Exercise 2.1
If (x/3 + 1, y − 2/3) = (5/3, 1/3), find the values of x and y.
Answer
Equating corresponding components of the ordered pair:
x/3 + 1 = 5/3
⇒ x/3 = 5/3 − 1 = 2/3
⇒ x = 2
y − 2/3 = 1/3 → y = 1/3 + 2/3 = 1
So x = 2, y = 1.
If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A × B).
Answer
n(A) = 3, n(B) = 3. n(A × B) = n(A) × n(B) = 3 × 3 = 9.
If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.
Answer
G × H = {(7,5), (7,4), (7,2), (8,5), (8,4), (8,2)}
H × G = {(5,7), (5,8), (4,7), (4,8), (2,7), (2,8)}
State whether each of the following statements is true or false. If false, rewrite correctly.
(i) If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (n, m)}.
(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.
(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ φ) = φ.
Answer
(i) False. P × Q = {(m,n), (m,m), (n,n), (n,m)} — all ordered pairs from P to Q.
(ii) True.
(iii) True. B ∩ φ = φ, and A × φ = φ.
If A = {−1, 1}, find A × A × A.
Answer
A × A × A = {(a, b, c) : a, b, c ∈ {−1, 1}}
= {(−1,−1,−1), (−1,−1,1), (−1,1,−1), (−1,1,1), (1,−1,−1), (1,−1,1), (1,1,−1), (1,1,1)}
If A × B = {(a,x), (a,y), (b,x), (b,y)}, find A and B.
Answer
The first elements of all ordered pairs form A, and the second elements form B.
A = {a, b}, B = {x, y}.
Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that:
Answer
(i) A × (B ∩ C) = (A × B) ∩ (A × C) (ii) A × C is a subset of B × D.
Ans.
(i) B ∩ C = φ (no common elements). LHS = A × φ = φ.
A × B = {(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}.
A × C = {(1,5),(1,6),(2,5),(2,6)}.
RHS = (A×B) ∩ (A×C) = φ (no common pairs). LHS = RHS = φ. ✓
(ii) A × C = {(1,5),(1,6),(2,5),(2,6)}.
B × D = {(1,5),(1,6),(1,7),(1,8),(2,5),…,(4,8)} — contains all pairs with first element in {1,2,3,4} and second in {5,6,7,8}.
All elements of A × C have first element in {1,2} ⊂ B and second in {5,6} ⊂ D. So A × C ⊂ B × D. ✓
Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.
Answer
A × B = {(1,3), (1,4), (2,3), (2,4)}. n(A × B) = 4, so it has 24 = 16 subsets.
Subsets: φ; {(1,3)}, {(1,4)}, {(2,3)}, {(2,4)}; {(1,3),(1,4)}, {(1,3),(2,3)}, {(1,3),(2,4)}, {(1,4),(2,3)}, {(1,4),(2,4)}, {(2,3),(2,4)}; {(1,3),(1,4),(2,3)}, {(1,3),(1,4),(2,4)}, {(1,3),(2,3),(2,4)}, {(1,4),(2,3),(2,4)}; {(1,3),(1,4),(2,3),(2,4)}.
Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B.
Answer
The second elements of ordered pairs in A × B form B. From the given pairs, B = {1, 2}. The first elements form A; since x, y, z are distinct, A = {x, y, z}.
The Cartesian product A × A has 9 elements among which are found (−1, 0) and (0, 1). Find the set A and the remaining elements of A × A.
Answer
n(A × A) = 9 means n(A) = 3. Since (−1, 0) and (0, 1) are in A × A, both −1 and 0 and 1 must be elements of A. So A = {−1, 0, 1}.
A × A = {(−1,−1),(−1,0),(−1,1),(0,−1),(0,0),(0,1),(1,−1),(1,0),(1,1)}.
Remaining elements (besides (−1,0) and (0,1)): (−1,−1), (−1,1), (0,−1), (0,0), (1,−1), (1,0), (1,1).
Exercise 2.2
Let A = {1, 2, 3, …, 14}. Define R from A to A by R = {(x, y) : 3x − y = 0, x, y ∈ A}. Write down its domain, codomain and range.
