{"id":170562,"date":"2024-05-28T15:47:41","date_gmt":"2024-05-28T10:17:41","guid":{"rendered":"https:\/\/icsesolutions.com\/?p=170562"},"modified":"2024-05-28T17:06:53","modified_gmt":"2024-05-28T11:36:53","slug":"ml-aggarwal-class-11-maths-solutions-section-a-chapter-2-chapter-test","status":"publish","type":"post","link":"https:\/\/icsesolutions.com\/ml-aggarwal-class-11-maths-solutions-section-a-chapter-2-chapter-test\/","title":{"rendered":"ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Chapter Test"},"content":{"rendered":"

Students can track their progress and improvement through regular use of Class 11 ISC Maths Solutions<\/a> Chapter 2 Relations and Functions Chapter Test.<\/p>\n

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

Question 1.
\nIf A = (1, 2, 3), B = {4, 5} and C = {5, 6}, then verify that
\n(i) A \u00d7 (B \u222a C) = (A \u00d7 B) \u222a (A \u00d7 C)
\n(ii) A \u00d7 (B \u2229 C) = (A \u00d7 B) \u2229 (A \u00d7 C)
\n(iii) A \u00d7 (B – C) = (A \u00d7 B) – (A \u00d7 C).
\nSolution:
\nGiven A = {1, 2, 3},
\nB = {4, 5}
\nand C = {5, 6)
\n(i) Now, B \u222a C = {4, 5, 6}
\nL.H.S. = A \u00d7 (B \u222a C)
\n= {1, 2, 3} \u00d7 {4, 5, 6}
\n= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
\nNowA \u00d7 B = ((1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}
\nA \u00d7 C = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6)}
\n\u2234 R.H.S.= (A \u00d7 B) \u222a (A \u00d7 C)
\n= {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
\nThus L.H.S = R.H.S
\n\u2234 A \u00d7 (B \u222a C) = (A \u00d7 B) \u222a (A \u00d7 C)<\/p>\n

(ii) Here, B \u2229 C = {5}
\nL.H.S. = A \u00d7 (B C)
\n= {1, 2, 3} \u00d7 (5)
\n= {(1, 5), (2, 5), (3, 5)}
\nNow, A \u00d7 B = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}
\nA \u00d7 C = {(1, 5), (1, 6), (2, 5), (2, 6), (3, 5), (3, 6)}
\nR.H.S. = (A \u00d7 B) \u2229 (A \u00d7 C)
\n= {(1, 5), (2, 5), (3, 5)}
\n\u2234 L.H.S = R.H.S.
\nThus, A \u00d7 (B \u2229 C) = (A \u00d7 B) \u2229 (A \u00d7 C)<\/p>\n

(iii) Now, B – C = {4}
\nL.H.S = A \u00d7 (B – C) = {1, 2, 3}
\n= {(1, 4), (2, 4), (3, 4)}
\nNow A \u00d7 B = {(1, 4), (1, 5), (2. 4), 2, 5), 2,, 4), (3. 5)}
\nA \u00d7 C = {(1, 5), (1, 6), (2, 5), (2, 6). (3, 5), (3, 6)}
\nR.H.S = (A \u00d7 B) – (A \u00d7 C)
\n= {(1, 4), (2, 4), (3, 4)}
\n\u2234 A \u00d7 (B – C) = (A \u00d7 B) – (A \u00d7 C)<\/p>\n

\"ML<\/p>\n

Question 2.
\nLet A = {2, 4, 6, 8} and B = {0, 6, 8, 9, 10}. Find the elements of (A \u2229 B) \u00d7 (A – B) corresponding to the relation \u2018is a multiple of\u2019.
\nSolution:
\nGiven A = (2, 4, 6, 8}
\nand B = {0, 6, 8, 9, 10}
\n\u2234 A \u2229 B = {6, 8}
\nand A – B = {2, 4}
\n(A \u2229 B) \u00d7 (A – B) = {(6, 2), (6, 4), (8, 2), (8, 4)}
\nNow elements corresponding to relation is \u2018a multiple of\u2019
\nsince 6 is a multiple of 2
\n\u21d2 (6, 2) \u2208 R
\n8 is a mu\u00cc!p!e of 2
\n\u21d2 (8, 2) \u2208 R
\nb a multiple of 4
\n\u21d2 (8, 4) \u2208 R
\nbut 6 is not a multiple of 4
\n\u21d2 (6, 4) \u2208 R
\n\u2234 required elements are (6, 2),(8, 2), (8, 4).<\/p>\n

