{"id":17602,"date":"2020-01-03T16:39:23","date_gmt":"2020-01-03T11:09:23","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=17602"},"modified":"2021-07-02T15:31:23","modified_gmt":"2021-07-02T10:01:23","slug":"2nd-puc-maths-question-bank-chapter-1-ex-1-4","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/2nd-puc-maths-question-bank-chapter-1-ex-1-4\/","title":{"rendered":"2nd PUC Maths Question Bank Chapter 1 Relations and Functions Ex 1.4"},"content":{"rendered":"

Students can Download Maths Chapter 1 Relations and Functions Ex 1.4 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 1 Relations and Functions Ex 1.4<\/h2>\n

2nd PUC Maths Relations and Functions NCERT Text Book Questions and Answers Ex 1.4<\/h3>\n

Question 1.
\nDetermine whether or not each of the definition of given below gives a binary operation. In the event that is not a binary operation, give justification for this.
\n(i) On Z+<\/sup>, * define by a * b = a – b
\nAnswer:
\n2 \u2208 Z+<\/sup>, 5\u00a0\u2208 Z+<\/sup>, 2 – 5 = -3 \u2209 Z+
\n<\/sup>hence not a binary operation.<\/p>\n

(ii) On Z+<\/sup>, define * by a * b =ab
\nAnswer:
\n\u2200 a \u2208 Z+<\/sup>, b \u2208 Z+<\/sup> ab \u2208 Z+<\/sup>, hence a * b \u2208 Z+
\n<\/sup>\u2234 * is a binary operation.<\/p>\n

(iii) On R, define * by a*b = ab2
\n<\/sup>Answer:
\na \u2208 R, b \u2208 R then ab2<\/sup> \u2208 R
\n\u2234 a*b \u2208 R
\n\u2234 * is a binary operation.<\/p>\n

(iv) On Z+<\/sup>, define * by a * 6 = |a – b|
\nAnswer:
\na \u2208 Z+<\/sup>, b \u2208 Z+<\/sup>, |a – b| \u2208 Z+
\n<\/sup>\u2234 a * b \u2208 Z+
\n\u2234 <\/sup>* is a binary operation.<\/p>\n

(v) On Z+<\/sup>, define* by a * b = a
\nAnswer:
\na \u2208 Z+<\/sup>, b \u2208 Z+<\/sup> then a \u2208 Z+
\n<\/sup>\u2234 a * b \u2208 Z+
\n<\/sup>\u2234 * is a binary operation.<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nFor each binary * operation defined below,determine * whether is commutative or associative.<\/p>\n

(i) On Z, define a * b = a – b
\nAnswer:
\na = 3, b = 2, a – b = 3 – 2= 1
\nb-a = 2-3 = -1
\na – b \u2260 b – a, hence not commutative
\na = 3, b = 2, c = 1
\n(a * b) * c = (3 – 2) * 1 = 1 * 1 = 0
\na*(b * c) = 3* (2 * 1) = 3*(2 – 1) = 3 * 1 = 3 – 1 = 2
\n(a * b) * c \u2260 a * (b * c), hence * is not associative<\/p>\n

(ii) On Q, define a*b = ab + 1
\nAns:
\na \u2208 Q, b \u2208 Q
\na * b = ab + 1
\nb * a = ab + 1
\n\u2234\u00a0 a * b = b * a, hence commutative ( a * b )*c = (ab+1) *c
\n= (ab + 1) c + 1 = abc + c + 1
\na*(b*c) = a* (bc + 1)
\n= a (bc + 1) + 1 = abc + a + 1
\n\u2234 (a * b) *c * a* (b * c)
\nhence * is not associative<\/p>\n

