The function f is defined by three values in intervals I, II, III as shown by the side.
For a given value of x = a, find out the interval at which the point a is located, there after find
f(a) using the particular value defined in that interval.
(i) First, we see that, x = 4 lie in the third interval.
Therefore, f(x) = 3x - 2; f(4) = 3(4) = 10
(ii) x = -2 lies in the second interval
Therefore, f(x) = x² - 2; f(-2) = (-2)² - 2 = 2
(iii) From (i), f(4) =10.
To find f(1) first we see that x = 1 lies in the second interval.
Therefore, f(x) = x²-2 ⇒ f(1) = 1² - 2 = -1
So, f(4) + 2f(1) = 10 + 2(-1) = 8
(iv) We know that f(1) = -1 and f(4) = 10
For finding f(-3), we see that x = −3, lies in the first interval.
Therefore, f(x) = 2x + 7; thus, f(-3) = 2(-3) + 7 = 1
Hence, \(\frac { f(1)-3f(4) }{ f(-3) } =\frac { -2-3(10) }{ 1 } \) = - 31

(i) To check if h is one – one, we assume that h(b₁) = h(b₂)
Then we get 2.47b₁ + 54.10 = 2.47b₂ + 54.10
2.47b₁ = 2.47b₂
⇒ b₁ = b₂
Thus, h(b₁) = h(b₂) ⇒ b₁ = b₂, So, the function h is one – one.
(ii) If the length of the thigh bone b = 50, then the height is
h(50) = (2.47 x 50) + 54.10 =177.6 cms
(iii) If the height of a person is 147.96 cms, then h(b) = 147.96 and so the length of the thigh bone is given by
2.47b + 54.10 = 147.96.
2. 47 = 147. 96 - 54. 10 = 93. 86
b = \(\frac { 93.86 }{ 2.47 } \)
Therefore, the length of the thigh bone is 38 cm.

A = {1, 2, 3, 4} ; B = {2, 5, 8,11,14}; f(x) = 3x − 1
f(1) = 3(1) –1 = 3 – 1 = 2; f(2) = 3(2) –1 = 6 –1 = 5
f(3) = 3(3) –1 = 9 –1 = 8; f(4) = 4(3) –1 = 12 –1 = 11
(i) Arrow diagram
Let us represent the function f :A ⟶ B by an arrow diagram  

(ii) Table form
The given function f can be represented in a tabular form as given below

x	1	2	3	4
f(x)	2	5	8	11

(iii) Set of ordered pairs
The function f can be represented as a set of ordered pairs as
f = {(1,2),(2,5),(3,8),(4,11)}
(iv) Graphical form
In the adjacent xy -plane the points
(1,2), (2,5), (3,8), (4,11) are plotted (Fig.1.20).

y = ax + b; x = forehand length; y = height

X	Y
35	56
45	65
50	69.5
55	74

For all the x-values, there is an image which is 'y
Moreover, the difference between two consecutive 'y' values is constant
In y = ax + b,
(i) The Relation 
R = { (35, 56), (45, 65), (50, 69.5), (55, 74) } is a function
(ii) In y = ax + b
when x = 35,y = 56
56 = 35a + b                 ..............(1)
when x = 45,y = 65
65 = 45a + b                ..............(2)
Solving (1) and (2), we get a = 0.90 and b = 24.5
(iii) Given, forehand length is 40 cm
i.e., when x = 40,y = ax + b
So, y = (0.90) (40) + 24.5 = 60.5
Height of person is 60.5 inches.
(iv) Given height is 53.3 inches
i.e. when y = 53.3, x = ?
53.3 = 0.9x + 24.5
53.3 - 24.5 = 0.9x
x = \(\frac{28.8}{0.9}\)
x = 32 cm
Length of fore hand is 32 cm.

From the diagram,
The solid is a cuboid' volume of cuboid = length x breadth x height
where l = 24 - 2x, b - 24 - 2x,. h = x
Volume V (x) = (24 - 2x) (24 - 2x) x
V(x) = x(24 - 2x)², x > 0
= 4x³ - 96x² + 576x, x > 0
So, the domain is 0 < x < 12

(i) From the given graph
(a) f(0) = 9
(b) f(7) = 6
(c) = f(2)
(d) = f(10) = 0
(ii) From the graph, it is known that
when x = 9.5, f(x) = 1
(iii) (a) Domain = {x|0 ≤ x ≤ 10, x \(\in \) R}
(b) Range = {x|0 ≤ x ≤ 9, x \(\in \) R}
(iv) The image of '6' under f is '5'.

Give the function 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)² - 5(x - 1) + 6
= x²- 2x + 1 - 5x + 5 + 6
= x²-7x + 12

(i) x = 1, we get
f(1) = 2(1) - (1)² = 2 - 1 = 1
(ii) x = x + 1, we get
f(x + 1) = 2(x + 1) - (x + 1)² = 2x + 2 - (x² + 2x + 1) = -x² + 1
(iii) f(x) + f(1) = (2x - x²) + 1= - x² + 2x + 1
[Note that f(x) + f(1) ≠ f(x + 1). In general f(a + b) is not equal to f(a) + f(b)]

Ordered Pair : The Domain of the relation is about the salaries given to person
Relation is R = {(10000, A₁), (10000, A₂), (10000, A₃),
(10000, A₄), (10000, A₅), (25000, C₁),
(25000, C₂), (25000, C₃), (25000, C₄),
(50000, M₁), (50000, M₂), (50000, M₃),
(100000, E₁), (100000, E₂)}
Relation R defined by x R y
'x' is the salary given to person y
Arrow diagram

