Chapter Solution

Chapter-3

EXERCISE 3.1 — SOLUTIONS

1. If (x − 5, y + 1) = (4, 6), find x and y.

Since two ordered pairs are equal, their corresponding components must be equal.

Therefore,
x − 5 = 4
x = 4 + 5
x = 9

Also,
y + 1 = 6
y = 6 − 1
y = 5

Answer: x = 9, y = 5

2. Let A = {1, 2} and B = {2, 3, 5}. List all elements of A × B and B × A.

The Cartesian product A × B consists of all ordered pairs (a, b), where a ∈ A and b ∈ B.

Therefore,

A × B = {
(1, 2), (1, 3), (1, 5),
(2, 2), (2, 3), (2, 5)
}

Similarly, B × A consists of all ordered pairs (b, a), where b ∈ B and a ∈ A.

B × A = {
(2, 1), (2, 2),
(3, 1), (3, 2),
(5, 1), (5, 2)
}

3. If n(A × B) = 20 and n(A) = 4, find n(B).

We use the formula:
n(A × B) = n(A) × n(B)

Given:
n(A × B) = 20
n(A) = 4

Therefore,
20 = 4 × n(B)
n(B) = 20 ÷ 4
n(B) = 5

4. If A = {1, 2, 3} and B = {x, y}, find A × B, B × A, A × A and B × B.

A × B:

A × B = { (1, x), (1, y), (2, x), (2, y), (3, x), (3, y) }

B × A:

B × A = { (x, 1), (x, 2), (x, 3), (y, 1), (y, 2), (y, 3) }

A × A:

A × A = { (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3) }

B × B:

B × B = { (x, x), (x, y), (y, x), (y, y) }

5. If A = {1, 2, 3} and B = {2, 3, 7}, find (A × B) ∩ (B × A).

First, observe that:
A ∩ B = {2, 3}

A pair belongs to both A × B and B × A only when its first and second components are common to both A and B.

Therefore,
(A × B) ∩ (B × A) = (A ∩ B) × (A ∩ B)

= {2, 3} × {2, 3}

= { (2, 2), (2, 3), (3, 2), (3, 3) }

Answer: (A × B) ∩ (B × A) = {(2, 2), (2, 3), (3, 2), (3, 3)}

6. Verify, A × (B ∪ C) = (A × B) ∪ (A × C) for A = {1, 2}, B = {2, 3}, C = {4, 5}.

Step 1: Find B ∪ C.

B ∪ C = {2, 3, 4, 5}

Step 2: Find A × (B ∪ C).

A × (B ∪ C) = {1, 2} × {2, 3, 4, 5}

= { (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5) }

Step 3: Find A × B.

A × B = {1, 2} × {2, 3}

= { (1, 2), (1, 3), (2, 2), (2, 3) }

Step 4: Find A × C.

A × C = {1, 2} × {4, 5}

= { (1, 4), (1, 5), (2, 4), (2, 5) }

Step 5: Find (A × B) ∪ (A × C).

(A × B) ∪ (A × C)

= { (1, 2), (1, 3), (2, 2), (2, 3), (1, 4), (1, 5), (2, 4), (2, 5) }

Rearranging the elements:

= { (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5) }

This is exactly the same as A × (B ∪ C).

Therefore, A × (B ∪ C) = (A × B) ∪ (A × C).

Hence, the result is verified.

EXERCISE 3.2 — SOLUTIONS

1. Let A = {1, 2, 3}, B = {4, 5, 6, 7}. Define a relation R from A to B by R = {(a, b) : a + b = 7; a ∈ A, b ∈ B}. Write the relation R in roster form and hence find its domain and range.

We need to find all ordered pairs (a, b) such that:
a + b = 7

For a = 1:
b = 7 − 1 = 6 ⇒ (1, 6)

For a = 2:
b = 7 − 2 = 5 ⇒ (2, 5)

For a = 3:
b = 7 − 3 = 4 ⇒ (3, 4)

Therefore,
R = {(1, 6), (2, 5), (3, 4)}

The domain is the set of first components:
Domain(R) = {1, 2, 3}

The range is the set of second components:
Range(R) = {4, 5, 6}

2. Given A = {2, 3, 4, 5}, B = {3, 6, 7, 10}. Define R = {(a, b) : a divides b; a ∈ A, b ∈ B}. Write R in roster form hence find its domain and range.