Question 3.
\nLet A = {6,7, 8, 10}, B = {2, 4, 5}, a \u2208 A, b \u2208 B and R be the relation from A to B defined by a R b if and only if a is divisible by b. Write R in the roster form.
\nSolution:
\nGiven A = {6, 7,8, 10}
\nand B = {2, 4, 5}
\nand R be the relation A to B defined by a R b if f a is divisible by b.
\nsince 6 is divisible by 2
\n\u21d2 6 R 2 \u21d2 (6, 2) \u2208 R
\n6 is not divisible by 4 and 5
\n\u21d2 (6, 4), (6, 5) \u2209 R
\nAlso, 8 is divisible by 2
\n\u21d2 (8, 2) \u2208 R
\n8 is divisible by 4
\n\u21d2 (8, 4) \u2208 R
\n10 is divisible by 2
\n\u21d2 (0, 2) \u2208 R
\nand 10 is divisible by 5
\n\u21d2 (10, 5) \u2208 R
\nHence, R = {(6. 2), (8. 2), (8, 4), (10, 2), (10, 5)}.<\/p>\n

\"ML<\/p>\n

Question 4.
\nLet R = {(x, y) : x + 2y < 6, x, y \u2208 N}
\n(i) Find the domain and the range of R
\n(ii) Write R as a set of ordered pairs.
\nSolution:
\nGiven R = {(x, y) : x + 2y < 6. x, y \u2208 N}
\nsince, x, y \u2208 N, x + 2y < 6
\n\u21d2 y < \\(\\frac{6-x}{2}\\)
\nWhen x = 1 ;
\ny < \\(\\frac{6-1}{2}=\\frac{5}{2}\\)
\n\u21d2 y = 1, 2 (\u2235 y \u2208 N)
\nThus, (1, 1), (1, 2) \u2208 R
\nWhen x = 2 ;
\ny < \\(\\frac{6-2}{2}=\\frac{4}{2}\\) = 2
\n\u21d2 y = 1 (\u2235 y \u2208 N)
\n\u2234 (2, 1) \u2208 R
\nWhen x = 3 ;
\ny < \\(\\frac{6-3}{2}=\\frac{3}{2}\\)
\n\u21d2 y = 1 (\u2235 y \u2208 N)
\n\u2234 (3, 1) \u2208 R
\nWhen x = 4 ;
\ny < \\(\\frac{6-4}{2}=\\frac{2}{2}\\) = 1
\nsince there is no natural number which is less than 1.
\nSo for other values of x, we donot get y \u2208 N
\nThus R = {(1, 1), (1, 2), (2, 1), (3, 1)}<\/p>\n

(i) Domain of R = {1, 2, 3}
\nand Range of R = {1, 2}<\/p>\n

Question 5.
\nLet R = {(x, y) ; y = x + 1 and y \u2208 {0, 1, 2, 3, 4, 5)}.
\n(i) List the elements of R..
\n(ii) Represent R by an arrow diagram.
\nSolution:
\nGiven R = {(x, y) : y = x + 1 and y \u2208 {0, 1, 2, 3, 4, 5}}
\n(i) Since y = x + 1, y \u2208 {0, 1, 2, 3, 4, 5}
\nWhen y = 0
\n\u21d2 0 = x + 1
\n\u21d2 x = – 1
\n\u2234 (- 1, 0) \u2208 R
\nWhen y = 1
\n\u21d2 1 = x + 1
\n\u21d2 x = 0
\n\u2234 (0, 1) \u2208 R
\nWhen y = 2
\n\u21d2 2 = x + 1
\n\u21d2 x = 1
\n\u2234 (1, 2) \u2208 R
\nWhen y = 3
\n\u21d2 3 = x + 1
\n\u21d2 x = 2
\n\u2234 (2, 3) \u2208 R
\nWhen y = 4
\n\u21d2 4 = x + 1
\n\u21d2 x = 3
\n(3, 4) \u2208 R
\nWhen y = 5
\n\u21d2 5 = x + 1
\n\u21d2 x = 3
\n\u2234 (4, 5) \u2208 R
\nThus, R = {(- 1, 0), (0, 1), (1, 2), (2, 3), (3, 4), (4, 5)}<\/p>\n