Answer
3x − y = 0 means y = 3x. For y to be in A (≤ 14): x = 1
⇒ y = 3; x = 2
⇒ y = 6; x = 3
⇒ y = 9; x = 4
⇒ y = 12; x = 5
⇒ y = 15 > 14 (excluded).
R = {(1,3), (2,6), (3,9), (4,12)}.
Domain = {1, 2, 3, 4}. Codomain = A = {1,2,…,14}. Range = {3, 6, 9, 12}.
Define R on N by R = {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈ N}. Write in roster form. Find domain and range.
Answer
x ∈ {1, 2, 3}: y = 1+5=6, 2+5=7, 3+5=8. R = {(1,6), (2,7), (3,8)}.
Domain = {1, 2, 3}. Range = {6, 7, 8}.
A = {1, 2, 3, 5} and B = {4, 6, 9}. R = {(x, y) : the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form.
Answer
Odd difference between x and y occurs when one is even and the other is odd.
Even elements of A: 2. Odd elements of A: 1, 3, 5. Even elements of B: 4, 6. Odd elements of B: 9.
Pairs where x is odd and y is even: (1,4),(1,6),(3,4),(3,6),(5,4),(5,6).
Pairs where x is even and y is odd: (2,9).
R = {(1,4),(1,6),(3,4),(3,6),(5,4),(5,6),(2,9)}.
The figure shows a relationship between sets P and Q. Write this relation (i) in set-builder form (ii) roster form. What is its domain and range?
Answer
(From the figure: P = {5, 6, 7}, Q = {3, 4, 5}, relation: x is related to (x − 2))
(i) R = {(x, y) : y = x − 2, x ∈ P, y ∈ Q}
(ii) R = {(5, 3), (6, 4), (7, 5)}
Domain = {5, 6, 7}. Range = {3, 4, 5}.
Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by {(a, b) : a, b ∈ A, b is exactly divisible by a}.
Answer
(i) Write R in roster form. (ii) Find the domain of R. (iii) Find the range of R.
Ans.
(i) Checking all pairs (a, b) where b is divisible by a:
(1,1),(1,2),(1,3),(1,4),(1,6),(2,2),(2,4),(2,6),(3,3),(3,6),(4,4),(6,6).
R = {(1,1),(1,2),(1,3),(1,4),(1,6),(2,2),(2,4),(2,6),(3,3),(3,6),(4,4),(6,6)}
(ii) Domain = {1, 2, 3, 4, 6}
(iii) Range = {1, 2, 3, 4, 6}
Determine the domain and range of R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}.
Answer
R = {(0,5),(1,6),(2,7),(3,8),(4,9),(5,10)}. Domain = {0,1,2,3,4,5}. Range = {5,6,7,8,9,10}.
Write R = {(x, x³) : x is a prime number less than 10} in roster form.
Answer
Primes less than 10: 2, 3, 5, 7. R = {(2,8),(3,27),(5,125),(7,343)}.
Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.
Answer
n(A × B) = 3 × 2 = 6. Number of relations = 26 = 64.
Let R be the relation on Z defined by R = {(a,b) : a, b ∈ Z, a − b is an integer}. Find the domain and range of R.
Answer
For any two integers a and b, a − b is always an integer. So every pair of integers is in R. Domain = Z. Range = Z.
Exercise 2.3
Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
(i) {(2,1),(5,1),(8,1),(11,1),(14,1),(17,1)}
(ii) {(2,1),(4,2),(6,3),(8,4),(10,5),(12,6),(14,7)}
(iii) {(1,3),(1,5),(2,5)}
Answer
(i) Function. Each first element (2,5,8,11,14,17) appears exactly once. Domain = {2,5,8,11,14,17}. Range = {1}.
(ii) Function. Each first element appears exactly once. Domain = {2,4,6,8,10,12,14}. Range = {1,2,3,4,5,6,7}.
(iii) Not a function. The element 1 in the domain maps to two different values (3 and 5).
Q2. Find the domain and range of the following real functions
(i) f(x) = −x (ii) f(x) = √(9 − x²)
Ans.
(i) f(x) = −x is defined for all real numbers. Domain = R. Since for every real x, f(x) = −x ∈ R, Range = R.
