{"id":8392,"date":"2024-08-24T09:38:59","date_gmt":"2024-08-24T04:08:59","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=8392"},"modified":"2024-08-24T09:38:59","modified_gmt":"2024-08-24T04:08:59","slug":"ncert-solutions-class-11-maths-chapter-2-relations-and-functions-ex-2-2","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-11-maths-chapter-2-relations-and-functions-ex-2-2\/","title":{"rendered":"NCERT Solutions Class 11 Maths Chapter 2 Relations and Functions – Ex 2.2"},"content":{"rendered":"

NCERT Solutions Class 11 Maths\u00a0<\/span><\/strong><\/span><\/h2>\n

The NCERT Solutions in English Language for Class 11 Mathematics Chapter – 2 Relations and Functions <\/strong>Exercise 2.2 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n

Chapter 2 (Relations and Functions)\u00a0<\/span><\/strong><\/span><\/h2>\n\n\n\n
Exercise – 2.2<\/span><\/strong><\/span><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

1. Let A = {1, 2, 3, \u2026 , 14}. Define a relation R from A to A by R = {(x<\/em>,\u00a0y<\/em>): 3x<\/em>\u00a0\u2013\u00a0y<\/em>\u00a0= 0, where\u00a0x<\/em>,\u00a0y<\/em> \u2208\u00a0A}. Write down its domain, codomain and range.
\n<\/strong>Solution –\u00a0<\/strong>Given, A = {1, 2, 3,\u2026,14}.\u00a0 <\/span>
\nHere, the relation R from A to A is given by, R = {(x, y): 3x \u2013 y = 0, where x, y \u2208 A}\u00a0 <\/span>
\nSo, relation R = {(1,3), (2,6), (3,9), (4,12)}\u00a0 <\/span><\/p>\n

Now, We know that, the domain of a relation R is the set of all the first elements of the ordered pairs in the relation.\u00a0<\/span>
\nSo, Domain of R = {1, 2, 3, 4}\u00a0 <\/span><\/p>\n

Now, Here the complete set A is the Codomain of relation R.\u00a0<\/span>
\nSo, Co-Domain of R = {1, 2, 3, 4,\u2026.,14}\u00a0 <\/span><\/p>\n

Now, We know that, the range of a relation R is the set of all the second elements of the ordered pairs in the relation.\u00a0<\/span>
\nSo, Range of R = {3, 6, 9, 12}<\/span><\/p>\n

2. Define a relation R on the set\u00a0N\u00a0of natural numbers by R = {(x<\/em>,\u00a0y<\/em>):\u00a0y<\/em>\u00a0=\u00a0x<\/em>\u00a0+ 5,\u00a0x<\/em>\u00a0is a natural number less than 4;\u00a0x<\/em>,\u00a0y<\/em> \u2208\u00a0N}. Depict this relationship using roster form. Write down the domain and the range.
\n<\/strong>Solution –\u00a0<\/strong>Here, the relation R is given by, R = {(x, y): y = x + 5, x is a natural number less than 4; x, y \u2208N}\u00a0 <\/span>
\nNow, As we know that the natural numbers less than 4 are 1, 2 and 3.\u00a0 <\/span>
\nSo, relation R = {(1,6), (2,7), (3,8)}\u00a0 <\/span><\/p>\n

Now, We know that, the domain of a relation R is the set of all the first elements of the ordered pairs in the relation.\u00a0<\/span>
\nSo, Domain of R = {1, 2, 3}\u00a0 <\/span><\/p>\n

Now, We know that, the range of a relation R is the set of all the second elements of the ordered pairs in the relation.\u00a0<\/span>
\nSo, Range of R = {6, 7, 8}<\/span><\/p>\n

