯) = cosβ‘γπ₯ (4 β cosβ‘π₯ )γ/(2 + cosβ‘π₯ )^2 Step 2:Putting fβ(π₯) = 0 cosβ‘π₯(4 β cosβ‘π₯ )/(2 + cosβ‘π₯ )^2 = 0 β΄ cos π₯ (4βcosβ‘π₯ ) = 0 β΄ cos π₯ = 0 π₯ = (2π+1) π/2 , n β Z Putting n = 0 π₯ = (2(0)+1) π/2 = π/2 Putting n = 1 π₯ = (2(1)+1) π/2 = 3π/2 cos π₯ = 0 4 β cos π₯ = 0 cos π₯ = 4 But β1" β€" cosβ‘π₯ β€ 1 So cos π₯ = 4 is not possible β΄ cos π₯ = 0 We know that cos ΞΈ = 0 at ΞΈ = (2π+1) Ο /2 , n β Z π₯ = (2π+1) π/2 , n β Z Putting n = 0 π₯ = (2(0)+1) π/2 = π/2 Putting n = 1 π₯ = (2(1)+1) π/2 = 3π/2 Putting n = 2 π₯ = (2(2)+1) π/2 = 5π/2 Since π₯ β (0, 2π) So value of π₯ are π/2 & 3π/2 Step 3: Plotting value of π₯ Thus, we divide the interval (0 , 2π) into three disjoint intervals (0 ,π/2), (π/2 ,3π/2) & (3π/2,2"Ο" ) f(π₯) ππ  strictly increasing on (π , π/π) & (ππ/π , ππ) & f(π₯) is strictly decreasing on (π/π ,ππ/π)

Miscellaneous