(ii)<\/p>\n

\"ML<\/p>\n

\"ML<\/p>\n

Question 6.
\nLet f be the subset of Q \u00d7 Z defined by f = {(\\(\\frac{m}{n}\\), m) : m, n \u2208 Z, n \u2260 0} function from Q to Z ? Justify your answer.
\nSolution:
\nGiven f \u2282 Q \u00d7 Z defined by
\nf = {(\\(\\frac{m}{n}\\), m) : m, n \u2208 Z, n \u2260 0}
\nHere f (\\(\\frac{1}{2}\\)) = 1 ;
\nf (\\(\\frac{2}{4}\\)) = 2
\nsince \\(\\frac{1}{2}=\\frac{2}{4}\\)
\nSo element \\(\\frac{1}{2}\\) \u2208 Q have two images 1 and 2 in Z.
\n\u2234 f is not a function from Q to Z.
\nsince each element in Q does not have unique image in Z.<\/p>\n

Question 7.
\nLet f : X \u2192 Y be defined byf(x) = x2<\/sup> for all x \u2208 X where X = {- 2, – 1, 0, 1, 2, 3} and Y = {0, 1, 4, 7, 9, 10}.
\nWrite the relation f in the roster form. Is f a function ?
\nSolution:
\nGiven f : X \u2192 Y be defined by
\nf(x) = x2<\/sup> \u2200 x \u2208 X
\nwhere X = {- 2, – 1, 0, 1, 2, 3}
\nand Y = {0, 1, 4, 7, 9, 10}
\nWhen x = – 2,
\nf (- 2) = (- 2)2<\/sup> = 4
\n\u21d2 (- 2, 4) \u2208 R
\nWhen x = 1,
\nf (- 1) = (- 1)2<\/sup> = 1
\n\u21d2 (- 1, 1) \u2208 R
\nWhen x = 0,
\nf (0) = 02<\/sup> = 0
\n\u21d2 (0, 0) \u2208 R
\nWhen x = 1,
\nf (1) = 12<\/sup> = 1
\n\u21d2 (1, 1) \u2208 R
\nWhen x = 2,
\nf (2) = 22<\/sup> = 4
\n\u21d2 (2, 4) \u2208 R
\nWhen x = 3,
\nf (3) = 32<\/sup> = 9
\n\u21d2 (3, 9) \u2208 R
\nThus, R {(- 2, 4), (- 1, 1), (0, 0), (1, 1), (2, 4), (3, 9)}
\nsince elements 1, – 1 have same image 1.
\n\u2234 f is many one function since every element in X has unique image in Y.<\/p>\n

\"ML<\/p>\n

Question 8.
\nIs g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function ? If this is described by the relation g (x) = \u03b1x + \u03b2, then what values should be assigned to \u03b1 and \u03b2.
\nSolution:
\nGiven g = {(1, 1), (2, 3), (3, 5), (4, 7)}
\nSince every element in domain of g has a unique image in codomain of g.
\nThus g be a function.
\nSince g (x) = \u03b1x + \u03b2 ……………….(1)
\nSince (1, 1) \u2208 g
\n\u2208 when x = 1, y = 1
\n\u2234 from (1) ; we have
\n1 = \u03b1 + \u03b2 ………………(2)
\nSince (2, 3) \u2208 g i.e. when x = 2, y = 2
\n\u2234 3 = 2\u03b1 + \u03b2
\n\u2234 from (2) and (3) ;
\n\u03b1 = 2 ; \u03b2 = – 1<\/p>\n

