Miscellaneous

Chapter 8 Class 12 Application of Integrals
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Misc 19

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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 β1" )) (B) β("2 β1" ) (C) β("2 " )+1 (D) β("2 " ) Finding point of intersection B Solving π¦=cosβ‘π₯ and π¦=sππβ‘π₯ cosβ‘π₯=sππβ‘π₯ At π₯=π/4 , both are equal Also, π¦=cosβ‘π₯ = cos π/4 = 1/β2 So, B =((π )/4 , 1/β2) Area Required Area Required = Area ABCO β Area BCO Area ABCO Area ABCO = β«_0^(π/4)βγπ¦ ππ₯γ Here, π¦=cosβ‘π₯ Thus, Area ABCO =β«_0^(π/4)βγcosβ‘π₯ ππ₯γ =[sinβ‘π₯ ]_0^(π/4) =[sinβ‘γπ/4βsinβ‘0 γ ] =1/β2β0 =1/β2 Area BCO Area BCO = β«_0^(π/4)βγπ¦ ππ₯γ Here, π¦=sinβ‘π₯ Thus, Area BCO =β«_0^(π/4)βγsinβ‘π₯ ππ₯γ =[γβcππ γβ‘π₯ ]_0^(π/4) =β[cosβ‘γπ/4βcosβ‘(0) γ ] =β[1/β2β1] =1β1/β2 Therefore Area Required = Area ABCO β Area BCO =1/β2β[1β1/β2] =1/β2+1/β2β1 =2/β2β1 =βπβπ β΄ Option B is Correct

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#### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.