(iii) On Q,define \\(a * b=\\frac{a b}{2}\\)
\nAnswer:
\n\"2nd<\/p>\n

(iv) On Z+<\/sup>, define a*b = 2\u201c*
\nAnswer:
\na g Z+<\/sup>, be Z+<\/sup>,
\na*b = 2ab<\/sup>
\nb * a = 2ba<\/sup> = 2ab<\/sup>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 a * b = b * a
\nhence * is commutative
\n(a * b) * c = 2ab<\/sup> * c = 2ab<\/sup> * c = 22al<\/sup>\u2019x<\/sup>\u2018
\na*(b*c) = a* 2ac<\/sup> = 2a<\/sup>–2<\/sup>\u201c
\nhence (a * b) * c * a * (b * c)
\n.: * is not Associative.<\/p>\n

(v) On Z+<\/sup>, define a * b = ab
\n<\/sup>Answer:
\na \u2208 Z+<\/sup>, b \u2208 Z+<\/sup>, a * b = ab
\n<\/sup>b * a = ba
\n<\/sup>a * b \u2260 b * a, hence * is not commutative
\n(a * b) * c = ab<\/sup> * c = (ab<\/sup>)c<\/sup> = abc
\n<\/sup>a* (b * c) = a * bc = (a)bc
\n<\/sup>(a * b) * c \u2260 a* (b * c)
\nhence * is not associative.<\/p>\n

\"KSEEB<\/p>\n

(vi) On R – {- 1}, define \\(a * b=\\frac{a}{b+1}\\)
\nAnswer:
\n\"2nd<\/p>\n

Question 3.
\nConsider the binary \u2227 operation on the set {1, 2, 3, 4, 5} defined by a \u2227 b = min {a, b}. Write the operation table of the operation \u2227
\nAnswer:<\/p>\n\n\n\n\n\n\n\n\n
\u2227<\/td>\n1<\/td>\n2<\/td>\n3<\/td>\n4<\/td>\n5<\/td>\n<\/tr>\n
1<\/td>\n1<\/td>\n1<\/td>\n1<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
2<\/td>\n1<\/td>\n2<\/td>\n2<\/td>\n2<\/td>\n2<\/td>\n<\/tr>\n
3<\/td>\n1<\/td>\n2<\/td>\n3<\/td>\n3<\/td>\n3<\/td>\n<\/tr>\n
4<\/td>\n1<\/td>\n2<\/td>\n3<\/td>\n4<\/td>\n4<\/td>\n<\/tr>\n
5<\/td>\n1<\/td>\n2<\/td>\n3<\/td>\n4<\/td>\n5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Question 4.
\nConsider a binary * operation on the set {1, 2, 3, 4, 5} given by the following multiplication table (Table 1.2).
\n(i) Compute (2 * 3) * 4 and 2 * (3 * 4)
\n(ii) * Is commutative?
\n(iii) Compute (2 * 3)*(4 * 5).
\n(Hint: use the following table)<\/p>\n

Table 1.2<\/p>\n\n\n\n\n\n\n\n\n
*<\/td>\n1<\/td>\n2<\/td>\n3<\/td>\n4<\/td>\n5<\/td>\n<\/tr>\n
1<\/td>\n1<\/td>\n1<\/td>\n1<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
2<\/td>\n1<\/td>\n2<\/td>\n1<\/td>\n2<\/td>\n1<\/td>\n<\/tr>\n
3<\/td>\n1<\/td>\n1<\/td>\n3<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
4<\/td>\n1<\/td>\n2<\/td>\n1<\/td>\n4<\/td>\n1<\/td>\n<\/tr>\n
5<\/td>\n1<\/td>\n1<\/td>\n1<\/td>\n1<\/td>\n5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Answer:
\n(i) (2 * 3) * 4 = 1 * 4= 1
\n2 *\u00a0 (3 * 4) = 2 * 1 = 1<\/p>\n

