{"id":20491,"date":"2020-01-21T15:48:54","date_gmt":"2020-01-21T10:18:54","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=20491"},"modified":"2021-07-02T15:28:07","modified_gmt":"2021-07-02T09:58:07","slug":"2nd-puc-maths-question-bank-chapter-8-ex-8-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-maths-question-bank-chapter-8-ex-8-2\/","title":{"rendered":"2nd PUC Maths Question Bank Chapter 8 Application of Integrals Ex 8.2"},"content":{"rendered":"

Students can Download Maths Chapter 8 Application of Integrals Ex 8.2 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers<\/a>\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.<\/p>\n

Karnataka 2nd PUC Maths Question Bank Chapter 8 Application of Integrals Ex 8.2<\/h2>\n

2nd PUC Maths Application of Integrals NCERT Text Book Questions and Answers Ex 8.2<\/h3>\n

Question 1.
\nFind the area of the circle 4x2<\/sup> + 4y2<\/sup> = 9 which is interior to the Parobla x2<\/sup> = 4y.
\nAnswer:
\nx2<\/sup> = 4y is up-ward parabola with vertex (0, 0)
\nPuting x2<\/sup> = 4y in 4x2<\/sup> + 4y2<\/sup> = 9
\n\"2nd
\n4 (4y) + 4y2<\/sup> = 9
\n\u21d2 4y2<\/sup> + 16 y – 9 = 0
\n\"2nd
\nSo required area is 2 x (shaded area is 1st<\/sup> quadrant)
\n= 2x[(area under circle from (x = o to x = \\(\\sqrt{2}\\) – area under parabola (from x = o to x = \\(\\sqrt{2}\\)]
\n\"2nd<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nFind the area bounded by curves (x – 1)2<\/sup> + y2<\/sup> = 1 and x2<\/sup> + y2<\/sup> = 1.
\nAnswer:
\n(1) (x – 1)2<\/sup> + y2<\/sup> =1 is a circle with centre (1,0) and radius 1
\n(2) x2<\/sup> + y2<\/sup> = 1 is circle with circle with centre (0,0) and radius (1)<\/p>\n

    \n
  • As shaded area is required one and is a symmetrical convex curve, so, 2 x area of half of shaded region<\/li>\n
  • It is symmentrical because radius of both are same<\/li>\n<\/ul>\n

    \"2nd<\/p>\n

    2x (area of curve x2<\/sup> + y2<\/sup> = 1 from x = \\(\\frac{1}{2}\\)to x = 1
    \n+ area curve (x -1)2<\/sup> + y2<\/sup> = 1 from x = 0 to x = \\(\\frac{1}{2}\\))
    \nAs if we do along \u2018x\u2019 axis or \u2018y\u2019 axis, we get area of 1st quadrant, so to get from 1st and 4th quadrant for a curve is this question we need to multiply by 2.
    \n\"2nd
    \n\"2nd
    \n\"2nd<\/p>\n

    Question 3.
    \nFind the area of the region bounded by the curves y = x2<\/sup> + 2, y = x, x = 0 and x = 3.
    \nAnswer:
    \nCurve y = x2<\/sup> + 2 is parabola with vertex (0, 2) (upward parabola)
    \ny = x, line points are (1, 1) (0, 0) (-1, -1) (2, 2) etc
    \nx = 0, line points are (0, 0) (0, 1) (0,-1), (0, 2) (0,-2) etc
    \nx = 3, line points are (3, 0) (3, 1), (3, 2), (3, 3) (3-1) etc…..
    \n\"2nd
    \n<\/sub>so lines meet parabola at :-
    \nline x = 0, y = 2 from y… y = x2<\/sup> + 2
    \nline x = 3, y = 11
    \nand line x = y, y2<\/sup> – y + 2 = 0 \u2192 y = not pts
    \nso does not meet
    \nRequired area is shaded area:-
    \n= (Area under curve y = x2<\/sup> + 2 from x = 0 to x = 3] – [Area under line x = y from x = 0 to x = 3]
    \n\"2nd<\/p>\n