(ii) f(x) = √(9 − x²) requires 9 − x² ≥ 0 → x² ≤ 9 → −3 ≤ x ≤ 3. Domain = [−3, 3]. For x ∈ [−3, 3], 9 − x² ranges from 0 to 9, so √(9−x²) ranges from 0 to 3. Range = [0, 3].
A function f is defined by f(x) = 2x − 5. Write down the values of:
Answer
(i) f(0) (ii) f(7) (iii) f(−3)
Ans.
(i) f(0) = 2(0) − 5 = −5
(ii) f(7) = 2(7) − 5 = 14 − 5 = 9
(iii) f(−3) = 2(−3) − 5 = −6 − 5 = −11
The function t which maps temperature in degree Celsius into Fahrenheit is defined by t(C) = 9C/5 + 32. Find:
Answer
(i) t(0) (ii) t(28) (iii) t(−10) (iv) The value of C when t(C) = 212
Ans.
(i) t(0) = 9(0)/5 + 32 = 32°F
(ii) t(28) = 9(28)/5 + 32 = 252/5 + 32 = 50.4 + 32 = 82.4°F
(iii) t(−10) = 9(−10)/5 + 32 = −18 + 32 = 14°F
(iv) 9C/5 + 32 = 212
⇒ 9C/5 = 180
⇒ C = 100°C
Q5. Find the range of each of the following functions
(i) f(x) = 2 − 3x, x ∈ R, x > 0
(ii) f(x) = x² + 2, x is a real number
(iii) f(x) = x, x is a real number
Ans.
(i) For x > 0: 3x > 0, so 2 − 3x < 2. As x → 0+, f → 2 (not included); as x → ∞, f → −∞. Range = (−∞, 2).
(ii) x² ≥ 0 for all x, so x² + 2 ≥ 2. Range = [2, ∞).
(iii) f(x) = x takes all real values. Range = R = (−∞, ∞).
Miscellaneous Exercise on Chapter 2
The relation f is defined by f(x) = x², 0 ≤ x ≤ 3 and f(x) = 3x, 3 ≤ x ≤ 10. The relation g is defined by g(x) = x², 0 ≤ x ≤ 2 and g(x) = 3x, 2 ≤ x ≤ 10. Show that f is a function and g is not a function.
Answer
For f: At x = 3 (the overlap point), f(3) = 3² = 9 and f(3) = 3 × 3 = 9. Both rules give the same value 9. So every x in [0, 10] has a unique image. f is a function.
For g: At x = 2 (the overlap point), g(2) = 2² = 4 and g(2) = 3 × 2 = 6. The two rules give different values. So x = 2 has two images. g is not a function.
If f(x) = x², find (f(1.1) − f(1)) / (1.1 − 1).
Answer
f(1.1) = (1.1)² = 1.21, f(1) = 1² = 1.
(f(1.1) − f(1)) / (1.1 − 1) = (1.21 − 1) / 0.1 = 0.21 / 0.1 = 2.1.
Find the domain of f(x) = (x² + 2x + 1) / (x² − 8x + 12).
Answer
f is defined when denominator ≠ 0. x² − 8x + 12 = (x−2)(x−6) = 0 at x = 2 and x = 6.
Domain = R − {2, 6}.
Find the domain and range of f(x) = √(x − 1).
Answer
x − 1 ≥ 0 → x ≥ 1. Domain = [1, ∞).
For x ≥ 1, √(x−1) ≥ 0. Range = [0, ∞).
Find the domain and range of f(x) = √(−x).
Answer
−x ≥ 0
⇒ x ≤ 0. Domain = (−∞, 0] = {x ∈ R : x ≤ 0}.
For x ≤ 0, √(−x) ≥ 0. Range = [0, ∞).
Let f: R → R be defined by f(x) = x / (1 + x²). Determine the range of f.
Answer
Let y = x / (1 + x²). Then yx² − x + y = 0. For real x, discriminant ≥ 0: 1 − 4y² ≥ 0
⇒ y² ≤ 1/4
⇒ −1/2 ≤ y ≤ 1/2. Range = [−1/2, 1/2].
Let f, g : R → R be defined by f(x) = x + 1, g(x) = 2x − 3. Find f + g, f − g and f/g.