(ii) Since the table is symmetric with main diagonal it is commutative.<\/p>\n

(iii) (2 * 3) * (4 * 5) = 1 * 1 = 1.<\/p>\n

\"KSEEB<\/p>\n

Question 5.
\n*’Let be the binary operation on the set {1, 2, 3, 4, 5} defined by a *’ b = H.C.F. of a and b. Is the *’ operation same as the * operation defined in Exercise 4 above? Justify your answer.
\nAnswer:
\na, b \u2208 {1,2, 3,4, 5}, a* b = H.C.F. of a and b we composition table for *’<\/p>\n\n\n\n\n\n\n\n\n
*’<\/td>\n1<\/td>\n2<\/td>\n3<\/td>\n4<\/td>\n5<\/td>\n<\/tr>\n
1<\/td>\n1<\/td>\n1<\/td>\n1<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
2<\/td>\n1<\/td>\n2<\/td>\n1<\/td>\n2<\/td>\n1<\/td>\n<\/tr>\n
3<\/td>\n1<\/td>\n1<\/td>\n3<\/td>\n1<\/td>\n1<\/td>\n<\/tr>\n
4<\/td>\n1<\/td>\n2<\/td>\n1<\/td>\n4<\/td>\n1<\/td>\n<\/tr>\n
5<\/td>\n1<\/td>\n1<\/td>\n1<\/td>\n1<\/td>\n5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Both the composition tables are exactly same hence the operation* and *’ are same.<\/p>\n

Question 6.
\nLet* be the binary operation on N given by a * b = L.C.M. of a and Find
\n(i) 5 * 7, 20 * 16
\n(ii) * Is commutative?
\n(iii) * Is associative?
\n(iv) Find the identity of * in N
\n(v) Which elements of N are invertible for the operation* ?
\nAnswer:
\n(i) 5 *7 = L.C.M. of 5, 7 = 35
\n20 *16 = L.C.M. of 20, 16 = 80.<\/p>\n

(ii) a * b = L.C.M. of (a, b) = L.C.M. (b,a)
\n= b * a
\nhence * is commutative.<\/p>\n

(iii) (a * b) * c = L.C.M. of a, b, c
\na* (b*c) = L.C.M. of a, b, c
\nhence * is associative.<\/p>\n

(iv) Let \u2018e\u2019 be the identity element, than
\na * e = e * a = a \u2200 a \u2208 N
\nL.C.M. (a, e) = L.C.M. (e, a) = a
\n\u21d2 e divides a \u2200a \u2208N
\n\u21d2 e = 1
\n\u2234 1 is the identity element.<\/p>\n

(v) Let a \u2208 N be invertible.
\n\u2234\u00a0 be N such that a * b = b * a = L.C.M. (a, b) = 1
\n\u21d2 a = 1, b = 1
\n\u2234 only invertible element in N is 1<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nIs * defined on the set {1, 2, 3, 4, 5} by a * b = L.C.M. of a and b a binary operation? Justify your answer.
\nAnswer:
\nL.C.M. of 3, 5 = 15
\n3 * 5 = L.C.M. of (3, 5) = 15 g {1,2, 3,4,5}
\nhence * is not a binary operation.<\/p>\n

Question 8.
\nLet * be the binary operation on N defined by a * b = H.C.F. of a and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N?
\nAnswer:
\na \u2208 N, b \u2208 N, a \u2208 b = H.C.F. of a and b
\n= H.C.F. of b and a = b * a
\nhence * is commutative
\na \u2208 N, b \u2208 N, e \u2208 N.
\na* (b * e) = H.C.F. a, b, c
\n= (a * b) * c * is associative.
\nLet e \u2208 N be the identity element
\na * e = e * a = a
\nH.C.F. a and e = a
\n\u2234\u00a0 a divides e \u2200 a \u2208 N
\nwhich is not possible as there does not exist such a number e, which is divisible by every number.
\n\u2234 identity element does not exist.<\/p>\n

Question 9.
\nLet * be a binary operation on the set Q of rational numbers as follows. Find which of the binary operations are commutative and which are associative.
\n(i) a * b = a – b
\nAnswer:
\na * b = a – b,\u2200 a<\/sup>> be Q
\na * b = a – b * a = b – a a * b \u2260 b * a
\nhence not commutative
\n(a * b) * c = (a – b) * c = a – b – c
\na* (b * c) = a* (b – c) = a – (b – c) = a – b + c \u2260 (a* b) * c
\nhence \u2260 is not associative<\/p>\n

