Chapter 3

Exercise 3.3 Solutions

Step-by-step solved questions on relations, domains, ranges, visual mapping of functions, and parameter evaluation.

Question 1 Relations, Dom & Range
For A = {1, 2, 3, 4}, find the following relations in A. State the domain and range of each relation:

(i) R1 = {(x, y) | y = x}

Solution
Find pairs in A × A where the second element equals the first element:
R1 = {(1, 1), (2, 2), (3, 3), (4, 4)}.
Domain (set of first elements): Dom(R1) = {1, 2, 3, 4}.
Range (set of second elements): Range(R1) = {1, 2, 3, 4}.

(ii) R2 = {(x, y) | y + x = 5}

Solution
Find pairs where the elements sum to 5:
R2 = {(1, 4), (2, 3), (3, 2), (4, 1)}.
Domain: Dom(R2) = {1, 2, 3, 4}.
Range: Range(R2) = {1, 2, 3, 4}.

(iii) R3 = {(x, y) | x + y < 5}

Solution
Find pairs whose sum is strictly less than 5:
- For x=1: 1+1=2, 1+2=3, 1+3=4. (Fits)
- For x=2: 2+1=3, 2+2=4. (Fits)
- For x=3: 3+1=4. (Fits)
- For x=4: 4+y ≥ 5 for all y ∈ A.
Relation: R3 = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)}.
Domain: Dom(R3) = {1, 2, 3}.
Range: Range(R3) = {1, 2, 3}.

(iv) R4 = {(x, y) | x + y > 5}

Solution
Find pairs whose sum is strictly greater than 5:
- For x=1: 1+y ≤ 5 for all y.
- For x=2: 2+4=6. (Fits)
- For x=3: 3+3=6, 3+4=7. (Fits)
- For x=4: 4+2=6, 4+3=7, 4+4=8. (Fits)
Relation: R4 = {(2, 4), (3, 3), (3, 4), (4, 2), (4, 3), (4, 4)}.
Domain: Dom(R4) = {2, 3, 4}.
Range: Range(R4) = {2, 3, 4}.
Question 2 Function Classification
Analyze if the following described mapping relations represent functions, and if so, determine their type:

(i) A = {1, 2, 3}, B = {a, b, c, d}; R1 = {(1, a), (1, b), (2, c), (3, d)}

Solution
Observe that the element 1 in the domain A is mapped to two different elements a and b in the codomain B.
Conclusion: This relation is not a function. (A function requires each element in the domain to map to exactly one element in the codomain).

(ii) A = {a, b, c}, B = {1, 3, 5}; R2 = {(a, 1), (b, 3), (c, 5)}

Solution
Each element of domain A maps to exactly one element of codomain B. This is a function.
Observe that:
- Every element in B has a unique pre-image (one-to-one or injective).
- The range is equal to the codomain, Range = {1, 3, 5} = B (onto or surjective).
Conclusion: The relation is a bijective function (both one-to-one and onto).

(iii) A = {1, 2, 3}, B = {a, b, c}; R3 = {(1, a), (2, b), (3, c)}

Solution
Each element in the domain maps to exactly one unique element in the codomain.
Since it is both one-to-one (injective) and onto (surjective), it is bijective.
Conclusion: The relation is a bijective function.

(iv) A = {l, m, n}, B = {x, y, z}; R4 = {(l, x), (m, x), (n, z)}

Solution
Each element in the domain maps to exactly one element in the codomain. This is a function.
Observe that:
- It is not one-to-one, because both l and m map to x.
- The range is {x, z}, which is a proper subset of B (since y is left out).
Conclusion: The relation is an into function (non-surjective function).
Question 3 Function Evaluation
If g(x) = 3x + 2 and h(x) = x2 + 1, find the values of:

(i) Find g(0):

Substitute x = 0 in g(x): g(0) = 3(0) + 2 = 2.


(ii) Find g(-3):

Substitute x = -3 in g(x): g(-3) = 3(-3) + 2 = -9 + 2 = -7.

(iii) Find g(1/3):

Substitute x = 1/3 in g(x): g(1/3) = 3(1/3) + 2 = 1 + 2 = 3.


(iv) Find h(1):

Substitute x = 1 in h(x): h(1) = (1)2 + 1 = 1 + 1 = 2.

(v) Find h(-4):

Substitute x = -4 in h(x): h(-4) = (-4)2 + 1 = 16 + 1 = 17.


(vi) Find h(-1/2):

Substitute x = -1/2 in h(x): h(-1/2) = (-1/2)2 + 1 = 1/4 + 1 = 5/4.

Question 4 Find Constants a & b
Given that f(x) = ax + b + 1, where a and b are constants. If f(3) = 8 and f(6) = 14, find the values of a and b.
Solution
Set up two equations by substituting x = 3 and x = 6 into f(x):
f(3) = a(3) + b + 1 = 8 ⇒ 3a + b = 7 (Eq. 1) f(6) = a(6) + b + 1 = 14 ⇒ 6a + b = 13 (Eq. 2)
Subtract Eq. 1 from Eq. 2 to eliminate b:
(6a + b) - (3a + b) = 13 - 7 3a = 6 a = 2
Substitute a = 2 back into Eq. 1:
3(2) + b = 7 6 + b = 7 b = 1
Conclusion: a = 2 and b = 1.
Question 5 Find Constants a & b (Ex 2)
Given that g(x) = ax + b + 5, where a and b are constants. If g(-1) = 0 and g(2) = 10, find the values of a and b.
Solution
Substitute the values to get two equations:
g(-1) = a(-1) + b + 5 = 0 ⇒ -a + b = -5 (Eq. 1) g(2) = a(2) + b + 5 = 10 ⇒ 2a + b = 5 (Eq. 2)
Subtract Eq. 1 from Eq. 2 to eliminate b:
(2a + b) - (-a + b) = 5 - (-5) 3a = 10 a = 103
Substitute a = 10/3 into Eq. 1:
-103 + b = -5 b = -5 + 103 b = -15 + 103 = -53
Conclusion: a = 103 and b = -53.
Question 6 Solve for x
Consider the function defined by f(x) = 5x + 1. If f(x) = 32, find the value of x.
Solution
Set up the equation:
5x + 1 = 32 5x = 31 x = 315 = 6.2
Conclusion: The value of x is 31/5 or 6.2.
Question 7 Quadratic Constants
Consider the function f(x) = cx2 + d, where c and d are constants. If f(1) = 6 and f(-2) = 10, find the values of c and d.
Solution
Substitute values to get two equations:
f(1) = c(1)2 + d = 6 ⇒ c + d = 6 (Eq. 1) f(-2) = c(-2)2 + d = 10 ⇒ 4c + d = 10 (Eq. 2)
Subtract Eq. 1 from Eq. 2 to eliminate d:
(4c + d) - (c + d) = 10 - 6 3c = 4 c = 43
Substitute c = 4/3 into Eq. 1:
43 + d = 6 d = 6 - 43 d = 18 - 43 = 143
Conclusion: c = 43 and d = 143.
Interactive Sandbox Relation & Function Mapping Drawer
Click nodes in Set A (Domain) and then nodes in Set B (Codomain) to draw arrows. The sandbox will analyze the relations in real-time to check if they form a function and determine the function type!
Controls

Click a node in Set A, then click a node in Set B to map them. You can map multiple nodes.

Analysis Output
Relation: {}
Is it a function? No
Details: Add connections to begin.
Set A
1
2
3
4
Set B
a
b
c
d