Question 9.
\nConsider the function f (x) = x + \\(\\frac{1}{x}\\), x \u2208 R, x \u2260 0. Is f one – one?
\nSolution:
\nGiven f (x) = x + \\(\\frac{1}{x}\\)
\nHere f (2) = 2 + \\(\\frac{1}{2}\\) = \\(\\frac{5}{2}\\)
\nand f (\\(\\frac{1}{2}\\)) = \\(\\frac{1}{2}+\\frac{1}{\\frac{1}{2}}\\)
\n= \\(\\frac{1}{2}+2=\\frac{5}{2}\\)
\nSince element 2 and \\(\\frac{1}{2}\\) in domain of f have same image \u00ef in codomain of f.
\nThus different elements have same image.
\n\u2234 f is not one-one.<\/p>\n

Question 10.
\nProve that the function f : N \u2192 N defined by f (x) = 3x – 2 is one – one but not onto.
\nSolution:
\nGiven f : N \u2192 N defined by
\nf (x) = 3x – 2 \u2200 x \u2208 N
\n\u2200 x, y \u2208 N s.t. f (x) = f (y)
\n\u21d2 3x – 2 = 3y – 2
\n\u21d2 3x = 3y
\n\u21d2 x = y
\n\u2234 f is one – one.
\nSince 2 \u2208 N (codomain of f)
\nLet x \u2208 N be any element such that f (x) = 2
\n\u21d2 3x – 2 = 2
\nHence the element 2 in codomain off has no pre-image in N (domain of f)
\n\u2234 f is not onto.<\/p>\n

P.Q. Find the domain of the function f given by f (x) = \\(\\frac{1}{\\sqrt{|x|-x}}\\).
\nSolution:
\nGiven f (x) = \\(\\frac{1}{\\sqrt{|x|-x}}\\)
\nFor Df<\/sub> : f (x) must be a real number .
\n\u21d2 \\(\\frac{1}{\\sqrt{|x|-x}}\\) must be a real number
\n\u21d2 |x| – x > 0
\n\u21d2 |x| > x
\n\u21d2 x < 0 [\u2235 |x| = x if x = 0 and |x| = x if x > 0
\n\u21d2 |x| – x = 0 if x \u2265 0]
\n\u2234 Df<\/sub> = (- \u221e, 0).<\/p>\n

\"ML<\/p>\n

Question 11.
\nDetermine a quadratic function \u2018f\u2019 defined by f (x) = ax2<\/sup> + bx + c if f (0) = 6, f (2) = 11 and f (- 3) = 6.
\nSolution:
\nGiven f (x) = ax2<\/sup> + bx + c ……………..(1)
\nNow f (0) = 6
\n\u21d2 6 = 0 + 0 + c
\n\u21d2 c = 6
\nf (2) = 11
\n\u21d2 11 = 4a + 2b + c
\n\u21d2 4a + 2b = 5 ……………(2)
\nand f (- 3) = 6
\n\u21d2 6 = 9a – 3b + c
\n\u21d2 9a – 3b = 0
\n\u21d2 3a – b = 0 …………………..(3)
\nFrom (2) and (3) ; we have
\n4a + 2 (3a) = 5
\n\u21d2 10a = 5
\n\u21d2 a = \\(\\frac{1}{2}\\)
\n\u2234 from (3) ;
\nb = 3a = \\(\\frac{3}{2}\\)
\n\u2234 from (1) ;
\nf (x) = \\(\\frac{1}{2}\\) x2<\/sup> + \\(\\frac{3}{2}\\) x + 6<\/p>\n

Question 12.
\nFind the domain and the range of the functionf(x) = 2 – 3x2<\/sup>. Also find f (- 2) and the numbers which are associated with the number – 25 in its range.
\nSolution:
\nGiven f (x) = 2 – 3x2<\/sup>
\nFor Df<\/sub> : (x) must be a real number
\n\u21d2 2 – 3x2<\/sup> must be a real number
\nwhich is a real number for all x \u2208 R
\nFor Rf<\/sub> :
\nLet y = f(x)
\n= 2 – 3x2<\/sup> \u2200 x \u2264 R
\n\u21d2 3x2<\/sup> – y
\n\u21d2 x2<\/sup> = \\(\\frac{2-y}{3}\\)
\nsince x2<\/sup> \u2265 0 \u2200 x \u2208 R
\n\u21d2 \\(\\frac{2-y}{3}\\) \u2265 0
\n\u21d2 2 – y \u2265 0
\n\u21d2 y \u2264 2
\n\u2234 Rf<\/sub> = (- \u221e, 2]
\nsince – 2 \u2208 R
\n:. f (- 2) = 2 – 3 (- 2)2<\/sup>
\n= 2 – 12
\n= – 10
\nsince – 25 \u2208 Rf<\/sub>,
\nlet x \u2208 Df<\/sub> s.t. f (x) = 25
\n\u21d2 2 – 3x2<\/sup> = – 25
\n\u21d2 3x2<\/sup> = 27
\n\u21d2 x2<\/sup> = 9
\n\u21d2 x = \u00b1 3 \u2208 R<\/p>\n