    \"KSEEB<\/p>\n

    Question 4.
    \nUsing integration find the area of region bounded by the triangle whose vertices are
    \n(-1, 0), (1, 3) and (3, 2).
    \nAnswer:
    \nWe need line eq: for line AB, BC and AC
    \n\"2nd
    \nso line AB,
    \n\"2nd
    \n\"2nd
    \nRequired area is shaded area
    \n\u21d2[Area under line AB from x = -1 to x = 1
    \n+ Area under BC from x = 1 to x = 3
    \n– Area under AC from x = -1 to x = 3]
    \n\"2nd<\/p>\n

    Question 5.
    \nUsing integration find the area of the triangular region whose sides have the equations y 2x + 1, y = 3x + 1 add x = 4.
    \nAnswer:
    \ny = 2x – 1
    \n\"2nd
    \npoints are (0, 1), (1, 3) (-1, -1) (4, 9) etc y = 3x + 1
    \npoints are (0, 1), (1,4), (-1, -2) (4, 13) etc x = 4
    \npoints are (4, 0), (4, 1), (4, 2) etc….
    \n\"2nd
    \nRequired area is shaded area:
    \n= [Area under line y = 3x + 1 from (x = 0 to x = 4)]
    \n– [Area under line y = 2x + 1 from (x = 0 to x = 4)]
    \n\"2nd<\/p>\n

    \"KSEEB<\/p>\n

    Choose the correct answer in the following exercises 6 and 7.<\/p>\n

    Question 6.
    \nSmaller area enclosed by the circle x2<\/sup> + y2<\/sup> = 4 and the line x + y = 2 is
    \n(A) 2 (\u03c0 – 2)
    \n(B) \u03c0- 2
    \n(C) 2\u03c0 – 1
    \n(D)2(\u03c0+2)
    \nAnswer:
    \nCircle x2<\/sup> + y2<\/sup> = 4 his centre (0, 0) and radius 2
    \nline x + y = 2 pts are (1,1) (2, 0) (0, 2) etc………
    \ncircle and line intersect at pts.
    \nIt can be found by solving the equation
    \nx2<\/sup> + y2<\/sup> = 4 put (y = 2 – x)
    \n\u21d2 x2<\/sup> + (2 -x)2<\/sup> = 4 \u21d2 2x2<\/sup> – 4x + 0 = 0
    \n\u21d2 2x2<\/sup> – 4x = 0 \u21d2 x2<\/sup> – 2x = 0
    \n\u21d2 (x -2) x = 0\u00a0 \u21d2 x = 0 or x = 2
    \nso y = 2 – x , x = 0 , y = 2\u00a0\u00a0\u00a0 (0, 2)
    \nx = 2 , y = 0 , (2,0)
    \nso they meet at (0, 2) and (2, 0)
    \n\"2nd
    \nRequired area is shaded area:\u00ad-
    \n(Area under circle x2<\/sup> + y2<\/sup> = 4 from x = 0 to x = 2)
    \n– (Area under line x+y = 2 from x = 0 to x = 2)
    \n\"2nd
    \n\"2nd<\/p>\n

    \"KSEEB<\/p>\n

    Question 7.
    \nArea lying between the curves y2<\/sup> = 4x and y = 2x is
    \n(A)\\(\\frac{2}{3}\\)
    \n(B)\\(\\frac{1}{3}\\)
    \n(C)\\(\\frac{1}{4}\\)
    \n(D)\\(\\frac{3}{4}\\)
    \nAnswer:
    \nCurve y2<\/sup> = 4x is right parabola with vertex (0, 0)
    \n1. y = 2x is line with pts.
    \n(0,0) (1, 2) (-1,-2) etc…
    \n\"2nd
    \nThey meet at pts
    \ny2<\/sup> = 4x but y = 2x
    \n\u21d2 (2x)2<\/sup> = 4x \u21d2 4x2<\/sup> = 4x \u21d2 x = 1, y = 2 (1, 2)
    \nx =0, y = 0 (0, 0)
    \nRequired area is shaded area:-
    \n= [Area under curve y2<\/sup> = 4x from x = 0 to x = 1]
    \n– [Area under line y = 2x from x = 0 to x = 1]
    \n\"2nd<\/p>\n","protected":false},"excerpt":{"rendered":"

    Students can Download Maths Chapter 8 Application of Integrals Ex 8.2 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 2nd PUC Maths Question Bank Chapter 8 Application of Integrals Ex 8.2 2nd PUC Maths …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[82],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20491"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=20491"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/20491\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=20491"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=20491"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=20491"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}