3. A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x<\/em>,\u00a0y<\/em>): the difference between\u00a0x<\/em>\u00a0and\u00a0y<\/em>\u00a0is odd;\u00a0x<\/em>\u00a0\u2208\u00a0A,\u00a0y\u00a0<\/em>\u2208 B}. Write R in roster form.
\n<\/strong>Solution –\u00a0<\/strong>Given, A = {1, 2, 3, 5} and B = {4, 6, 9}
\nHere, the relation from A to B is given by, R = {(x, y): the difference between x and y is odd; x \u2208 A, y \u2208 B}
\nSo, relation R = {(1,4), (1,6), (2,9), (3,4), (3,6), (5,4), (5,6)} <\/span><\/p>\n

4. The figure shows a relationship between the sets P and Q. Write this relation
\n<\/strong>(i) <\/strong>in set-builder form
\n(ii) <\/strong>in roster form
\nWhat is its domain and range?
\n\"NCER <\/strong><\/span><\/p>\n

Solution –\u00a0<\/strong>From the given figure, we can see that \u2013\u00a0 <\/span>
\nP = {5, 6, 7} and Q = {3, 4, 5}\u00a0 <\/span>
\nNow, The relation between sets P and Q \u2013\u00a0 <\/span><\/p>\n

(i) In set-builder form
\n<\/strong>R = {(x, y): y = x \u2013 2; x \u2208 P} \u2018or\u2019 R = {(x, y): y = x \u2013 2 for x = 5, 6, 7}\u00a0 <\/span><\/p>\n

(ii) In roster form
\n<\/strong>R = {(5,3), (6,4), (7,5)}\u00a0 <\/span>
\nNow, We know that, the domain of a relation R is the set of all the first elements of the ordered pairs in the relation.\u00a0<\/span>
\nSo, Domain of R = {5, 6, 7} = P.\u00a0 <\/span><\/p>\n

Now, We know that, the range of a relation R is the set of all the second elements of the ordered pairs in the relation.\u00a0<\/span>
\nSo, Range of R = {3, 4, 5} = Q.<\/span><\/p>\n

5. Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by <\/strong>{(a<\/em>,\u00a0b<\/em>):\u00a0a<\/em>,\u00a0b<\/em>\u00a0\u2208\u00a0A,\u00a0b<\/em>\u00a0is exactly divisible by\u00a0a<\/em>}.
\n<\/strong>(i) <\/strong>Write R in roster form
\n(ii) <\/strong>Find the domain of R
\n(iii) <\/strong>Find the range of R <\/span><\/p>\n

Solution – <\/strong>Given, A = {1, 2, 3, 4, 6} and relation R = {(a<\/em>,\u00a0b<\/em>):\u00a0a<\/em>,\u00a0b<\/em>\u00a0\u2208\u00a0A,\u00a0b<\/em>\u00a0is exactly divisible by\u00a0a<\/em>}
\nHence, <\/span><\/p>\n

(i)<\/strong> The relation R in roster form will be \u2013\u00a0 <\/span>
\nR = {(1,1), (1,2), (1,3), (1,4), (1,6), (2,2), (2,4), (2,6), (3,3), (3,6), (4,4), (6,6)}\u00a0<\/span><\/p>\n

(ii)\u00a0<\/strong>We know that, the domain of a relation R is the set of all the first elements of the ordered pairs in the relation.\u00a0<\/span>
\nSo, Domain of R = {1, 2, 3, 4, 6}\u00a0 <\/span><\/p>\n

(iii)<\/strong>\u00a0We know that, the range of a relation R is the set of all the second elements of the ordered pairs in the relation.\u00a0<\/span>
\nSo, Range of R = {1, 2, 3, 4, 6}<\/span><\/p>\n

6. Determine the domain and range of the relation R defined by R = {(x<\/em>,\u00a0x<\/em>\u00a0+ 5):\u00a0x<\/em> \u2208 {0, 1, 2, 3, 4, 5}}.
\n<\/strong>Solution – <\/strong>Given,\u00a0Relation R = {(x<\/em>,\u00a0x<\/em>\u00a0+ 5):\u00a0x<\/em> \u2208\u00a0{0, 1, 2, 3, 4, 5}}
\nThus,
\nR = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}
\nSo,
\nDomain of R = {0, 1, 2, 3, 4, 5} and,
\nRange of R = {5, 6, 7, 8, 9, 10}<\/span><\/p>\n

