NCERT Solutions for Class 12 Maths<\/a> Chapter 8 Application of Integrals Ex 8.2 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-8-ex-8-2\/<\/p>\nNCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Exercise 8.2<\/h2>\n <\/p>\n
Ex 8.2 Class 12 NCERT Solutions Question 1.<\/strong> \nFind the area of the circle 4x\u00b2 + 4y\u00b2 = 9 which is interior to the parabola x\u00b2 = 4y. \nSolution: \n \n4x\u00b2 + 4y\u00b2 = 9 \n\u2234 x\u00b2 + y\u00b2 = \\(\\frac { 9 }{ 4 }\\) … (1) is a circle with \ncentre (0,0) and radius = \\(\\frac { 3 }{ 2 }\\) \nx\u00b2 = 4y … (2) \nSubstituting (2) in (1), we get 4y + y\u00b2 \n\u21d2 4y\u00b2 + 16y – 9 = 0 \n\u21d2 4y\u00b2 – 2y + 18y – 9 = 0 \n\u21d2 2y(2y – 1) + 9(2y – 1) = 0 \n\u21d2 (2y – 1)(2y + 9) = 0 \n\u21d2 y = \\(\\frac { 1 }{ 2 }\\), y = \\(\\frac { – 9 }{ 2 }\\) \n\u2234 y = \\(\\frac { 1 }{ 2 }\\) \nSubstitute y = \\(\\frac { 1 }{ 2 }\\) in (2) we get x = \u00b1 \\(\\sqrt{2}\\) \n\u2234 The point of intersection of the curves are (- \\(\\sqrt{2}\\), \\(\\frac { 1 }{ 2 }\\)) and (\\(\\sqrt{2}\\), \\(\\frac { 1 }{ 2 }\\)) \nSince the shaded region is symmetric with respect to y-axis, \nRequired Area = 2[Area under the circle – Area under the parabola] lying in the first quadrant between the lines x = 0 and x = \\(\\sqrt{2}\\) \n <\/p>\nExercise 8.2 Class 12 NCERT Solutions Question 2.<\/strong> \nFind the area bounded by curves (x – 1)\u00b2 + y\u00b2 = 1 and x\u00b2 + y\u00b2 = 1. \nSolution: \n \nx\u00b2 + y\u00b2 = 1 is a circle with centre at origin and intersecting positive x-axis at (1, 0). \n(x – 1)\u00b2 + y = 1 is a circle with centre at the origin and passing through the origin. \nSolving x\u00b2 + y\u00b2 = 1 and (x – 1)\u00b2 + y\u00b2 = 1, \nwe get x\u00b2 – (x – 1)\u00b2 = 0 \n(x – x + 1)(x + x – 1) = 0 \n2x – 1 = 0 \n\u2234 x = \\(\\frac { 1 }{ 2 }\\) \nWhen x = \\(\\frac { 1 }{ 2 }\\), we get (\\(\\frac { 1 }{ 2 }\\))\u00b2 + y\u00b2 = 1 \ny\u00b2 = \\(\\frac { 3 }{ 4 }\\) \n\u2234 y = \\(\\frac{\\pm \\sqrt{3}}{2}\\) \nHence the points of intersection of the two circles are (\\(\\frac { 1 }{ 2 }\\), \\(\\frac{\\pm \\sqrt{3}}{2}\\)) and (\\(\\frac { 1 }{ 2 }\\), \\(\\frac{\\pm \\sqrt{- 3}}{2}\\)). \nThe area bounded by the two curves is sym-metric w.r.t. x-axis. \n\u2234 Required area = 2 x shaded area … (1) \nArea of shaded region = Area under arc OA + Area under arc AC \n <\/p>\n <\/p>\n
8.2 Class 12 NCERT Solutions Question 3.<\/strong> \nFind the area of the region bounded by the curves y = x\u00b2 + 2, y = x, x = 0 and x = 3. \nSolution: \nEquation of the parabola is y = x\u00b2 + 2 or x\u00b2 = (y – 2) \nIts vertex is (0,2) axis is y-axis. \nBoundary lines are y = x, x = 0, x = 3. \nGraphs of the curve and lines have been shown in the figure. \nArea of the region PQRO = Area of the region OAQR – Area of region OAP \n <\/p>\nEx 8.2 Class 12 Maths Ncert Solutions Question 4.