\"ML<\/p>\n

Question 13.
\nFind the domain and the range of the following functions :
\n(i) \\(\\sqrt{x-3}\\)
\n(ii) \\(\\sqrt{25-x^2}\\)
\n(iii) 5 – |x + 1|
\nSolution:
\n(i) Given, f (x) = \\(\\sqrt{x-3}\\)
\nFor Df<\/sub> : f (x) must be a real number
\n\u21d2 \\(\\sqrt{x-3}\\) must be a real number
\n\u21d2 x – 3 \u2265 0
\n\u21d2 x \u2265 3
\nThus D = [3, \u221e)
\nFor Rf<\/sub> :
\nLet y = f (x)
\n= \\(\\sqrt{x-3}\\) \u2200 x \u2208 Df<\/sub>
\nas x \u2265 3
\n\u21d2 x – 3 \u2265 0
\n\u21d2 4x – 3 \u2265 0
\n\u2234 Rf<\/sub> = [0, \u221e)<\/p>\n

(ii) Given f (x) = \\(\\sqrt{25-x^2}\\)
\nFor Df<\/sub> : f (x) must be a real number
\n\u21d2 \\(\\sqrt{25-x^2}\\) must be a real number.
\n\u21d2 25 – x2<\/sup> \u2265 0
\n\u21d2 x2<\/sup> \u2264 25
\n\u21d2 |x| \u2264 5
\n\u21d2 – 5 \u2264 x \u2264 5
\n\u2234 Df<\/sub> = [- 5, 5]
\nFor Rf<\/sub> :
\nLet y = f (x)
\n= \\(\\sqrt{25-x^2}\\)
\nSince square root of real number is always non-negative
\n\u2234 y \u2265 0
\nOn squaring ; we get
\ny2<\/sup> = 25 – x2<\/sup>
\n\u21d2 x2<\/sup> = 25 – y2<\/sup>
\nsince x2<\/sup> \u2265 0
\n\u21d2 25 – y2<\/sup> \u2265 0
\n\u21d2 25 \u2265 y2<\/sup>
\n\u21d2 y2<\/sup> \u2264 25
\n\u21d2 |y| \u2264 5
\n\u21d2 – 5 \u2264 y \u2264 5 and y \u2265 0
\n\u21d2 0 \u2264 y \u2264 5
\n\u2234 Rf<\/sub> = [0, 5]<\/p>\n

(iii) Given f (x) = 5 – \\x + 1|
\nFor Df<\/sub> : f (x) must be a real number
\n\u21d2 5 – |x + 1| must be a real number
\nwhich is a real number for all x \u2208 R
\n\u2234 Df<\/sub> = R
\nFor Rf<\/sub> :
\nLet y = f (x) = 5 . |x + 1|
\nsince x \u2265 0 \u2200 x \u2208 R
\n\u21d2 |x + 1| \u2265 0 \u2200 x \u2208 R
\n\u21d2 – |x + 1| \u2264 0
\n\u21d2 5 – |x + 1| \u2264 5
\n\u21d2 y \u2264 5
\n\u2234 Rf<\/sub> = (- \u221e, 5].<\/p>\n

\"ML<\/p>\n

Question 14.
\nDraw the graph of the function f (x) = \\(\\begin{cases}1+2 x, & x<0 \\\\ 3+5 x, & x \\geq 0\\end{cases}\\). Hence, find its range.
\nSolution:
\nGiven f (x) = \\(\\begin{cases}1+2 x, & x<0 \\\\ 3+5 x, & x \\geq 0\\end{cases}\\)
\nThe graph consists of two parts.
\nCase – I :
\nWhen x < 0, f (x) = 1 + 2x<\/p>\n