7. Write the relation R = {(x<\/em>,\u00a0x<\/em>3<\/sup>):\u00a0x\u00a0<\/em>is a prime number less than 10} in roster form.
\n<\/strong>Solution –\u00a0<\/strong>Given, Relation R = {(x<\/em>,\u00a0x<\/em>3<\/sup>):\u00a0x\u00a0<\/em>is a prime number less than 10}
\nThe prime numbers less than 10 are 2, 3, 5, and 7.
\nTherefore,
\nR = {(2, 8), (3, 27), (5, 125), (7, 343)} <\/span><\/p>\n

8. Let A = {x<\/em>,\u00a0y<\/em>, z} and B = {1, 2}. Find the number of relations from A to B.
\n<\/strong>Solution –\u00a0<\/strong>Given, A = {x<\/em>,\u00a0y<\/em>, z} and B = {1, 2}
\nNow,
\nA \u00d7 B = {(x<\/em>, 1), (x<\/em>, 2), (y<\/em>, 1), (y<\/em>, 2), (z<\/em>, 1), (z<\/em>, 2)}
\nAs\u00a0n<\/em>(A \u00d7 B) = 6, the number of subsets of A \u00d7 B will be 26<\/sup>.
\nThus, the number of relations from A to B is 26<\/sup>. <\/span><\/p>\n

9. Let R be the relation on\u00a0Z\u00a0defined by R = {(a<\/em>,\u00a0b<\/em>):\u00a0a<\/em>,\u00a0b<\/em>\u00a0\u2208\u00a0Z,\u00a0a\u00a0<\/em>\u2013\u00a0b<\/em> is an integer}. Find the domain and range of R.
\n<\/strong>Solution –\u00a0<\/strong>Given, Relation R = {(a<\/em>,\u00a0b<\/em>):\u00a0a<\/em>,\u00a0b<\/em>\u00a0\u2208\u00a0Z,\u00a0a\u00a0<\/em>\u2013\u00a0b<\/em> is an integer}
\nWe know that the difference between any two integers is always an integer.
\nTherefore,
\nDomain of R =\u00a0Z and Range of R =\u00a0Z<\/span><\/p>\n

Go Back To Chapters<\/i><\/b><\/a><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"

NCERT Solutions Class 11 Maths\u00a0 The NCERT Solutions in English Language for Class 11 Mathematics Chapter – 2 Relations and Functions Exercise 2.2 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter 2 (Relations and Functions)\u00a0 Exercise – 2.2 1. Let A = {1, 2, 3, \u2026 , […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1674],"tags":[1675,1679,1687,1689,1678,1680,1681],"class_list":["post-8392","post","type-post","status-publish","format-standard","hentry","category-class-11-maths","tag-class-11-ncert-mathematics-solutions","tag-class-11-ncert-solutions","tag-ncert-class-11-mathematics-chapter-2-relations-and-functions-solutions","tag-ncert-class-11-mathematics-exercise-2-2-solutions","tag-ncert-class-11-mathematics-solutions","tag-ncert-solutions-class-11-mathematics","tag-ncert-solutions-class-11-maths"],"yoast_head":"\nNCERT Class 11 Maths Solutions - Exercise 2.2 | TheExamPillar NCERT<\/title>\n<meta name=\"description\" content=\"NCERT Solutions Class 11 Maths\u00a0The NCERT Solutions in English Language for Class 11 Mathematics Chapter - 2 Relations and Functions Exercise 2.2 has been provided here to help the students in solving the questions from this exercise.\u00a0Chapter 2 (Relations and Functions)\u00a0\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-11-maths-chapter-2-relations-and-functions-ex-2-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions Class 11 Maths Chapter 2 Relations and Functions - 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