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).

Given 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) and (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 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-thirds 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.