\"KSEEB<\/p>\n

(ii) a * b = a2<\/sup> + b2
\n<\/sup>Answer:
\na, b \u2208 Q \u21d2 a * b = a2<\/sup> + b2
\n<\/sup>= b2<\/sup> + a2<\/sup> = b * a
\nhence * is commutative
\n(a * b) * c = (a2<\/sup> + b2<\/sup>) * c2<\/sup> = (a2<\/sup> + b2<\/sup>)2<\/sup> + c2
\n<\/sup>a*(b * c) = a* (b2 <\/sup>+ c2<\/sup>) = a2<\/sup> + (b2<\/sup> + c2<\/sup>)2
\n<\/sup>hence * is not associative<\/p>\n

(iii) a * b = a + ab
\nAnswer:
\na * b = a + ab
\nb * a = b + ab\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u2234 a * b \u2260 b * a,
\nhence * is not commutative
\n(a * b) * c = (a + ab) * c = (a + ab) + (a + ab) c
\n= a + ab + ac + abc
\na* (b * c) = a * (b + bc)
\n= a + a (b + bc)
\n= a + ab + bca
\n\u2260 (a * b) * c
\nhence * is not associative<\/p>\n

(iv) a * b = (a – b)2
\n<\/sup>Answer:
\na * b = (a – b)2<\/sup> = (b – a)2<\/sup>= b * a
\nhence * is not commutative
\n(a * b) * c = (a – b)2<\/sup> * c = ((a-b)2<\/sup> – c)2<\/sup> = (a – b)4<\/sup>+ c2<\/sup> – 2c (a – b)2
\n<\/sup>a*(b * c) = a* (b – c)2<\/sup> = [a -(b – c)]2<\/sup> = a2<\/sup> + (b – c)2<\/sup> – 2a (b – c)2
\n<\/sup>hence not associative<\/p>\n

\"KSEEB<\/p>\n

(v) \\(a * b=\\frac{a b}{4}\\)
\nAnswer:
\n\"2nd
\n\"2nd<\/p>\n

(vi) a*b = ab2
\n<\/sup>Answer:
\na \u2208 Q, be Q , a * b = ab2
\n<\/sup>b * a = ba2<\/sup> \u2234 a * b\u2260 b *\u00a0a
\nhence not commutative.
\n(a * b) * c = (ab2<\/sup>) * c = ab2<\/sup>c2
\n<\/sup>a * (b * c) = a * (bc2<\/sup>) = ab2<\/sup>c4
\n<\/sup>(a * b) * c = a * (b)
\n\u2234 * is not associative.<\/p>\n

Question 10.
\nShow that none of the operations given above has identity.
\n(i) Let \u2018e\u2019 be the identity in Q such that
\nAnswer:
\na * e = e * a = a
\na – e = a = e – a
\n\u21d2 e = 0 or e = 2a \u2200 a \u2208 Q
\nwhich is not possible. Hence no identity.<\/p>\n

(ii) a * b = a2<\/sup> + b2
\n<\/sup>Answer:
\nLet e \u2208 Q be the identity element
\na * e = a2<\/sup> + e2 <\/sup>=a e * a = c2<\/sup> + a2<\/sup> = a
\nwhich is not possible as a2<\/sup> \u2260 a \u2200 a \u2208 Q
\nhence identity element doesn\u2019t exist.<\/p>\n

(iii) a*b = a + ab
\nAnswer:
\nLet e \u2208 Q be the identity element
\na * e = a + ae = a \u21d2 ae = 0, \u21d2 e = 0
\ne * a = e + ae = a \u21d2 \\(e=\\frac{a}{1+a}\\)
\n\u2234\\(\\mathrm{e}=0 \\text { and } \\mathrm{e}=\\frac{\\mathrm{a}}{1+\\mathrm{a}} \\forall \\mathrm{a} \\in \\mathrm{Q} \\) which is not possible. Hence identity element does not exist.<\/p>\n

