{"id":170508,"date":"2024-05-25T15:29:28","date_gmt":"2024-05-25T09:59:28","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=170508"},"modified":"2024-05-27T11:23:35","modified_gmt":"2024-05-27T05:53:35","slug":"ml-aggarwal-class-11-maths-solutions-section-a-chapter-2-ex-2-3","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-11-maths-solutions-section-a-chapter-2-ex-2-3\/","title":{"rendered":"ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Ex 2.3"},"content":{"rendered":"

Effective ML Aggarwal Class 11 Solutions ISC<\/a> Chapter 2 Relations and Functions Ex 2.3 can help bridge the gap between theory and application.<\/p>\n

ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Ex 2.3<\/h2>\n

Question 1.
\nWhich of the following relations are functions ? Give reasons. If it is a function, determine its domain and range.
\n(i) {(1, 1), (1, 2), (1, 3), (1, 4)}
\n(ii) {(a, b), (b, e), (e, d), (d, e)}
\n(iii) {(1, 2), (3, 1), (1, 3), (4, i)}
\n(iv) {(1, 2), (2, 2), (3, 2), (4, 2)}
\n(v) {(2, 1), (0, – 1), (3, 1), (5, 4), (- 1, 0), (3, 4), (1, 0)}
\n(vi) ((2, 1), (4, 2), (6, 3), (8, 4), (10, 3), (14, 7)}.
\nSolution:
\n(i) Given R = {(1, 1), (1, 2), (1, 3), (1, 4)}
\nSince element I has distinct images 1, 2, 3 and 4 respectively.
\nThus element 1 has no unique image.
\nThus given relation is not a function.<\/p>\n

(ii) Given R = {(a, b), (b, e), (c, d), (d, e)}
\nSince every element has a unique image and hence R be a function,
\n\u2234 Domain of R = {a, b, e, d}
\nand Range of R = {b, e, d, e}<\/p>\n

(iii) Given Relation R = {(1, 2), (3, 1), (1, 3), (4, 1)}
\nSince element 1 has two images 2 and 3.
\nHence relation R is not a function.<\/p>\n

(iv) Given relation R = {(1, 2), (2, 2), (3, 2), (4, 2)}
\nClearly different elements have same image i.e. 2.
\nand no element has more than one image.
\nThus given relation is a function.
\n\u2234 domain = {1, 2, 3, 4)
\nand Range = {2}<\/p>\n

(v) Given relation R = {(2, 1), (0, – 1), (3, 1), (5, 4), (- 1, 0), (3, 4), (1, 0)}
\nHere, element 3 has two images 1 and 4.
\n\u2234 given relation is not a function.<\/p>\n

(vi) Given relation R = {(2, 1), (4, 2), (6, 3), (8, 4), (10, 3), (14, 7)}.
\nThus, different elements have unique \u00a1mage
\nand hence given relation be a function.
\n\u2234 domain = {2, 4, 6, 8, 10, 14}
\nand Range = {1, 2, 3, 4, 7}.<\/p>\n

\"ML<\/p>\n

Question 2.
\nIf X = {- 4, 1, 2, 3) and Y = {a, b, c), which of the following relations \u00a1s a function from X to Y?
\n(j) {(- 4, a), (1, a), (2, b)}
\n(ii) {(- 4, b), (1, b), (2, a), (3, c)}
\n(iii) {(- 4, a), (1, a), (2, b), (3, c), (1, b)}
\nSolution:
\nGiven X = {- 4, 1, 2, 3}
\nand Y = {a, b, c}
\n(i) Let R = {(- 4, a), (1, a), (2, b)}
\nbut element 3 has no image
\n\u2234 given relation is not a function from X to Y.<\/p>\n

(ii) Given R = {(- 4, b), (1, b), (2, a), (3, c)}
\nSince every element has a unique image.
\n\u2234 given relation be a function from X to Y.<\/p>\n

(iii) Let R {(- 4, a), (t. a), (2, b), (3, c), (1, b))
\nSince element 1 have two images a and b in Y
\n\u2234 R is not a function from X to Y.<\/p>\n

Question 3.
\nLet A = {1, 2, 3, 4}, B = {1, 5,9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}.
\nAre the following true?
\n(i) f is a relation from A to B.
\n(ii) f is a function from A to B?
\nJustify your answer in each case.
\nSolution:
\n(i) Given A = {1, 2, 3, 4},
\nB = {1, 5, 9, 11, 15, 16}
\nandf {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}
\nClearly f \u2286 A \u00d7 B and hence f be a relation from A to B.<\/p>\n

