Relations and Functions

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5 months ago

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Payal Gupta

Contributor-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 ago

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Payal Gupta

Contributor-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 ago

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Payal Gupta

Contributor-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}

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5 months ago

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Payal Gupta

Contributor-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 ago

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Payal Gupta

Contributor-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 ago

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Payal Gupta

Contributor-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

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New answer posted

5 months ago

0 Follower 2 Views

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Payal Gupta

Contributor-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 ago

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Payal Gupta

Contributor-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

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P
Payal Gupta

Contributor-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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