Misc 19 (MCQ) - Chapter 8 Class 12 Application of Integrals (Term 2)
Last updated at Aug. 11, 2021 by Teachoo
Last updated at Aug. 11, 2021 by Teachoo
Transcript
Misc 19 The area bounded by the π¦-axis, π¦=cosβ‘π₯ and π¦=sinβ‘π₯ when 0β€π₯β€π/2 is (A) 2 ( β("2 β1" )) (B) β("2 β1" ) (C) β("2 " )+1 (D) β("2 " ) Finding point of intersection B Solving π¦=cosβ‘π₯ and π¦=sππβ‘π₯ cosβ‘π₯=sππβ‘π₯ At π₯=π/4 , both are equal Also, π¦=cosβ‘π₯ = cos π/4 = 1/β2 So, B =((π )/4 , 1/β2) Area Required Area Required = Area ABCO β Area BCO Area ABCO Area ABCO = β«_0^(π/4)βγπ¦ ππ₯γ Here, π¦=cosβ‘π₯ Thus, Area ABCO =β«_0^(π/4)βγcosβ‘π₯ ππ₯γ =[sinβ‘π₯ ]_0^(π/4) =[sinβ‘γπ/4βsinβ‘0 γ ] =1/β2β0 =1/β2 Area BCO Area BCO = β«_0^(π/4)βγπ¦ ππ₯γ Here, π¦=sinβ‘π₯ Thus, Area BCO =β«_0^(π/4)βγsinβ‘π₯ ππ₯γ =[γβcππ γβ‘π₯ ]_0^(π/4) =β[cosβ‘γπ/4βcosβ‘(0) γ ] =β[1/β2β1] =1β1/β2 Therefore Area Required = Area ABCO β Area BCO =1/β2β[1β1/β2] =1/β2+1/β2β1 =2/β2β1 =βπβπ β΄ Option B is Correct
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