We check which elements of A divide elements of B.

For a = 2:
2 divides 6 and 10.
Therefore, (2, 6) and (2, 10) belong to R.

For a = 3:
3 divides 3 and 6.
Therefore, (3, 3) and (3, 6) belong to R.

For a = 4:
4 does not divide 3, 6, 7 or 10.

For a = 5:
5 divides 10.
Therefore, (5, 10) belongs to R.

Hence,
R = {(2, 6), (2, 10), (3, 3), (3, 6), (5, 10)}

Therefore, the domain is:
Domain(R) = {2, 3, 5}

The range is:
Range(R) = {3, 6, 10}

3. Let R = {(a, b) : a + 2b = 12, a, b ∈ N}. Write R in roster form and hence find its domain and range.

We have:
a + 2b = 12

Taking b = 1, 2, 3, 4, 5, we get:

For b = 1:
a + 2(1) = 12 ⇒ a = 10
⇒ (10, 1)

For b = 2:
a + 4 = 12 ⇒ a = 8
⇒ (8, 2)

For b = 3:
a + 6 = 12 ⇒ a = 6
⇒ (6, 3)

For b = 4:
a + 8 = 12 ⇒ a = 4
⇒ (4, 4)

For b = 5:
a + 10 = 12 ⇒ a = 2
⇒ (2, 5)

Therefore,
R = {(10, 1), (8, 2), (6, 3), (4, 4), (2, 5)}

Domain:
Domain(R) = {2, 4, 6, 8, 10}

Range:
Range(R) = {1, 2, 3, 4, 5}

4. Write R = {(x, x²) : x is a prime number less than 10} in roster form. Also find the range of R.

The prime numbers less than 10 are:
2, 3, 5, 7

For x = 2:
x² = 4 ⇒ (2, 4)

For x = 3:
x² = 9 ⇒ (3, 9)

For x = 5:
x² = 25 ⇒ (5, 25)

For x = 7:
x² = 49 ⇒ (7, 49)

Therefore,
R = {(2, 4), (3, 9), (5, 25), (7, 49)}

The range consists of the second components:
Range(R) = {4, 9, 25, 49}

5. Let A = {p, q, r, s} and B = {1, 2}. How many relations can be defined from set A to set B? List any four of them.

A relation from A to B is any subset of A × B.

Since:
n(A) = 4 and n(B) = 2,
n(A × B) = 4 × 2 = 8

The number of possible relations is the number of subsets of A × B:
2⁸ = 256

Therefore, 256 relations can be defined from A to B.

Now,
A × B = {(p, 1), (p, 2), (q, 1), (q, 2), (r, 1), (r, 2), (s, 1), (s, 2)}

Any four possible relations are:

R₁ = {(p, 1), (q, 2)}

R₂ = {(p, 1), (p, 2), (r, 1)}

R₃ = {(q, 1), (r, 2), (s, 1)}

R₄ = {(p, 2), (q, 1), (r, 2), (s, 1)}

6. Let A = {1, 2, 3, 4, 5}. Define a relation R on A by R = {(a, b) : |a − b| = 2}. Write R in roster form and hence find its domain and range.

We need all ordered pairs (a, b) for which:
|a − b| = 2

Check each element of A:

For a = 1:
|1 − 3| = 2
⇒ (1, 3)

For a = 2:
|2 − 4| = 2
⇒ (2, 4)

For a = 3:
|3 − 1| = 2 and |3 − 5| = 2
⇒ (3, 1), (3, 5)

For a = 4:
|4 − 2| = 2
⇒ (4, 2)

For a = 5:
|5 − 3| = 2
⇒ (5, 3)

Therefore, the relation in roster form is:
R = {(1, 3), (2, 4), (3, 1), (3, 5), (4, 2), (5, 3)}

The domain consists of all first components:
Domain(R) = {1, 2, 3, 4, 5}

The range consists of all second components:
Range(R) = {1, 2, 3, 4, 5}

EXERCISE 3.3 — SOLUTIONS

1. Which of the following relations are functions? Justify your answer.

Definition: A relation is a function if every element of the domain has exactly one image in the codomain.