(ii) Since element 2 has two images 9 and 11.
\n\u2234 every element in A does not have unique image in B.
\n\u2234 f is not a function from A to B.<\/p>\n

Question 4.
\nIs the given relation a function ? Justify your answer.
\n(1) f = {(x, x) : x \u2208 R}
\n(ii) g = {(n, n2<\/sup>) : n is a positive integer}
\n(iii) h = {(x, 3) : x \u2208 R)
\nSolution:
\n(i) Given relation R\u2019 = {(x, x) : x \u2208 R}
\nSo every element of R has a unique image x in R.
\nThus R\u2019 be a function.<\/p>\n

(ii) Given g = {(n, n2<\/sup>) : n is a positive integer}
\nsince every element n \u2208 N has a unique
\nimage n2<\/sup> \u2208 N s.t. (n, n2<\/sup>) \u2208 g
\nThus g be a function from N to N.<\/p>\n

(iii) Given h = {(x, 3) : x \u2208 R)
\nSince different elements of R has a unique image 3.
\n\u2234 given relation be a function.<\/p>\n

\"ML<\/p>\n

Short and Long answer questions (5 to 24) :<\/p>\n

Question 5.
\n(i) A function \u2018\u2018 is defined by f(x) = x2<\/sup> + 1 where x \u2208 {- 1, 0, 1, 3). List the elements of f.
\n(ii) Is a func(ionf is defined byf(x) = 2x – 1 where x \u2208 {- 2, 0, 3, 5}, then find its range.
\nSolution:
\n(i) Given a function f is defined by
\nf(x) = x2<\/sup> + 1
\nwhere x \u2208 {- 1, 0, 1, 3}
\nWhen x = – 1
\n\u2234 f(x) = (- 1)2<\/sup> + 1 = 2
\n\u21d2 (- 1, 2) \u2208 f
\nWhen x = 0
\n\u21d2 (0, 1) \u2208 f
\nWhen x = 1
\n\u2234 f(x) = 12<\/sup> + 1 = 2
\n\u21d2 (1, 2) \u2208 f
\nWhen x = 3
\n\u2234 f(x) = 32<\/sup> + 1 = 10
\n\u21d2 (3, 10) \u2208 f
\n\u2234 f = {(- 1, 2), (0, 1), (1, 2), (3, 10)}.<\/p>\n

(ii) Given a functionf is defined by
\nf(x) = 2x – 1
\nwhere x \u2208 {- 2, 0, 3, 5}
\nWhen x = – 2
\n\u2234 f(x) = 2 \u00d7 (- 2) -1 = – 5
\n\u21d2 (- 2, – 5) \u2208 f
\nWhen x = 0
\n\u2234 f(x) = 0 – 1 = – 1
\n\u21d2 (0, – 1) \u2208 f
\nWhen x = 3
\n\u2234 f(x) = 6 – 1 = 5
\n\u21d2 (3, 5) \u2208 f
\nWhen x = 5
\n\u2234 f(x) = 10 – 1 = 9
\n\u21d2 (5, 9) \u2208 f
\n\u2234 f = {(- 2, – 5), (0,- 1), (3, 5), (5, 9)}
\nThus range of f = {- 5, – 1, 5, 9}.<\/p>\n

Question 6.
\nDoes the adjacent arrow diagram represent a function? If so, write its range.
\nSolution:
\nSince every element of (domain) set A has a unique image in (codomain) set B.
\nThus given arrow diagram represents a function.
\nRange = {0, 2, 7}.<\/p>\n

Question 7.
\nThe adjacent arrow diagram represents a relation. Represent the relation in roster form. Is this relation a function ? Give reasons for your answer.<\/p>\n

\"ML<\/p>\n

Solution:
\nSince 1, 3 and 5 has image 4.
\n\u2234 (1, 4), (3, 4), (5, 4) \u2208 R
\nelements 2 and 4 has images 1 and 3 respectively.
\n\u2234 (2, 1), (4, 3) \u2208 R.
\nThus, R = {(1, 4), (3, 4), (5, 4), (2, 1),(4, 3)}
\nThus every element in domain has a unique image in codomain hence the given relation be a function.<\/p>\n

\"ML<\/p>\n

Question 8.
\nThe adjacent arrow diagram represents a relation. List the pairs that satisfy the relation. Is this relation a function ? Also find the rule (relation) for the above correspondence.<\/p>\n

