CBSE Class 12 Sample Paper for 2021 Boards

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

## Find the area of the region bounded by the curves π₯ 2 + π¦ 2 = 4, y = β3π₯ πππ π₯ β ππ₯ππ  ππ π‘βπ ππππ π‘ ππ’ππππππ‘

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Question 34 (Choice 1) Find the area of the region bounded by the curves π₯^2+π¦^2=4, π¦=β3 π₯ πππ π₯ β ππ₯ππ  ππ π‘βπ ππππ π‘ ππ’ππππππ‘ Given Equation of Circle π₯2+π¦2=4 π₯2+π¦2=2^2 So, Radius = 2 β΄ Point A (2, 0) and B is (0, 2) Let point where line and circle intersect be point M Required Area = Area of shaded region = Area OMA Finding Point M Point M is intersection of line and circle Putting π=βπ π in Equation of Circle π₯2+π¦2=4 π₯2+(β3 π₯)2=4 4π₯2=4 π₯2=1 β΄ π₯=Β±1 When π=π π¦=β3 π₯=β3 So, point is (1, β3) When π=βπ π¦=β3 π₯=ββ3 So, point is (β1, ββ3) As Point M is in 1st Quadrant β΄ M = (1, β3) β΄ Required Area = Area OMP + Area PMA = β«1_0^1βπ¦1ππ₯+β«1_1^2βπ¦2ππ₯ Finding y1 and y2 For y1 β Equation of Line y = β3x So, y1 = β3x For y2 β Equation of Circle π₯^2+π¦^2=4 π¦^2=4βπ₯^2 β΄ π¦=Β±β(2^2βπ₯^2 ) As Required Area is in first Quadrant β΄ π¦2=β(2^2βπ₯^2 ) Thus, Required Area = β«1_0^1βγβ3 π₯ ππ₯γ+β«1_1^2βγβ(2^2βπ₯^2 ) ππ₯γ =β3 β«1_0^1βγπ₯ ππ₯γ+β«1_1^2βγβ(2^2βπ₯^2 ) ππ₯γ =β3 [π₯^2/2]_0^1+β«1_1^2βγβ(2^2βπ₯^2 ) ππ₯γ =β3 [1^2/2β0^2/2]_0^1+β«1_1^2βγβ(2^2βπ₯^2 ) ππ₯γ =β3/2+β«1_1^2βγβ(2^2βπ₯^2 ) ππ₯γ It is of form β«1βγβ(π^2βπ₯^2 ) ππ₯=1/2 π₯β(π^2βπ₯^2 )γ+π^2/2 γπ ππγ^(β1)β‘γπ₯/π+πγ Replacing a by 2 , we get γ=β3/2+[π₯/2 β((2)^2βπ₯^2 )+(2)^2/2 sin^(β1)β‘γπ₯/2γ ]γ_1^2 =β3/2+2/2 β((2)^2β(2)^2 )+2 sin^(β1)β‘γ2/2γβ1/2 β((2)^2β("1" )^2 )β2 sin^(β1)β‘γ1/2γ =β3/2+0+2 sin^(β1)β‘γ(1)β1/2 β(4β1)βγ 2 sin^(β1)β‘(1/2) =β3/2+2 Γπ/2ββ3/2β2Γπ/6 =π βπ/3 =ππ/π square units