(a)

From the mapping diagram:
2 → 0
−1 → −1
−2 → 0
3 → 1

Every element of the domain has exactly one image.
Therefore, (a) is a function.

(b)

From the mapping diagram:
6 → 2
2 → 0
3 → 1

The element −1 in the domain has no image. Therefore, every element of the domain does not have an image.
Hence, (b) is not a function.

(c)

From the mapping diagram:
1 → 1
1.2 → 1
1.3 → 1
1.9 → 1
2.5 → 2

Every element of the domain has exactly one image. It is perfectly acceptable for different elements of the domain to have the same image.
Therefore, (c) is a function.

(d)

The element 5 has arrows going to 7, 5 and 2. Thus, one element of the domain has more than one image.

Therefore, (d) is not a function.

Final Answer:
(a) Function    (b) Not a function    (c) Function    (d) Not a function

2. Which of the following relations from A = {3, 5, 7, 9} to B = {1, 2, 3, 4, 5} are functions from A to B?

A relation from A to B is a function from A to B if:

1. Every first component must belong to A.
2. Every second component must belong to B.
3. Every element of A must occur exactly once as a first component.

(a) R₁ = {(3, 2), (5, 4), (7, 5), (9, 5)}

The first components are:
{3, 5, 7, 9} = A

Each element of A occurs exactly once as a first component. Also, all second components 2, 4 and 5 belong to B.

Therefore, R₁ is a function from A to B.

(b) R₂ = {(1, 3), (3, 5), (5, 7)}

The first components are {1, 3, 5}, but 1 is not an element of A and 7 is not an element of B.

Also, the elements 7 and 9 of A have no images.
Therefore, R₂ is not a function from A to B.

(c) R₃ = {(2, 3), (2, 5), (2, 7), (3, 5), (3, 7), (5, 7)}

The first component 2 is not in A, and 7 is not in B. Moreover, the element 2 has three different images:
2 → 3, 2 → 5, 2 → 7

Thus, it does not satisfy the definition of a function from A to B.
Therefore, R₃ is not a function from A to B.

(d) R₄ = {(3, 3), (5, 5)}

Although both ordered pairs are valid elements of A × B, the elements 7 and 9 of A have no images.

Hence, not every element of A has exactly one image.
Therefore, R₄ is not a function from A to B.

Final Answer:
Only R₁ is a function from A to B.

EXERCISE 3.4 — SOLUTIONS

1. What is the domain and range of each of the relations given below? Which of these relations are functions?

(a) R = {(5, 1), (4, 1), (3, 1), (2, 0)}

The domain is the set of first components:
Domain(R) = {2, 3, 4, 5}

The range is the set of second components:
Range(R) = {0, 1}

Every element of the domain has exactly one image. Therefore, R is a function.

(b) R = {(1, −1), (2, −2), (3, −3), (4, −4), (5, −5)}

Domain(R) = {1, 2, 3, 4, 5}

Range(R) = {−5, −4, −3, −2, −1}

Every element of the domain has exactly one image. Therefore, R is a function.

(c) R = {(3, −1), (3, 0), (3, 1), (3, 2)}

Domain(R) = {3}

Range(R) = {−1, 0, 1, 2}

The element 3 has four different images: −1, 0, 1 and 2. Therefore, R is not a function.

Final Answer:
(a) Function    (b) Function    (c) Not a function

2. Draw a rough sketch of each of the following relations. Also write their domain and range.

(a) R = {(x, y) : xy = 8, where x, y ∈ Z}

Since xy = 8, the possible integer factor pairs are:

(1, 8), (2, 4), (4, 2), (8, 1),
(−1, −8), (−2, −4), (−4, −2), (−8, −1)

Therefore,
R = {(1, 8), (2, 4), (4, 2), (8, 1), (−1, −8), (−2, −4), (−4, −2), (−8, −1)}

A rough sketch consists of these eight points on the curve y = 8/x in the first and third quadrants.

