X std mathematics - Relations and functions (Ex 1.3), Maths, IX std Maths, Samacheerkalvi maths, II year B.Ed., Pedagogy, Mathematics, functions, definition of functions, representation by arrow diagram,

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

2. X STD
Ex 1.3

15. Solution:
F = {(x, y)|x, y ∈ N and y = 2x}
x = {1, 2, 3,…}
y = {1 × 2, 2 × 2, 3 × 2, 4 × 2, 5 × 2 …}
R = {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10),…}
Domain of R = {1, 2, 3, 4,…},
Co-domain = {1, 2, 3…..}
Range of R = {2, 4, 6, 8, 10,…}
Yes, this relation is a function.

16. Solution:
x = {3,4, 6, 8}
R = ((x, f(x))|x ∈ X, f(x) = X2 +
1}
f(x) = x2 + 1
f(3) = 32 + 1 = 10
f(4) = 42 + 1 = 17
f(6) = 62 + 1 = 37
f(8) = 82 + 1 = 65
R = {(3, 10), (4, 17), (6, 37), (8,
65)}
R = {(3, 10), (4, 17), (6, 37), (8,
65)}

17. Answer:
f(x) = x2 – 5x + 6
(i) f (-1) = (-1)2 – 5 (-1) + 6 = 1 + 5 + 6 = 12
(ii) f (2a) = (2a)2 – 5 (2a) + 6 = 4a2 – 10a +
6
(iii) f(2) = 22 – 5(2) + 6 = 4 – 10 + 6 = 0
(iv) f(x – 1) = (x – 1)2 – 5 (x – 1) + 6
= x2 – 2x + 1 – 5x + 5 + 6
2

18. Solution:
From the graph
(a) f(0) = 9

19. (b) f(7) = 6
(c) f(2) = 6
(d) f(10) = 0
(ii) At x = 9.5, f(x) = 1
(iii) Domain = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
= {x |0 < x < 10, x ∈ R}
Range = {x|0 < x < 9, x ∈ R}
= {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
(iv) The image of 6 under f is 5.

20. Solution:
Given f(x) = 2x + 5, x ≠ 0.

21. Solution:
Given f(x) = 2x – 3

22. Solution:
Volume of the box = Volume of the cuboid
= l × b × h cu. units
Here l = 24 – 2x
b = 24 – 2x
h = x
∴ V = (24 – 2x) (24 – 2x) × x
= (576 – 48x – 48x + 4x2)x
V = 4x3 – 96x2 + 576x

23. Solution:
f(x) = 3 – 2x
f(x2) = 3 – 2x2

24. Answer:
Speed of the plane = 500 km/hr
Distance travelled in “t” hours
= 500 × t (distance = speed × time)
= 500 t

26. Solution:
(i) Given y = ax + b …………. (1)
The ordered pairs are R = {(35, 56) (45, 65) (50, 69.5) (55,
74)}
∴ Hence this relation is a function.

27. Substituting a = 0.9 in (2) we get
⇒ 65 = 45(.9) + b
⇒ 65 = 40.5 + b
⇒ b = 65 – 40.5
⇒ b = 24.5
∴ a = 0.9, b = 24.5
∴ y = 0.9x + 24.5
(iii) Given x = 40 , y = ?
∴ (4) → y = 0.9 (40) + 24.5
⇒ y = 36 + 24.5
⇒ y = 60.5 inches

28. (iv) Given y = 53.3 inches, x = ?
(4) → 53.3 = 0.9x + 24.5
⇒ 53.3 – 24.5 = 0.9x
⇒ 28.8 = 0.9x
⇒ x = 28.80.9 = 32 cm
∴ When y = 53.3 inches, x = 32
cm