{"id":22276,"date":"2020-02-03T16:10:54","date_gmt":"2020-02-03T10:40:54","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=22276"},"modified":"2021-07-02T15:25:53","modified_gmt":"2021-07-02T09:55:53","slug":"1st-puc-basic-maths-question-bank-chapter-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-basic-maths-question-bank-chapter-2\/","title":{"rendered":"1st PUC Basic Maths Question Bank Chapter 2 Sets, Relations and Functions"},"content":{"rendered":"

Students can Download Basic Maths Chapter 2 Sets, Relations and Functions Questions and Answers, Notes Pdf, 1st PUC Basic 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 1st PUC Basic Maths Question Bank Chapter 2 Sets, Relations and Functions<\/h2>\n

sets<\/span><\/p>\n

Question 1.
\nIf A = {1, 2, 3, B = {2, 3, 4}, C = {3, 4, 5, 6} and U = {1, 2, 3, 4, 5, 6, 7}. Find
\n(i) (A \u222a B)’
\n(ii) Verity (A \u2229 B)’ = A\u2019 \u222a B’
\nAnswer:
\n(i) A \u222a B = {1,2, 3, 4} \u2234 (A \u222a B)’ = \u222a – (A \u222a B) = {5, 6, 7}
\n(ii) (A \u2229 B) = {2,3}
\n(A \u2229 B)’ = \u222a – (A \u2229 B) = {1,4, 5,6} …… (1)
\n\u21d2 A’ = \u222a – A = {4, 5, 6, 7}
\nB’ = \u222a – B = {1, 5, 6, 7}
\n\u2234 A’ \u222a B\u2019 = (1,4, 5, 6, 7} ……. (2)
\nFrom (1) and (2) (A \u2229 B)’ = A’ \u222a B’<\/p>\n

Question 2.
\nn(u) = 700, n(A) = 200, n(B).= 300. n(A \u2229 B) = 100. Findn(A’ \u2229 B\u2019)
\nAnswer:
\nn(A’ \u2229 B’) = n(A \u222a B)’
\n= n(U) – n(A \u222a B) = 700 – n(A \u222a B)
\nNow n(A \u222a B) = n(A) + n(B) – n(A \u2229 B)
\n= 200 + 300 – 100 = 400
\n\u2234 n(A’ \u2229 B’) = 700 – 400 = 300<\/p>\n

\"1st<\/p>\n

Question 3.
\nIn a town three daily news papers X, Y, Z are published. 52% of the people read paper X, 61 % read Y and 78% read Z. 40% read X and Y; 38% read Y and Z, 46% read X and Z; 18% do not read any of the three news papers. Find the percentage of persons who read all the three papers.
\nAnswer:
\nLet the number of person is the town be 100.
\nNow it is given n(X) = 52, n(Y) = 61, n(Z) = 78, n(X \u2229 Y) = 40, n(Y \u2229 Z) = 38
\nn(X \u2229 Z) = 46, n(X \u222a Y \u222a Z) = 100 – 18 = 82
\nWe have
\nn(X \u222a Y \u222a Z) = n(X) + n(Y) + n(Z) – n(X \u2229 Y) – n(Y \u2229 Z) – n(X \u2229 Z) + n(X \u2229 Y \u2229 Z).
\n\u21d2 82 = 52 + 61 + 78 – 40 – 38 – 46 + n(X \u2229 Y \u2229 Z)
\n= 191 – 124 + n(X \u2229 Y \u2229 Z) = 67 + (X \u2229 Y \u2229 Z)
\n\u2234 n(X \u2229 Y \u2229 Z) = 82 – 67 = 15
\nHere 15% of the people read all the three papers.<\/p>\n

Question 4.
\nHow many integers between 1 and 500 are divisible by 2, 3, or 5.
\nAnswer:
\n\\(\\frac { 500 }{ 2 }\\) = 250 No. of integer divisible by 2 is 250.
\n\u2234 n(A) = 250
\n\\(\\frac { 500 }{ 3 }\\) = 16666 \u2234 No. of integer divisible by 3 is 166. i.e., n(B) = 166
\n\\(\\frac { 500 }{ 5 }\\) = 100 i.e., No. of integer divisibility 5 is 100 is n(C) = 100
\nNow \\(\\frac{500}{2 \\times 3}=\\frac{500}{6}\\) = 83.33 divisible 2 and 3 is 83.
\nn(A \u2229 B) = 83.
\n\"1st
\nWe have
\nn(A \u2229 B \u2229 C) = n(A) + n(B) + n(c) – n(A \u2229 B) – n(B \u2229 C) – n(A \u2229 C) + n(A \u2229 B \u2229 C)
\n= 250 + 166 + 100 – 83 – 33 – 50 + 17 = 533 – 166 = 367<\/p>\n

