X std mathematics - Relations and functions (Ex 1.2)

PEDAGOGY OF
MATHEMATICS – PART II
BY
Dr. I. UMA MAHESWARI
Principal
Peniel Rural College of Education,Vemparali,
Dindigul District
iuma_maheswari@yahoo.co.in

Solution:
R1 = {(2,1), (7,1)}
It is not a relation there is no element as 1 in B.

(ii) R2 = {(-1, 1)}
It is not [∵ -1 ∉ A, 1 ∉ B]
(iii) R3 = {(2,-1), (7, 7), (1,3)}
It is a relation.
R4 = {(7,-1), (0, 3), (3, 3), (0, 7)}
It is also not a relation. [∵ 0 ∉ A]

Solution:
A = {1, 2, 3, 4, . . . 45}, A × A = {(1, 1), (2, 2) ….. (45,45)}
R – is square of’
R = {(1,1), (2,4), (3, 9), (4, 16), (5,25), (6,36)}
R ⊂ (A × A)
Domain of R = {1, 2, 3, 4, 5, 6}
Range of R = {1,4, 9, 16, 25, 36}

Solution:
x = {0,1,2,3,4,5}
y = x + 3
⇒ y = {3, 4, 5, 6, 7, 8}
R = {(x,y)}
= {(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}

Solution:
(i){(x,y)|x = 2y,x ∈ {2,3,4,5},y ∈ {1,2,3,4}} R = (x = 2y)
2 = 2 × 1 = 2
4 = 2 × 2 = 4
(c) {(2,1), (4, 2)}

(ii) {(x, v)[y = x + 3, x,+ are natural numbers <10}
x = {1,2, 3, 4, 5, 6, 7, 8,9} R = (y = x + 3)
y = {1,2, 3, 4,5,6, 7, 8,9}
R = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}
(c) R = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}

Solution:
A – Assistants → A1, A2, A3, A4, A5
C – Clerks → C1, C2, C3, C4
D – Managers → M1, M2, M3
E – Executive officer → E1, E2

(a) R = {(10,000, A1), (10,000, A2), (10,000, A3),
(10,000, A4), (10,000, A5), (25,000, C1),
(25,000, C2), (25,000, C3), (25,000, C4),
(50,000, M1), (50,000, M2), (50,000, M3),
(1,00,000, E1), (1,00,000, E2)}

X std mathematics - Relations and functions (Ex 1.2)

X std mathematics - Relations and functions (Ex 1.2)

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