(iv) a * b = (a – b)2
\n<\/sup>Answer:
\nLet e be the identity element
\na * e = (a – e)2<\/sup> = a
\ne * a = (e – a)2<\/sup> = a
\n\u2234 a – e = \u2260 a, e – a = \u00b1 a
\n\u21d2 e = 0 and e = 2a \u2200 a \u2208 Q which is not possible.
\nHence no identity element.<\/p>\n

\"KSEEB<\/p>\n

(v) \\(a * b = \\frac{a b}{4}\\)
\nAnswer:
\nLet e be the identity element
\n\\(a * e=\\frac{a e}{4}=a \\text { and } e * a=\\frac{e a}{4}=a\\)
\n\u2234 e = 4
\n\u2234 4 is the identity element.<\/p>\n

(vi) a*b = ab2
\n<\/sup>Answer:
\nLet e be the identity element
\na * e = ae2<\/sup> = a
\ne * a = ea2<\/sup> = a
\n\u2234\\( \\mathrm{e}^{2}=1 \\text { and } \\mathrm{e}=\\frac{1}{\\mathrm{a}} \\quad \\forall \\mathrm{a} \\in \\mathrm{Q}\\)
\nwhich is not possible. Hence identity element does not exist.<\/p>\n

Question 11.
\nLet A = N x N and * be the binary operation on A defined by (a, b)* (c, d) = (a + c, b + d) Show that is commutative and associative. Find the identity element for * on A, if any. Answer:
\n(a, b) * (c,d) = (a + c, b + d)
\n= (c + a, d + b) = (c, d) * (a, b)
\nhence commutative
\n[(a,b) * (c,d)] * (e, f) = (a + c, b + d) * (e, f) = (a + c + e, b + d + f) = (a, b) * [(c,d) * (e, f)] hence * is associative.
\nLet (e, f) be the identity element of A
\n(e, f) * (a, b) = (a,b) * (c, f) = (a, b)
\n(a + e, b + 0 = (a, b) \u21d2 (e = 0, f = 0) \u2209 N
\nhence no identity element.<\/p>\n

Question 12.
\nState whether the following statements are true or false. Justify.<\/p>\n

(i) For an arbitrary binary operation * on a set N, a * a = a \u2200 a \u2208 N.
\nAnswer:
\nIf a * b = a+ b, then
\na * a = a + a = 2a \u2260 a
\n\u2234\u00a0 statement is false<\/p>\n

(ii) If * is a commutative binary operation N, then a*(b*c) = (c*b) * a
\nAnswer:
\nSince commutative a* (b*c) = (b*c) *a
\n= (c *b) * a
\nhence true.<\/p>\n

(iii) For every binary operation defined on a set having exactly one element a is necessarily commutative and associative .with a as the identity element and a being the inverse of a.
\n\"2nd
\nAnswer:
\nA = {a}
\n* : A x A \u2192 A defined by a* a = a \u2200 a \u2208 A Also from the table it is commutative and associative. Also a is the identity element and also inverse of a .<\/p>\n

\"KSEEB<\/p>\n

Question 13.
\nConsider a binary operation * on N defined as a * b = a3<\/sup> + b3<\/sup>. Choose the
\n(A) *Is both associative and commutative?
\n(B) *Is commutative but not associative?
\n(c) *Is associative but not commutative?
\n(D) *Is neither commutative nor associative?
\nAnswer:
\na*b = a3<\/sup> + b3<\/sup> = b3<\/sup> + a3<\/sup> = b*a
\nhence commutative
\n(a*b) * c = (a3<\/sup> + b3<\/sup>) *c = (a3<\/sup> + b3<\/sup>)3<\/sup> + c3
\n<\/sup>a * (b * c) = a * (b3<\/sup> + c3<\/sup>) = a3<\/sup> + (b3<\/sup> + c3<\/sup>)3
\n<\/sup>hence not associative
\n\u201cB \u201c is the correct answer.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Maths Chapter 1 Relations and Functions Ex 1.4 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 1 Relations and Functions Ex 1.4 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\/17602"}],"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=17602"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/17602\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=17602"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=17602"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=17602"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}