<\/strong> \nUsing integration find the area of region bounded by the triangle whose vertices are (-1, 0), (1, 3) and (3, 2). \nSolution: \nLet the vertices of the triangle are A(- 1, 0), B(1, 3) and C(3, 2). \n \nEquation of AB is y – 0 = \\(\\frac { 3 – 0 }{ 1+1 }\\)(x+1) \ni.e, y = \\(\\frac { 3 }{ 2 }\\)(x + 1) \nEquation of BC is y – 3 = \\(\\frac { 2 – 3 }{ 3 – 1 }\\)(x – 1) \ni.e, y – 3 = \\(\\frac { – 1 }{ 2 }\\)(x – 1) \u2234 y = \\(\\frac { 7 – x }{ 2 }\\) \nEquation of AC is y – 0 = \\(\\frac { 2 – 0 }{ 3 + 1 }\\) (x + 1) \ni.e., y = \\(\\frac { 1 }{ 2 }\\)(x + 1) \nArea of \u2206 ABC = Area of \u2206 ADB + Area of trapezium BDEC – Area of \u2206 AEC \n <\/p>\n <\/p>\n
Exercise 8.2 Class 12 Maths Ncert Solutions Question 5.<\/strong> \nUsing integration find the area of the triangular region whose sides have the equations y = 2x + 1,y = 3x + 1 and x = 4. \nSolution: \nThe equations of sides of the triangle are \ny = 2x+ 1 … (1), y = 3x + 1 … (2), x = 4 … (3) \nSolving equations (1) and (2), we get x = 0 and y = 1 \nLet A be the point (0,1) \nSolving equations (1) and (3), we get x = 4 and y = 9. \nLet B be the point (4, 9) Solving equations (2) and (3) we get x = 4 and y = 13 \nLet C be the point (4, 13) \n <\/p>\nEx 8.2 Class12 NCERT Solutions Question 6.<\/strong> \nSmaller area bounded by the circle x\u00b2 + y\u00b2 = 4 and the line x + y = 2 \n(a) 2 (\u03c0 – 2) \n(b) \u03c0 – 2 \n(c) 2\u03c0 – 1 \n(d) 2(\u03c0 + 2) \nSolution: \nx\u00b2 + y\u00b2 = 4 intersects the positive x-axis at (2, 0) and the positive y-axis at (0, 2). The line x + y = 2 passes through (2, 0) and (0, 2). \n \nRequired area is shaded in the figure. Required Area = (Area of circle – Area under the line) lying in the first quadrant \n <\/p>\n <\/p>\n
Ex 8.2 Class 12 Ncert Solutions Question 7.<\/strong> \nArea lying between the curves y\u00b2 = 4x and y = 2x. \n(a) \\(\\frac { 2 }{ 3 }\\) \n(b) \\(\\frac { 1 }{ 3 }\\) \n(c) \\(\\frac { 1 }{ 4 }\\) \n(d) \\(\\frac { 3 }{ 4 }\\) \nSolution: \n(b) \\(\\frac { 1 }{ 3 }\\) \n \nThe given curves are y\u00b2 = 4x … (1) \ny = 2x … (2) \n4x(x – 1) = 0 \u21d2 x = 0, x = 1 \nWhen x = 0, y = 0 \nWhen x = 1, y = 2 \nThe points of intersection are (0, 0) and (1, 2) \nRequired area is shaded in the figure. \n <\/p>\n","protected":false},"excerpt":{"rendered":"These NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2 Questions and Answers are prepared by our highly skilled subject experts. https:\/\/mcq-questions.com\/ncert-solutions-for-class-12-maths-chapter-8-ex-8-2\/ NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Exercise 8.2 Ex 8.2 Class 12 NCERT Solutions Question 1. Find the area of the circle 4x\u00b2 + …<\/p>\n
NCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"default","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":""},"categories":[3],"tags":[],"yoast_head":"\nNCERT Solutions for Class 12 Maths Chapter 8 Application of Integrals Ex 8.2 - MCQ Questions<\/title>\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n\t \n\t \n\t \n