\"ML<\/p>\n

Table of values are given as under :<\/p>\n\n\n\n\n
x<\/td>\n– 1<\/td>\n– 2<\/td>\n– 3<\/td>\n<\/tr>\n
y<\/td>\n– 1<\/td>\n– 3<\/td>\n\u00a0– 5<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The part of graph is a straight line as shown in adjoining figure.<\/p>\n

Case – II :
\nWhen x \u2265 0, f (x) = 3 + 5x
\nTable of values is given as under:<\/p>\n\n\n\n\n
x<\/td>\n0<\/td>\n1<\/td>\n\\(\\frac{1}{5}\\)<\/td>\n<\/tr>\n
y<\/td>\n3<\/td>\n8<\/td>\n\u00a04<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

The part of graph is a straight line as shown in adjoining figure.
\nFrom graph it is clear that Rf<\/sub> = (- \u221e, 0) \u222a [3, \u221e).<\/p>\n

\"ML<\/p>\n

Question 15.
\nIf f (x) = 2x + 5 an g (x) = x2<\/sup> – 1 are two real valued functions, find tht following functions :
\n(i) f + g
\n(ii) f – g
\n(iii) fg
\n(iv) \\(\\frac{f}{g}\\)
\n(v) \\(\\frac{g}{f}\\)
\n(vi) 3g + 2f2<\/sup>
\nSolution:
\nGiven f (x) = 2x + 5
\nand g (x) = x2<\/sup> – 1
\nHere Df<\/sub> = R ; Dg<\/sub> = R
\nSo f and g have same domains.
\n(i) (f + g) (x) = f (x) + g (x)
\n= 2x + 5 + x2<\/sup> – 1
\n= x2<\/sup> + 2x + 4 \u2200 x \u2208 R<\/p>\n

(ii) (f – g) (x) = f (x) – g (x)
\n= 2x + 5 – x2<\/sup> + 1
\n= 2x + – x t I
\n= 2x – x2<\/sup> + 6 \u2200 x \u2208 R<\/p>\n

(iii) (fg) (x) = f (x) g (x)
\n= (2x + 5) (x2<\/sup> – 1)
\n= 2x3<\/sup> + 5x2<\/sup> – 2x – 5 \u2200 x \u2208 R<\/p>\n

(iv) \\(\\left(\\frac{f}{g}\\right)(x)=\\frac{f(x)}{g(x)}=\\frac{2 x+5}{x^2-1}\\)
\nwith domain D
\nwhere D = R except g (x) = 0
\ni.e. x2<\/sup> – 1 = 0
\n\u21d2 x = \u00b1 1
\n\u21d2 D = R – {- 1}<\/p>\n

(v) \\(\\left(\\frac{g}{f}\\right)\\) (x) = \\(\\frac{g(x)}{f(x)}\\)
\n= \\(\\frac{x^2-1}{2 x+5}\\)
\nwith domain D1<\/sub>
\nwhere D1<\/sub> = R exccpt f (x) = 0
\ni.e. 2x + 5
\ni.e. x = – \\(\\frac{5}{2}\\)<\/p>\n

(vi) (3g + 2f2<\/sup>) (x) = 3 g (x) – 2f2<\/sup> (x)
\n= 3g (x) + 2 (f (x))2<\/sup>
\n= 3 (x2<\/sup> – 1) + 2 (2x + 5)2<\/sup>
\n= 3x2<\/sup> – 3 + 2 (4x2<\/sup> + 20x + 25)
\n= 11x2<\/sup> + 40x + 47 \u2200 x \u2208 R<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can track their progress and improvement through regular use of Class 11 ISC Maths Solutions Chapter 2 Relations and Functions Chapter Test. ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Chapter Test Question 1. If A = (1, 2, 3), B = {4, 5} and C = {5, 6}, …<\/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\/170562"}],"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=170562"}],"version-history":[{"count":3,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170562\/revisions"}],"predecessor-version":[{"id":170567,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/posts\/170562\/revisions\/170567"}],"wp:attachment":[{"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/media?parent=170562"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/categories?post=170562"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/icsesolutions.com\/wp-json\/wp\/v2\/tags?post=170562"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}