Relations and Functions
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New answer posted
5 months agoContributor-Level 10
16. Given, R = { (x, x+5): x {0,1,2,3,4,5}
= { (0,0+5), (1,1+5), (2,2+5), (3,3+5), (4,4+5), (5,5+5)}
= { (0,5), (1,6), (2,7), (3,8), (4,9), (5,10)}
So, domain of R= {0,1,2,3,4,5}
range of R= {5,6,7,8,9,10}
New answer posted
5 months agoContributor-Level 10
15. Given, A= {1,2,3,4,6}
R= { (a, b): a, b A, b is exactly divisible by a}
(i) R= { (1,1), (1,2), (1,3), (1,4), (1,6), (2,2), (2,4), (2,6), (3,3), (3,6), (4,4), (6,6)}
(ii) Domain of R= {1,2,3,4,6}
(iii) Range of R= {1,2,3,4,6}
New answer posted
5 months agoContributor-Level 10
14. As R is a relation from set P to Q.
(i) R = { (x, y): x – 2 = y ; 5 ≤ x ≤ 7}
(ii) R = { (5,3), (6,4), (7,5)}
Domain of R= {5,6,7}
range of R= {3,4,5}
New answer posted
5 months agoContributor-Level 10
13. Given, A= {1,2,3,5}
B= {4,6,9}
R= { (x, y) : the difference of x & y is odd; x A, y B}.
= { (x, y):|x – y| is odd and x A, y B}
= { (1,4), (1,6), (2,9), (3,4), (3,6), (5,4), (5,6)}.

New answer posted
5 months agoContributor-Level 10
12. Given, R = { (x, y): y = x + 5, x is a natural number less than 4; x, y N}
= { (x, y): y = x + 5; x, y N and x < 4}.
= { (1,1+5), (2,2+5), (3,3+5)}
= { (1,6), (2,7), (3,8)}
So, domain of R = {1,2,3}
range of R = {6,7,8}
New answer posted
5 months agoContributor-Level 10
11. Given, A = {1,2,3, …, 14}
R = { (x, y): 3x – y = 0; x, y A}
= { (x, y): 3x = y; x, y A}.
= { (1,3), (2,6), (3,9), (4,12)}
Domain of R is the set of all the first elements of the ordered pairs in R
So, domain of R= {1,2,3,4}
Codomain of R is the whole set A.
So, codomain of R= {1,2,3, …, 14}
Range of R is the set of all the second elements of the ordered pains in R.
So, range of R= {3,6,9,12}
New question posted
5 months agoNew answer posted
5 months agoContributor-Level 10
10. Given, n (A * A)=9
n (A) *n (A) = 9.
n (A)2 = 32.
n (A) = 3 .
And (–1,0), (0,1) A * A i.e., A * A = { (x, y), x A, y B}
? A= {–1,0,1}
And A * A= {–1,0,1} * {–1,0,1}
= { (–1, –1), ( –1,0), ( –1,1), (0, –1), (0,0), (0,1), (1, –1), (1,0), (1,1)}
New answer posted
5 months agoContributor-Level 10
9. Given, n (A)=3
n (B)= 2
So, n (Ax B)=n (A).n (B)=3x 2=6
as (x, 1), (y, 2), (z, 1) ∈Ax B= { (x, y), x∈Aand y∈B}.
A= {x, y, z} and B= {1,2}
As n (A) = 3as n (B) = 2
New answer posted
5 months agoContributor-Level 10
8. A Given, A= {1,2}
B= {3,4}
So, A* B= { (1,3), (1,4), (2,3), (2, 4)}
i.e., n (A *B)=4
A *B will have subset =24=16.They are,
Φ, { (1,3)}, { (1,4)}, { (2, 3)}, { (2,4)}, { (1,3), (1,4)}, { (1,3), (2,3)},
{ (1,3), (2, 4)}, { (1,4), (2, 3)}, { (1,4), (2, 4)}, { (2,3), (2, 4)},
{ (1,3), (1,4), (2, 3)}, { (1,3), (1,4), (2,4)}, { (1,3), (2,3), (2, 4)}, { (1,4), (2,3), (2,4)},
and { (1,3), (1,4), (2,3), (2,4)}
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