\"ML<\/p>\n

Clearly the relation of from A to B is given by {(- 1, 1), (0, 0), (1, 1), (2, 4), (- 2, 4)}
\nSolution:
\nClearly every element in domain A has a unique image in codomain B.
\n\u2234 Given relation be a function.
\nClearly f : A \u2192 B is defined by
\nf(x) = x2<\/sup> \u2200 x \u2208 A<\/p>\n

Question 9.
\nWrite the relation represented by the adjoining diagram, by listing the ordered pairs. State the domain, the codomain and the range of the relation. Is the relation a function ?<\/p>\n

\"ML<\/p>\n

Solution:
\nClearly the set of ordered pairs {(p, 3), (q, 7), (r, 7), (s, 1)}
\nrepresents a relation R since R \u2286 A \u00d7 B.
\ndomain (R) = {p, q, r s} ;
\ncodomain (R) = {1, 3, 5, 7}
\nand range (R) = {1, 3, 7}<\/p>\n

Question 10.
\nA = {- 2, – 1, 1, 2} and f = {(x, \\(\\frac{1}{x}\\)) : x \u2208 A}.
\n(i) List the domain of f.
\n(ii) List the range of f.
\n(iii) Is f a function ?
\nSolution:
\nGiven A = {- 2, – 1, 1, 2}
\nand f = {(x, \\(\\frac{1}{x}\\)) : x \u2208 A}
\n\u2234 f = {(- 2, – \\(\\frac{1}{2}\\)), (- 1, – 1), (1, 1), (2, \\(\\frac{1}{2}\\))}<\/p>\n

(i) Thus domain of f = {- 2, – 1, 1, 2)<\/p>\n

(ii) Range of f = (- \\(\\frac{1}{2}\\), – 1, 1, \\(\\frac{1}{2}\\)}<\/p>\n

(iii) Since every element of A has a unique image in A.
\nThus given relationf represents a function.<\/p>\n

\"ML<\/p>\n

Question 11.
\nExpress the following function as a set of ordered pairs and find its range :
\nf : X \u2192 R defined by f(x) = x3<\/sup> + 1, where X = {- 1, 0, 3, 9, 7}.
\nSolution:
\nGiven f : X \u2192 R defined by
\nf(x) = x3<\/sup> + 1, where x = {- 1, 0, 3, 9, 7}
\nWhen x = – 1
\n\u2234 f(x) = (- 1)3<\/sup> + 1 = 0
\n\u21d2 (- 1, 0) \u2208 f
\nWhen x = 0
\n\u2234 f(x) = 0 + 1 = 1
\n\u21d2 (0, 1) \u2208 f
\nWhen x = 3
\n\u2234 f(x) = 33<\/sup> + 1 = 28
\n\u21d2 (3, 28) \u2208 f
\nWhen x = 9
\n\u2234 f(x) = 93<\/sup> + 1 = 730
\n\u21d2 (9, 730) \u2208 f
\nWhen x = 7
\n\u2234 f(x) = 73<\/sup> + 1 = 344
\n\u21d2 (7, 344) \u2208 f
\n\u2234 f = {(- 1, 0), (0, 1), (3, 28), (9, 730), (7, 344)}
\n\u2234 range of f = {0, 1, 28, 730, 344}.<\/p>\n

Question 12.
\nIf f(x) = ax + b, where a and b are integers, f (- 1) – 5 and f (3) = 3, find a and b.
\nSolution:
\nGiven f(x) = ax + b
\nNow f (- 1) = – 5
\n\u21d2 – 5 = – a + b ……………….(1)
\nNow f (3) = 3
\n\u21d2 3 = 3a + b ………………….(2)
\neqn. (2) – eqn. (1) gives ;
\n8 = 4a
\n\u21d2 a = 2
\n\u2234 from (1) ;
\nb = – 5 + a = – 3
\nHence f(x) = 2x – 3.<\/p>\n

Question 13.
\nIf (a, 8) and (2, b) are ordered pairs which belong to the mapping f : x \u2192 3x + 4 where x \u2208 R, find a and b.
\nSolution:
\nGiven a function f : R \u2192 R defined by
\nf(x) = 3x + 4 \u2200 x \u2208 R
\nNow (a, 8) \u2208 f
\n\u2234 8 = 3a + 4
\n\u21d2 3a = 4
\n\u21d2 a = 4\/3
\nand (2, b) \u2208 f
\n\u2234 b = 6 + 4b = 10<\/p>\n