Domain(R) = {−8, −4, −2, −1, 1, 2, 4, 8}

Range(R) = {−8, −4, −2, −1, 1, 2, 4, 8}

(b) R = {(x, y) : x = |y|, where x ∈ Z and 0 ≤ x ≤ 5}

Since x = |y|, we have:
y = ±x

For x = 0:
(0, 0)

For x = 1:
(1, 1), (1, −1)

For x = 2:
(2, 2), (2, −2)

For x = 3:
(3, 3), (3, −3)

For x = 4:
(4, 4), (4, −4)

For x = 5:
(5, 5), (5, −5)

Thus, the graph consists of the points:
(0,0), (1,1), (1,−1), (2,2), (2,−2), (3,3), (3,−3), (4,4), (4,−4), (5,5), (5,−5)

Domain(R) = {0, 1, 2, 3, 4, 5}

Range(R) = {−5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5}

(c) R = {(x, y) : y = −√x, where x ∈ (0, ∞)}

Since y = −√x, x must be positive. The graph is the lower half of the square-root curve, beginning just to the right of the y-axis and extending downward as x increases.

Some points on the curve are:
(1, −1), (4, −2), (9, −3), (16, −4)

Domain(R) = (0, ∞)

Since x > 0, we have y < 0.
Range(R) = (−∞, 0)

3. Draw the graph of the functions f, g and h on the same coordinate axes. You may fill the tables given below to draw the graphs.

Given:
f(x) = |x|
g(x) = |x| − 1
h(x) = |x| + 1

Table for f(x) = |x|

x	y = |x|
−2	2
−1	1
0	0
1	1
2	2

Table for g(x) = |x| − 1

x	y = |x| − 1
−2	1
−1	0
0	−1
1	0
2	1

Table for h(x) = |x| + 1

x	y = |x| + 1
−2	3
−1	2
0	1
1	2
2	3

The three graphs are V-shaped graphs.

Relationship among the graphs:

f(x) = |x| has its vertex at (0, 0).
g(x) = |x| − 1 is obtained by shifting the graph of f(x) 1 unit downward. Its vertex is (0, −1).
h(x) = |x| + 1 is obtained by shifting the graph of f(x) 1 unit upward. Its vertex is (0, 1).

4. Draw the graphs of the functions f, g and h on the same coordinate axes. You may fill the tables given below to draw the graph.

Given:
f(x) = x²
g(x) = (x − 1)²
h(x) = (x + 2)²

Table for f(x) = x²

x	y = x2
−2	4
−1	1
0	0
1	1
2	4

Table for g(x) = (x − 1)²

x	y = (x − 1)2
−2	9
−1	4
0	1
1	0
2	1

Table for h(x) = (x + 2)²

x	y = (x + 2)2
−2	0
−1	1
0	4
1	9
2	16

All three graphs are upward-opening parabolas.

Relationship among the graphs:

f(x) = x² has vertex at (0, 0).
g(x) = (x − 1)² is obtained by shifting the graph of f(x) 1 unit to the right. Its vertex is (1, 0).
h(x) = (x + 2)² is obtained by shifting the graph of f(x) 2 units to the left. Its vertex is (−2, 0).

Domains:
Domain(f) = Domain(g) = Domain(h) = R

Ranges:
Range(f) = Range(g) = Range(h) = [0, ∞)

Therefore, their domains are equal and their ranges are also equal.

5. Determine the domain and range of the following functions:

(a) y = 1/x²

The denominator cannot be zero.
Therefore, x ≠ 0.

Domain = R − {0}

Since x² is always positive for x ≠ 0,
1/x² > 0.

Range = (0, ∞)

(b) y = 2 − |x|

|x| is defined for every real number x.
Therefore:
Domain = R

Since |x| ≥ 0:
2 − |x| ≤ 2

The maximum value of y is 2, obtained when x = 0. As |x| can become arbitrarily large, y can become arbitrarily negative.

Range = (−∞, 2]

(c) y = (x − 1)³

A cubic expression is defined for every real value of x.
Therefore:
Domain = R

As x takes all real values, (x − 1)³ also takes all real values.
Range = R

(d) y = √(−x)

For a square root to be real, the expression inside the square root must be non-negative.

−x ≥ 0
x ≤ 0

Therefore:
Domain = (−∞, 0]

A square root is always non-negative, so:
y ≥ 0

As x becomes more negative, √(−x) can become arbitrarily large.
Therefore:
Range = [0, ∞)