f(x) = 3x - 5 can be written as f{(x, 3x - 5) |x \(\in \) R }
(a, 4) means the image of a is 4. That is, f(a) = 4
3a - 5 = 4 ⇒ a = 3
(1,b) means the image of 1 is b. That is, f(1) = b ⇒ b =-2
3(1) - 5 = b ⇒ b = -2

R₁ = {(–5, a), (1, a), (3, b)}
We may represent the relation R₁ in an arrow diagram 
R₁ is not a function as 4 \(\in\) X does not have an image in y.

Given Set =  {(x, y) / y = x + 3, x \(\in \) {0, 1, 2, 3, 4, 5}}
When x = 0, y = 0 + 3 = 3
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
Relation 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}

A = {1, 2, 3, 4,  ..., 45}
Relation is "is square of a number" and A \(\rightarrow\) A on A
A x A = {(1, 1), (1,2), (1, 3), (1, 4).....( 45, 45)}
The square of 1 is 1 ∈ A and (1, 1) ∈ A x A 
The square of 2 is 4 ∈ A and (4, 2) ∈ A x A
The square of 3 is 9 ∈ A and (9, 3) ∈ A x A  
The square of 4 is 16 ∈ A and (16, 4) ∈ A x A 
The square of 5 is 25 ∈ A and (25, 5) ∈ A x A 
The square of 6 is 36 ∈ A and (36, 6) ∈ A x A 
The square of 7 is 49 \(\notin\) A.
R = {(1, 1), (4, 2),(9, 3), (16, 4),(25, 5), (36, 6)}
Domain of R = {1, 4, 9,16, 25, 36 }
Range of R = {1,2, 3, 4, 5, 6}

(i) Set builder form of R = ((x, y) | y = x - 2, x \(\in \) P, y \(\in \) Q}
(ii) Roster form R = {(5 , 3),(6 , 4)(7 , 5)}
(iii) Domain of R = {5, 6, 7} and range of R = {3, 4, 5}

A x B = {(3,2), (3,4), (5,2), (5,4)}
We have A = {set of all first coordinates of elements of A x B}. Therefore, A = {3,5}
B = {set of all second coordinates of elements of A x B}. Therefore, B = {2,4}
Thus A = {3,5} and B = {2,4}.

Given t (C) = F where \(F=\frac{9 C}{5}+32\)
C - Celsius, F - Fahrenheit
\(\therefore t(C)=\frac{9 C}{5}+32 \)
(i) \(t(0) =\frac{9(0)}{5}+32=0+32=32^{\circ} \mathrm{F} \)
(ii) \(t(28) =\frac{9(28)}{5}+32=\frac{252}{5}+32 \)
= 50.4 + 32 = 82.4^oF
(iii) \(t(-10)=\frac{9(-10)}{5}+32\) = -18 + 32 - 14^oF
(iv) Given t (C) = 212
\(\therefore \frac{9 C}{5}+32 =212 \Rightarrow \frac{9 C}{5}=212-32 \)
\(C =180 \times \frac{5}{9}=100^{\circ} C \)
(v) The temperature when the Celsius value is equal to the Fahrenheit value.
F = C
\(\frac{9 C}{5}+32=C \)
\(\frac{9 C}{5}-C=-32 \Rightarrow \frac{9 C-5 C}{5}=-32 \)
\(4 C=-32 \times 5 \Rightarrow C=-\frac{160}{4} \)
^oC = -40

\(f(x)= \begin{cases}x+2 & \text { if } x>1 \\ 2 & \text { if }-1 \leq x \leq 1 \\ x-1 & \text { if }-3<x<-1\end{cases}\)
i) f(3) = 3 + 2 = 5
ii) f(0) = 2
iii) f(-1.5) = -1.5 - 1 = -2.5
iv) f(2) + f( -2) = ( 2 + 2 ) + ( -2 -1)
= 4 - 3 = 1

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).

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

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 triangle PQR, we are required to construct another triangle whose sides are \(\frac { 3 }{ 5 } \) of 5 the corresponding sides of the triangle PQR

 
Steps of construction:
1. Constructed a PQR with any measurement.
2. Drawn a ray QX making an acute angle with QR on the side opposite to the vertex p.
Located 3 points Q₁, Q₂ and Q₃ on QX so that Q Q₁ = Q₁ Q₂ = Q₂ Q₃
4. Joined Q₃R and drawn a line through Q₂ parallel to Q₃R to intersect QR at R,
5. Drawn a line through R' parallel to the line RP to intersect QP at p,. Then PQR is the required triangle each of whose sides is two third of the corresponding sides of PQR.

Given a triangle PQR, we are required to construct another triangle whose sides are \(\frac { 7 }{ 4 } \) of the corresponding sides of the triangle PQR.
Steps of construction
1. Construct a DPQR with any measurement.
2. Draw a ray QX making an acute angle with QR on the side opposite to vertex P.
3. Locate 7 points (the greater of 7 and 4 in \(\frac { 7 }{ 4 } \))
Q_1,Q_2,Q_3,Q_4,Q_5,Q₆ and Q₇ on QX so that 
QQ₁ = Q₁Q₂ = Q₂Q₃ = Q₄Q₅ = Q₅Q₆ = Q₆Q₇
4. Join Q₄ (the 4th point, 4 being smaller of 4 and 7 in \(\frac { 7 }{ 4 } \)) to R and draw a line through Q₇ parallel to Q₄R, intersecting the extended line segment QR at R'.
5. Draw a line through R' parallel to RP intersecting the extended line segment QP at P'.
Then \(\triangle\)P'QR' is the required triangle each of whose sides is seven-fourths of the corresponding sides of \(\triangle\)PQR.