\"ML<\/p>\n

Question 14.
\nIf a function f from R to R is defined by f = {(x, 3x – 5) : x \u2208 R}, find the values of a and b given that (a, 4) and (1, b) belong to f.
\nSolution:
\nGiven a function f : R \u2192 R defined by
\nf(x) = 3x – 5 \u2200 x \u2208 R
\nSince (a, 4) \u2208 f
\n\u2234 4 = 3a – 5
\n\u21d2 3a = 9
\n\u21d2 a = 3
\nand (1, b) \u2208 f
\n\u2234 b = 3 x 1 – 5 = – 2<\/p>\n

Question 15.
\nLet A = {1, 2, 3}, B = {4, 5, 6, 7} and f : {(1, 4), (2, 5), (3, 6)} be a function from A to B. Show thatf is one-one but not onto.
\nSolution:
\nGiven A = {1, 2, 3} ;
\nB = {4, 5, 6,7}
\nand f = {(1, 4), (2, 5), (3, 6)} be a function from A to B
\nsince different elements 1, 2, 3 in A have different images 4, 5 and 6 in B respectively.
\n\u2234 f is one-one.
\nSince the element 7 in B has no pre-image in A.
\n\u2234 f is not onto.
\nMoreover codoinain B \u2260 Rf<\/sub>.<\/p>\n

Question 16.
\nShow that the function f : Q \u2192 Q defined by f(x) = 3x – 2 \u00a1s one-one.
\nSolution:
\nGiven a function f : Q \u2192 Q defined by
\nf(x) = 3x – 2 \u2200 x \u2208 Q
\n\u2200 x, y \u2208 Q he such that
\nf(x) = f(y)
\n\u21d2 3x – 2 = 3y – 2
\n\u21d2 x = y
\n\u2234 f is one-one.<\/p>\n

\"ML<\/p>\n

Question 17.
\nShow that the function f : N \u2192 N defined by f(x) = 2x – 1 is one-one but not onto.
\nSolution:
\nGiven the function f : N \u2192 N defined by
\nf (x) = 2x – 1 \u2200 x \u2208 N
\n\u2200 x, y \u2208 N be such that
\nf(x) = f(y)
\n\u21d2 2x – 1 = 2y – 1
\n\u21d2 2x = 2y
\n\u21d2 x = y
\n\u2234 f is one-one.
\nLet n \u2208 N be any arbitrary element such that f (n) = 2
\n\u21d2 2n – 1 = 2
\n\u21d2 2n = 3
\n\u21d2 n = \\(\\frac{3}{2}\\) \u2209 N
\nHence element 2 in codoniain of f has no pre-imag in N.
\n\u2234 f is not onto.
\nHencef is one-one and not onto.<\/p>\n

Question 18.
\nShow that the function f : N \u2192 N defined by f (n) = n2<\/sup> is injective.
\nSolution:
\nGiven, f : N \u2192 N defined by
\nf(x) = n2<\/sup> \u2200 n \u2208 N
\n\u2200 m, n \u2208 N be s.t.
\nf(n) = f(m)
\n\u21d2 n2<\/sup> = m2<\/sup>
\n\u21d2 (n – m) (n + m) = 0
\n\u21d2 n – m = 0
\n[\u2235 n, m \u2208 N
\n\u2234 n + m \u2260 0]
\n\u21d2 n = m
\nHence f is one-one i.e. injective.<\/p>\n

Question 19.
\nShow that the function f : N \u2192 N defined by f(m) = m2<\/sup> + 2m + 3 is one-one but not onto.
\nSolution:
\nGiven function f : N \u2192 N defined by
\nf (m) = m2<\/sup> + 2m + 3 \u2200 m \u2208 N
\n\u2200 m, n \u2208 N be s.t
\nf(m) = f(n)
\n\u21d2 m2<\/sup> + 2m + 3 = n2<\/sup> + 2n + 3
\n\u21d2 m2<\/sup> – n2<\/sup> + 2 (m – n) = 0
\n\u21d2 (m – n) (mn + 2) = 0
\n\u21d2 m – n = 0
\n[\u2235 m + n + 2 \u2260 0 as m, n \u2208 N]
\nThus, f is one-one.
\nLet n \u2208 N be any arbitrary element such that f(n) = 2
\n\u21d2 n2<\/sup> + 2n + 3 = 2
\n\u21d2 n2<\/sup> + 2n + 1 = 0
\n\u21d2 (n + 1)2<\/sup> = 0
\n\u21d2 n = – 1, – 1 \u2209 N
\nHence element 2 in codomain of f has no pre-image in N (domain of f).
\n\u2234 f is not onto.
\nThus, f is one-one but not onto.<\/p>\n

