CBSE Class 12 Sample Paper for 2025 Boards
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CBSE Class 12 Sample Paper for 2025 Boards
Last updated at June 23, 2025 by Teachoo
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Transcript
Question 32 Draw the rough sketch of the curve π¦=20 cos 2π₯; (where π/6β€π₯β€π/3) Using integration, find the area of the region bounded by the curve y=20 cos2x from the ordinates π₯=π/6 to π₯=π/3 and the π₯-axis.Now, π¦=20 cos 2π₯; Since π/6β€π₯β€π/3, We find value of y at key points At x = π /π π¦=20 cos 2(π/6) = 20 cos π/3 = 20 Γ1/2 = ππ At x = π /π π¦=20 cos 2(π/3) = 20 cos 2π/3 = 20 cos(πβπ/3) = 20 Γ β cos π/3 = 20 Γ(β1)/2 = βππ At x = π /π π¦=20 cos 2(π/4) = 20 cos π/2 = 20 Γ0 = π Thus, graph of π¦=20 cos 2π₯ is Now, Area Required = Area ADB + Area BEC + Area DEF Area ADB Area ADB = β«_(π/6)^(π/( 4))βγπ¦ ππ₯γ π¦β20 cosβ‘2π₯ = β«_(π/6)^(π /( π))βγππ πππβ‘ππ π πγ = 20[sinβ‘2π₯/2]_(π/6)^(π/4) =10[sinβ‘2(π/4)βsinβ‘2(π/6) ] =10[sinβ‘(π/2)βsinβ‘(π/6) ] =10[1ββ3/2] =10[(2 β β3)/2] =5(2 β β3) =10β5(2 β β3) =10[(2 β β3)/2] =5(2 β β3) =ππβπβπ Area BEC Area BEC = β«_(π/4)^(π/( 3))βγπ¦ ππ₯γ π¦β20 cosβ‘2π₯ = β«_(π/4)^(π /( π))βγππ πππβ‘ππ π πγ = 20[sinβ‘2π₯/2]_(π/4)^(π/3) =10[sinβ‘2(π/3)βsinβ‘2(π/4) ] =10[sinβ‘(2π/3)βsinβ‘(π/2) ] =10[sinβ‘(πβπ/3)βsinβ‘(π/2) ] =10[sinβ‘(π/3)βsinβ‘(π/2) ] =10[β3/2β1] =10 Γβ3/2β10 =5β3β10 Since β3 = 1.73, 5β3β10 is negative And, area cannot be negative β΄ Area BEC = ππβπβπ Therefore Area Required = Area ADB + Area BEC = (10β5β3)+(10β5β3) = 2 Γ(10β5β3) = ππβππβπ square unit