(i) Given Set - Builder form
{(x, y)|x = 2y, x \(\in \) {2,3,4,5}, y \(\in \) {1,2,3,4}
x = 2y
y = 1 ⇒ x = 2
y = 2 ⇒ x = 4
y = 3 ⇒ x = 6
y = 4 ⇒ x = 8
Relation R - {(2, 1), (4,2)}
(a) Arrow diagram

b) Graph

(c) Roster form R =  {(2, 1), (4, 2)}
ii. Given set
{(x, y)|y = x + 3, x, y are natural numbers < 10}.
When x = 1, y = 1 + 3 = 4
When x = 2, y = 2 + 3 = 5
When x = 3, y = 3 + 3 = 6
When x = 4, y = 4 + 3 = 7
When x = 5, y = 5 + 3 = 8
When x = 6, y = 6 + 3 = 9
When x = 7, y = 7 + 3 = 10 is not possible
Since x and y are less than 10
Relation R = {(1,4), (2,5), (3,6), (4,7), (5,8), (6,9)}
(a) Arrow diagram

(b) Graph

(c) Roster form
{(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)}

Given: A is the set of all natural numbers less than 8 A = {1,2,3,4,5,6,7}
B is the set of all prime numbers less than 8 B = {2,3,5,7}
'C' is the set of all even prime number
C = {2}
verify
(A ∩ B) x C = (A x C) ∩ (B x C)
\(A\cap B\) = {2,3,5,7}
\((A \cap B) \times C=\{2,3,5,7\} \times\{2\}\)
= {(2,2),(3,2),(5,2),(7,2}}    ......(1)
A X C = {1,2,3,4,5,6,7} x {2}
= {(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),(7,2)}
B x C = {2,3,5,7} x {2}
= {(2,2),(3,2),(5,2),(7,2)}
\((A\times C)\cap (B\times C)\) = {(2,2) ,(3,2),(5,2),(7,2)}  ...(2)
From (1) and (2) it is clear that
(A ∩ B) x C = (A x C) ∩ (B x C)
Hence verified.

Given A = {x \(\in \) W| x < 2} A = {0,1}
B = {x \(\in \) N| 1 < x ≤ 4} B = {2,3,4}
C = {3,5}
A x (B U C) = (A x B) U (A x C)
\(B\cup C\) = {2,3,4,5}
A x (B U C) = {0,1} x {2,3,4,5}
= {{0,2},(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)}          ...(1)
A x B = {0,1} x {2,3,4}
= {(0,2),(0,3),(0,4),(1,2),(1,3),(1,4)}
A x C = {0,1} x {3,5}
= {{0,3},(0,5),(1,3),(1,5)}
\((A\times B)\cup (A\cup C)\) = {(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)} ...(2)
From (1) x (2),it is clear that
\(A\times (B\cup C)=(A\times B)\cup (A\times C)\)
Hence verified

Given A = {5,6} , B = {4,5,6} , C = {5,6,7}
L.H.S: A x A = {5,6} x {5,6}
= {(5,6),(5,6),(6,5),(6,6)}          
R.H.S: B x B = {4,5,6} x {4,5,6}
= {(4,4),(4,5),(4,6),(5,4),(5,5),(5,6),(6,4),(6,5),(6,6)}
C x C = {5,6,7} x {5,6,7}
= {(5,5),(5,6),(5,7),(6,5),(6,6),(6,7),(7,5),(7,5),(7,6),(7,7)}
(B x B) ∩ (C x C) = {(5,5),(5,6),(6,5),(6,6)}
LHS = RHS
A x A = (B x B) ∩ (C x C)
Hence proved

A =  {x \(\in \) N| 1 < x < 4} = {2,3), B = {x \(\in \) W| 0 ≤ x < 2) = (0,1), C = {x \(\in \) N| x < 3} = (1,2)
(i)  A x (B U C) = (A x B) U (A x C)
B U C = (0,1) U (1,2) = {0,1,2}
A x (B U C) = {2,3) x {0,1,2} = {(2,0),(2,1)(2,2)(3,0)(3,1),(3,2)         ..(1)
A x B = {2,3} x {0,1} = {(2,0),(2,1),(3,0),(3,1)}
A x C = {2,3} x {1,2} = {(2,1),(2,2),(3,1)(3,2)}
(A x B) U (A x C) = {(2,0),(2,1),(3,0),(3,1)} U {(2,1),(2,2),(3,1),(3,2)}
= {(2,0),(2,1),(2,2),(3,0),(3,1),(3,2)}                ...(2)
From (1) and (2), A x (B U C) = (A x B) U  (A x C) is verified.
(ii)  A x (B ∩ C) = (A x B) ∩ (A x C)
(B ∩ C) = {0,1} ∩ {1,2} =  {1}
A x (B ∩ C) = {2,3} x {1} = {(2,1),(3,1)}     .... (3)
A x B = {2,3} x {0,1} = {(2,0),(2,1),(3,0),(3,1)}
A x C = {2,3} x {1,2} = {(2,1),(2,2),(3,1),(3,2)
(A x B) ∩ (A x C) = {(2,0),(2,1),(3,0),(3,1)} ∩ {(2,1),(2,2),(3,1),(3,2)}
= {(2,1),(3,1)}                .... (4)
From (3) and (4), A x (B ∩ C) = (A x B) ∩ (A x C) is verified.

From B x A, All the first entries belong to the set B and all the second entries belong to A.
A = {3,4} and
B = {-2,0,3}