Question 5.
\nIf A, B and C are three subsets of the universal set U, draw the venn diagram of (A\u2019 \u2229 B’)\u2229C’
\nAnswer:
\n\"1st
\n(A’ \u2229 B’)\u2229C’ – A’ \u2229 B’ \u2229 C’ = {x\/x\u2208A’ and x\u2208B’ and x\u2208C’}
\n= {x\/x\u2209A, x\u2209B, x\u2209C}
\nThe shaded region is the compliment of the set.<\/p>\n

\"1st<\/p>\n

Relations and Functions<\/span><\/p>\n

Question 1.
\nIf A = {a, b}, B = {b, c}, C = {c, d). Find (i) (A \u00d7 B) \u222a (A \u00d7 C) (ii) A \u00d7 (B \u2229 C) .
\nAnswer:
\n(i) A \u00d7 B = {(a, b), (a, c),(b, b), (b, c)}
\nA \u00d7 C = {(a,c), (a,d), (b,c) (b,d)}
\n(A \u00d7 B) \u222a (A \u00d7 C) = {(a, b) (a, c) (b, b) (b, c) (a, d) (b, d)}
\n(ii) B \u2229 C = {c}, A \u00d7 (B \u2229 C) = {a, b} \u00d7 {c} = {(a, c), (b, c}<\/p>\n

Question 2.
\nIf (x, x + y) = (6,2). Find x and y.
\nAnswer:
\n2x = 6 x + y = 2
\n\u2234 x = 3 y = 2 – 3 = – 1.<\/p>\n

Question 3.
\nIf A = {x\/x \u2208 N and x < 3} and B = {x\/x2<\/sup> – 16 = 0 and x < 0}. FindB x A,
\nAnswer:
\n\u21d2 A = {1,2), since n2<\/sup> – 16 = 0, x = + 4, B = {-4}
\n\u2234 B \u00d7 A = {(-4,1) (-4,2)}<\/p>\n

Question 4.
\nIf f(n) = 7x + 10. Find f-1<\/sup>(x) and f(4), f-1<\/sup>(5).
\nAnswer:
\nBy data f(x) = 7x + 10
\nLet f-1<\/sup>(x) = y \u21d2 f(y) = x \u21d2 7y + 10 = x
\n\u21d2 y = \\(\\frac{x-10}{7}\\)
\nThen f-1<\/sup>(x) = \\(\\frac{x-10}{7}\\) Now f-1<\/sup>(5) = \\(\\frac{5-10}{10}=\\frac{5}{7}\\)
\nAlso, (x) = 7(4) + 10 = 28 + 10 = 38
\nf(4) . f-1<\/sup>(5) = 38. \\(\\left(\\frac{-5}{7}\\right)=\\frac{-190}{7}\\)<\/p>\n

\"1st<\/p>\n

Question 5.
\nIf f: R \u2192 R, 5 defined by f(n) = 9x – 11. Show that f is bijective and find the inverse function.
\nAnswer:
\nWe have f(x) \u21d2 9x – 11
\nLet f(x1<\/sub>) = f(x2<\/sub>) = x1<\/sub> = x2<\/sub> \u21d2 9x1<\/sub> – 1 = 9x2<\/sub> – 11
\n\u21d2 9x1<\/sub> = 9x2<\/sub> \u21d2 x1<\/sub> = x2<\/sub> f is one-one
\nTo show that f is on to, we have to show that for every y \u2208 Q (codomain) there exists x \u2208 Q (domain) such that f(x) = y.
\nLet f(x) = y \u21d2 9x – 11 = y
\n\u21d2 9x = y + 11 \u2234 x = \\(\\frac{b+11}{9}\\)
\nThen for any y \u2208 Q, we have x \u2208 Q, such that f(x) = y. ie., x = \\(\\frac{b+11}{9}\\). Hence f is on to.<\/p>\n

Question 6.
\nIf f(x) = x2<\/sup> + 2x + 1, g(x) = 2x – 3. Find (i) fog, (ii) gof.
\nAnswer:
\nBy data f(x) = x2<\/sup>(2x + 1), gx = 2x – 3
\n(i) fog fg(x) = f(2x – 3) where k = 2x – 3
\n= k2<\/sup> + 2k + 1 = (2x – 3)2<\/sup> + 2(2x – 3) + 1
\n= 4x2<\/sup> 9 – 12x + 4x – 6 + 1 = 4x2<\/sup> – 8x + 4<\/p>\n

(ii) gof = gf(x) = g(n2<\/sup> + 2x + 1) = g(k)
\n= 2k – 3 = 2(n2<\/sup> + 2x + 1) – 3
\n= 2x2<\/sup> + 4x + 2 – 3 = 2x2<\/sup> + 4x – 1
\n\u2234 gof(2) = 2(22<\/sup>) + 4(2) – 1 = 8 + 8 – 1 = 15<\/p>\n

\"1st<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Basic Maths Chapter 2 Sets, Relations and Functions Questions and Answers, Notes Pdf, 1st PUC Basic Maths Question Bank with Answers\u00a0helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations. Karnataka 1st PUC Basic Maths Question Bank Chapter 2 Sets, Relations and Functions sets Question …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/22276"}],"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=22276"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/22276\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=22276"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=22276"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=22276"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}