\"ML<\/p>\n

Question 20.
\nIn each of the following cases, state whether the function is bijective or not. Justify your answer.
\n(i) f : R \u2192 R defined by f(x) = 2x + 1
\n(ii) f : R \u2192 R defined by f(x) = 3 – 4x
\nSolution:
\n(i) Given f : R \u2192 R defined by
\nf(x) = 2x + 1 \u2200 x \u2208 R
\n\u2200 x, y \u2208 R s.t. f(x) = f (y)
\n\u21d2 2x + 1 = 2y + 1
\n\u21d2 2x = 2y
\n\u21d2 x = y
\nThus, f is one-one.<\/p>\n

Onto:
\nLet y \u2208 R be any arbitrary element
\nand let y = f(x) = 2x + 1
\n\u21d2 x = \\(\\frac{y-1}{2}\\) \u2208 R
\n[\u2235 y \u2208 R \u2203x \u2208 R]
\n\u2200 y \u2208 R \u2203 x \u2208 R s.t.
\nf(x) = f \\(\\left(\\frac{y-1}{2}\\right)\\)
\n= 2 \\(\\left(\\frac{y-1}{2}\\right)\\) + 1
\n= y – 1 + 1 = y
\n\u2234 f is onto.
\nHence f is 1 – 1 and onto.
\n\u2234 f is bijective.<\/p>\n

(ii) Given f : R \u2192 R defined by
\nf(x) = 3 – 4x \u2200 x \u2208 R
\nOne – one:
\n\u2200 x, y \u2208 R s.t. f(x) = f(y)
\n\u21d2 3 – 4x = 3 – 4y
\n\u21d2 – 4x = – 4y
\n\u21d2 x = y
\n\u2234 f is one – one.
\nOnto :
\nConsider any y \u2208 R (codomain of f)
\nand let y = f(x) = 3 – 4x
\n\u21d2 4x = 3 – y
\n\u21d2 x = \\(\\frac{3-y}{4}\\)
\nsince y \u2208 R = 3 – y \u2208 R
\n\u21d2 \\(\\frac{3-y}{4}\\) \u2208 R
\n\u21d2 x \u2208 R
\nThus, \u2200 y \u2208 R \u2203
\nx = \\(\\frac{3-y}{4}\\) \u2208 R s.t.
\nf(x) = f(\\(\\frac{3-y}{4}\\))
\n= 3 – 4 (\\(\\frac{3-y}{4}\\)) = y
\n\u2234 f is onto.
\nThus f is one-one, onto and therefore f is bijective.<\/p>\n

\"ML<\/p>\n

Question 21.
\nShow that the function f : R – {3} \u2192 R defined by f(x) = \\(\\frac{x-2}{x-3}\\) is one-one but not onto.
\nSolution:
\nGiven a function f : R – {3) \u2192 R defined by
\nf(x) = \\(\\frac{x-2}{x-3}\\)
\none – one:
\n\u2200 x, y \u2208 R – {3} s.t.f(x) = f(y)
\n\u21d2 \\(\\frac{x-2}{x-3}=\\frac{y-2}{y-3}\\)
\n\u21d2 (x – 2) (y – 3) = (x – 3) (y – 2)
\n\u21d2 xy – 3x – 2y + 6 = xy – 2x – 3y + 6
\n\u21d2 – x + y = 0
\n\u21d2 x = y
\n\u2234 f is one – one
\nSince I \u2208 R
\nLet x \u2208 R – {3) s.t.f(x) = 1
\n\u21d2 \\(\\frac{x-2}{x-3}\\) = 1
\n\u21d2 x – 2 = x – 3
\n\u21d2 2 = 3
\nwhich is impossible
\nThus element I \u2208 codomain of f has no pre-image.
\n\u2234 f is not onto.
\nHence f is 1 – 1 but not onto.<\/p>\n

Question 22.
\nShow that the function f : R \u2192 R defined by f(x) = |x + 2| is neither one – one nor onto.
\nSolution:
\nGiven the function f : R \u2192 R defined by
\nf(x) = |x + 2| \u2200 x \u2208 R
\none-one :
\nsince f(0) = |0 + 2| = 2
\nand f (- 4) = |- 4 + 2| = 2
\nThus elements 0 and – 4 have same image 2.
\n\u2234 f is many one i.e. not one-one.
\nNow, |x + 2| \u2265 0 \u2200 x \u2208 R
\n\u2234 range of f = [0, \u221e) which is a proper subset of R.
\nThus codomain of f \u2260 R
\nThus, f is not onto.
\nHence, f is neither 1 – 1 nor onto.<\/p>\n

