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### Relations and Functions, Ex. No. 1.2, 10th New Syllabus 2019 - 2020.

Relations and Functions,
Ex. No. 1.2

(1)  Let A = {1, 2, 3, 7} and B = {3, 0, –1, 7}, which of the following are relation from A to B ?
(i)  R1 = {(2, 1), (7, 1)}
(ii) R2 =  {(–1,1)}
(iii) R3 = {(2, –1), (7, 7), (1, 3)}
(iv) R4= {(7,–1), (0,3), (3,3), (0,7)}
(i)  R1 = {(2, 1), (7, 1)}
Solution :
Cartesian product of A and B
A x B  = {(1, 3) (1, 0) (1, -1) (1, 7) (2, 3) (2, 0) (2, -1) (2, 7)(3, 3) (3, 0) (3   , -1)(3, 7)(7, 3)(7, 0)(7, -1)(7, 7)}
Note : We don't have to list out all these things. In order to understand, we have listed out.
R1 ∈ (2, 1) ∉ A x B. Hence R1 is not a relation.
(ii) R2 =  {(–1,1)}
Solution :
(-1, 1) is a element of R2, but in Cartesian product, we have (1, -1).
(1, -1) (-1, 1). Hence R2 is not a relation.
(iii) R3 = {(2, –1), (7, 7), (1, 3)}
Solution :
R3 is a subset of A x B, Hence it is a relation from A to B.
(iv) R4= {(7,–1), (0,3), (3,3), (0,7)}
Solution :
The element (7,–1) ∈ A x B
But (0,3), (3,3), (0,7) ∉ A x B
Hence it is not a relation.

(2)  Let A = {1, 2, 3, 4,..., 45} and R be the relation defined as “is square of ” on A. Write R as a subset of A x A. Also, find the domain and range of R.
Solution :
A = {1, 2, 3, 4,..., 45}
Let "A" be a set which contains some of elements of A
A = {1, 2, 3, 4, 5, 6}
Now, we have to write the other set of elements from A. The elements in this set denotes the square value of previous set.
Square values of set A  = {1, 4, 9, 16, 25, 36}
A x A  = {(1, 1)(1,4)(1, 9)(1, 16) (1, 25)  (1, 36)..................}
Domain = {1, 2, 3, 4, 5, 6}
Range  = {1, 4, 9, 16, 25, 36}

(3)  A Relation R is given by the set {(x, y) /y = x + 3, x ∈  {0, 1, 2, 3, 4, 5}}. Determine its domain and range.
Solution :
Given that :
y = x + 3

 y = x + 3 x = 0 y = 0 + 3 y = 3 y = x + 3 x = 1 y = 1 + 3 y = 4 y = x + 3 x = 2 y = 2 + 3 y = 5 y = x + 3 x = 3 y = 3 + 3 y = 6 y = x + 3 x = 4 y = 4 + 3 y = 7 y = x + 3 x = 5 y = 5 + 3 y = 8

R = {(0, 3) (1, 4) (2, 5) (3, 6) (4, 7) (5, 8)}
Domain of R = {0, 1, 2, 3, 4, 5}
Range of R = {3, 4, 5, 6, 7, 8}

(4)  Represent each of the given relations by (a) an arrow diagram, (b) a graph and (c) a set in roster form, wherever possible.
(i) {(x, y)| x = 2y, x {2, 3, 4, 5}, y {1, 2, 3, 4}   Solution
Solution :
Given that :
x = 2y
y = x/2

 if x = 2 y = 2/2 y = 1 f x = 3 y = 3/2 if x = 4 y = 4/2 y = 2 if x = 5 y = 5/2

Hence the required relation is {(2, 1) (3, 3/2) (4, 2) (5, 5/2)}
(i)  Arrow diagram (ii)  A graph (iii)  in roster form
= {(2, 1) (4, 2)}

(ii) {(x, y) | y = x + 3, x, y are natural numbers < 10}    Solution

Solution :
Given that :
y = x + 3

 if x = 1 y = 4 if x = 2 y = 5 if x = 3 y = 6 if x = 4 y = 7 if x = 5 y = 8 if x = 6 y = 9

R = {(1, 4) (2, 5) (3, 6) (4, 7) (5, 8) (6, 9)}
An arrow diagram : (ii)  A graph In Roster form :
= {(1, 4) (2, 5) (3, 6) (4, 7) (5, 8) (6, 9)}

(5)  A company has four categories of employees given by Assistants (A), Clerks (C), Managers (M) and an Executive Officer (E). The company provide ₹10,000, ₹25,000, ₹50,000 and ₹1,00,000 as salaries to the people who work in the categories A, C, M and E respectively. If A1, A2, A3, A4and A5 were Assistants; C1, C2, C3, C4 were Clerks; M1, M2, M3 were managers and E1, E2 were Executive officers and if the relation R is defined by xRy, where x is the salary given to person y, express the relation R through an ordered pair and an arrow diagram.
Solution :
"x" be the salary given to person
"y" be the set of employees ## PDF FILE TO YOUR EMAIL IMMEDIATELY PURCHASE NOTES & PAPER SOLUTION. @ Rs. 50/- each (GST extra)

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