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Miscellaneous
Last updated at March 16, 2023 by Teachoo
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