Question 23.
\nShow that the function f : R \u2192 R defined by f(x) = 3x2<\/sup> – 2 is neither one- one nor onto.
\nSolution:
\nGiven, f : R \u2192 R defined by
\nf(x) = 3x2<\/sup> – 2 \u2200 x \u2208 R
\n\u2200 x, y \u2208 R s.t. f(x) = f(y)
\n\u21d2 3x2<\/sup> – 2 = 3y2<\/sup> – 2
\n\u21d2 3x2<\/sup> = 3y2<\/sup>
\n\u21d2 x = \u00b1 y
\n\u2234 x \u2260 y (always)
\n\u2234 f is not one – one.
\nMoreover,
\nf (1) = 3 – 2 = 1 ;
\nf (- 1) = 3 – 2 = 1
\nHence, elements 1 and – 1 have same image 1.
\n\u2234 f is not one – one.
\nSince – 5 \u2208 R.
\nLet x \u2208 R be any arbitrary element s.t. f(x) = – 5
\n\u21d2 3x2<\/sup> – 2 = – 5
\n\u21d2 3x2<\/sup> = – 3
\nThus, element – 5 has no pre-image in R.
\n\u2234 f is not onto.
\nHence f is neither one-one nor onto.<\/p>\n

\"ML<\/p>\n

Question 24.
\nIf A = R – {- 2}, B = R – {1}and f : A \u2192 B is a function defined by f(x) = \\(\\frac{x-3}{x+2}\\) for all x \u2208 A, then show that f is a one- one correspondence.
\nSolution:
\nGiven A = R – {- 2} ;
\nB = R – {1}
\nand f : A \u2192 B defined by
\nf(x) = \\(\\frac{x-3}{x+2}\\) \u2200 x \u2208 A
\nOne – one:
\n\u2200 x, y \u2208 A s.t. f(x) = f(y)
\n\u21d2 \\(\\frac{x-3}{x+2}=\\frac{y-3}{y+2}\\)
\n\u21d2 (x – 3) (y + 2) = (y – 3) (x + 2)
\n\u21d2 xy + 2x – 3y – 6 = xy + 2y – 3x – 6
\n\u21d2 5x = 5y
\n\u21d2x = y
\n\u2234 f is one-one.
\nOnto:
\nLet y \u2208 B = R – { 1 } be any arbitrary element
\nand let y = \\(\\frac{x-3}{x+2}\\)
\n\u21d2 xy + 2y = x – 3
\n\u21d2 x (y – 1) = – 2y – 3
\n\u21d2 x = \\(\\frac{2 y+3}{1-y}\\)
\nsince y \u2260 1
\n\\(\\frac{2 y+3}{1-y}\\) \u2208 R, as y \u2208 R
\nAlso \\(\\frac{2 y+3}{1-y}\\) \u2260 – 2
\nif \\(\\frac{2 y+3}{1-y}\\) = – 2
\n\u21d2 2y + 3 = – 2 + 2y
\n\u21d2 3 = – 22
\nwhich is impossible.
\nThus x = \\(\\frac{2 y+3}{1-y}\\) \u2208 R – {- 2} = A
\n\u2234 \u2200 y \u2208 B \u2203x \u2208 A s.t.
\nf(x) = f \\(\\left(\\frac{2 y+3}{1-y}\\right)\\)
\n= \\(\\frac{\\frac{2 y+3}{1-y}-3}{\\frac{2 y+3}{1-y}+2}\\)
\n= \\(\\frac{2 y+3-3+3 y}{2 y+3+2-2 y}\\) = y
\n\u2234 f is onto.
\nThus f is one-one onto.
\n\u2234 f is one-one correspondence.<\/p>\n","protected":false},"excerpt":{"rendered":"

Effective ML Aggarwal Class 11 Solutions ISC Chapter 2 Relations and Functions Ex 2.3 can help bridge the gap between theory and application. ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Ex 2.3 Question 1. Which of the following relations are functions ? Give reasons. If it is a function, …<\/p>\n","protected":false},"author":8,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6768],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170508"}],"collection":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/users\/8"}],"replies":[{"embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/comments?post=170508"}],"version-history":[{"count":3,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170508\/revisions"}],"predecessor-version":[{"id":170517,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170508\/revisions\/170517"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=170508"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=170508"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=170508"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}