Answer
(f + g)(x) = f(x) + g(x) = (x+1) + (2x−3) = 3x − 2
(f − g)(x) = f(x) − g(x) = (x+1) − (2x−3) = −x + 4
(f/g)(x) = f(x)/g(x) = (x+1)/(2x−3), provided 2x − 3 ≠ 0, i.e., x ≠ 3/2
Let f = {(1,1),(2,3),(0,−1),(−1,−3)} be a function from Z to Z defined by f(x) = ax + b for some integers a, b. Determine a and b.
Answer
Using f(0) = −1: a(0) + b = −1
⇒ b = −1. Using f(1) = 1: a(1) + (−1) = 1
⇒ a = 2.
So a = 2, b = −1. Check: f(2) = 2(2)−1 = 3 ✓, f(−1) = 2(−1)−1 = −3 ✓.
Let R be a relation from N to N defined by R = {(a, b) : a, b ∈ N and a = b²}. Are the following true?
Answer
(i) (a, a) ∈ R for all a ∈ N (ii) (a,b) ∈ R implies (b,a) ∈ R (iii) (a,b) ∈ R and (b,c) ∈ R implies (a,c) ∈ R
Ans.
(i) False. (a, a) ∈ R requires a = a², i.e., a(a−1) = 0, so a = 1 only (not all natural numbers).
(ii) False. (9, 3) ∈ R since 9 = 3². But (3, 9) would require 3 = 9² = 81, which is false.
(iii) False. (16, 4) ∈ R (16 = 4²) and (4, 2) ∈ R (4 = 2²), but (16, 2) ∈ R would require 16 = 2² = 4, which is false.
Let A = {1,2,3,4}, B = {1,5,9,11,15,16} and f = {(1,5),(2,9),(3,1),(4,5),(2,11)}. Are the following true?
Answer
(i) f is a relation from A to B (ii) f is a function from A to B
Ans.
(i) True. f ⊂ A × B since all first elements belong to A and all second elements belong to B.
(ii) False. The element 2 in A has two images (9 and 11), violating the definition of a function.
Let f be the subset of Z × Z defined by f = {(ab, a + b) : a, b ∈ Z}. Is f a function from Z to Z? Justify.
Answer
f is not a function. Consider: (2, 1) and (2, −1) give f contains (2×1, 2+1) = (2, 3) and (2×(−1), 2+(−1)) = (−2, 1). More directly: a = 2, b = 2 gives (4, 4); a = 1, b = 4 gives (4, 5). So the pair (4, 4) and (4, 5) would both be in f — the first element 4 maps to two different values. Hence f is not a function.
Let A = {9,10,11,12,13} and f : A → N be defined by f(n) = the highest prime factor of n. Find the range of f.
Answer
Highest prime factor: f(9) = 3 (9 = 3²); f(10) = 5 (10 = 2×5); f(11) = 11 (prime); f(12) = 3 (12 = 2²×3); f(13) = 13 (prime).
Range = {3, 5, 11, 13}.
Important Extra Questions
If A has m elements and B has n elements, how many functions are possible from A to B?
Answer
Each of the m elements of A can independently map to any of the n elements of B. So the number of possible functions = nm.
What is the difference between a relation and a function?
Answer
A relation R from A to B is any subset of A × B. A function is a special relation where every element of A has exactly one image in B (no element of A is left unmapped, and no element maps to two different values).
Find the domain and range of f(x) = 1/(x − 3).
Answer
f is undefined at x = 3. Domain = R − {3}. As x varies over R − {3}, f(x) can take any real value except 0. Range = R − {0}.
Define the modulus function and give its domain and range.
Answer
The modulus function f : R
⇒ R is defined by f(x) = |x| = x if x ≥ 0, and −x if x < 0. Domain = R. Range = [0, ∞) = {x ∈ R : x ≥ 0}.
If f(x) = x² − 1 and g(x) = x + 1, find (f/g)(x) and state its domain.
Answer
(f/g)(x) = (x²−1)/(x+1) = (x−1)(x+1)/(x+1) = x−1, provided x ≠ −1. Domain = R − {−1}. (f/g)(x) = x − 1 for x ≠ −1.