Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
How to Use This Page
Solve each Samacheer Kalvi exercise question on your own first, then compare your method and answer with the step-by-step solutions below. Every question from Chapter 1 (Relations and Functions) is worked out in full. Read the concept, apply the definition, and verify your answer. Pay special attention to the Common Mistakes section to avoid losing marks.
Key Concepts (Quick Recap)
- Ordered Pair: A pair (a, b) where order matters; (a, b) is not equal to (b, a) unless a = b
- Cartesian Product A x B: Set of all ordered pairs (a, b) where a is in A and b is in B; n(A x B) = n(A) times n(B)
- Relation from A to B: Any subset of A x B; total possible relations = 2^(n(A) x n(B))
- Domain of a Relation R: Set of all first elements of ordered pairs in R
- Range of a Relation R: Set of all second elements of ordered pairs in R
- Co-domain: The entire set B (range is always a subset of co-domain)
- Function f: A to B: A relation where every element of A maps to exactly one element of B
- One-one (Injective): f(a1) = f(a2) implies a1 = a2; different inputs give different outputs
- Onto (Surjective): Range = Co-domain; every element of B has a pre-image in A
- Bijective: Both one-one and onto
- Composition (g o f): (g o f)(x) = g(f(x)); apply f first, then g
- Vertical Line Test: A graph represents a function if no vertical line cuts it more than once
- Horizontal Line Test: A function is one-one if no horizontal line cuts its graph more than once
Exercise 1.1 — Ordered Pairs and Cartesian Product
Q1. Find A x B, A x A and B x A for the following: (i) A = {2, -2, 3} and B = {1, -4} (ii) A = B = {p, q} (iii) A = {m, n}; B = empty set
(i) A = {2, -2, 3}, B = {1, -4}
A x B = {(2,1), (2,-4), (-2,1), (-2,-4), (3,1), (3,-4)}
n(A x B) = 3 x 2 = 6
A x A = {(2,2), (2,-2), (2,3), (-2,2), (-2,-2), (-2,3), (3,2), (3,-2), (3,3)}
n(A x A) = 3 x 3 = 9
B x A = {(1,2), (1,-2), (1,3), (-4,2), (-4,-2), (-4,3)}
n(B x A) = 2 x 3 = 6
(ii) A = B = {p, q}
A x B = {(p,p), (p,q), (q,p), (q,q)}
A x A = {(p,p), (p,q), (q,p), (q,q)}
B x A = {(p,p), (p,q), (q,p), (q,q)}
Here A x B = A x A = B x A since A = B.
(iii) A = {m, n}, B = empty set
A x B = empty set (no elements in B to pair with)
A x A = {(m,m), (m,n), (n,m), (n,n)}
B x A = empty set (no elements in B to start a pair)
Q2. Let A = {1, 2, 3} and B = {x | x is a prime number less than 10}. Find A x B and B x A.
B = {2, 3, 5, 7} (primes less than 10)
A x B = {1,2,3} x {2,3,5,7}
= {(1,2), (1,3), (1,5), (1,7),
(2,2), (2,3), (2,5), (2,7),
(3,2), (3,3), (3,5), (3,7)}
n(A x B) = 3 x 4 = 12
B x A = {2,3,5,7} x {1,2,3}
= {(2,1), (2,2), (2,3),
(3,1), (3,2), (3,3),
(5,1), (5,2), (5,3),
(7,1), (7,2), (7,3)}
n(B x A) = 4 x 3 = 12
Q3. If B x A = {(-2, 3), (-2, 4), (0, 3), (0, 4), (3, 3), (3, 4)}, find A and B.
In B x A, the first component is from B and the second from A.
Second components (these are elements of A): 3, 4, 3, 4, 3, 4
A = {3, 4}
First components (these are elements of B): -2, -2, 0, 0, 3, 3
B = {-2, 0, 3}
Verification: B x A = {-2,0,3} x {3,4}
= {(-2,3), (-2,4), (0,3), (0,4), (3,3), (3,4)} matches given set.
Q4. If A = {5, 6}, B = {4, 5, 6}, C = {5, 6, 7}, show that A x A = (B x B) intersection (C x C).
A x A = {5,6} x {5,6} = {(5,5), (5,6), (6,5), (6,6)} ...(1)
B x B = {4,5,6} x {4,5,6}
= {(4,4), (4,5), (4,6), (5,4), (5,5), (5,6),
(6,4), (6,5), (6,6)}
C x C = {5,6,7} x {5,6,7}
= {(5,5), (5,6), (5,7), (6,5), (6,6), (6,7),
(7,5), (7,6), (7,7)}
(B x B) intersection (C x C) = common pairs in both
= {(5,5), (5,6), (6,5), (6,6)} ...(2)
From (1) and (2): A x A = (B x B) intersection (C x C) Verified.
Q5. Given A = {1, 2, 3}, B = {2, 3, 5}, C = {3, 4} and D = {1, 3, 5}, check if (A intersection C) x (B intersection D) = (A x B) intersection (C x D) is true.
LHS:
A intersection C = {1,2,3} intersection {3,4} = {3}
B intersection D = {2,3,5} intersection {1,3,5} = {3,5}
(A intersection C) x (B intersection D) = {3} x {3,5}
= {(3,3), (3,5)} ...(1)
RHS:
A x B = {(1,2), (1,3), (1,5), (2,2), (2,3), (2,5),
(3,2), (3,3), (3,5)}
C x D = {(3,1), (3,3), (3,5), (4,1), (4,3), (4,5)}
(A x B) intersection (C x D) = {(3,3), (3,5)} ...(2)
From (1) and (2): LHS = RHS. The statement is true.
Q6. Let A = {x in W | x < 2}, B = {x in N | 1 < x < 4} and C = {3, 5}. Verify: (i) A x (B union C) = (A x B) union (A x C) (ii) A x (B intersection C) = (A x B) intersection (A x C)
A = {0, 1}, B = {2, 3, 4}, C = {3, 5}
(i) B union C = {2, 3, 4, 5}
A x (B union C) = {(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5)} ...(1)
A x B = {(0,2), (0,3), (0,4), (1,2), (1,3), (1,4)}
A x C = {(0,3), (0,5), (1,3), (1,5)}
(A x B) union (A x C) = {(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1,5)} ...(2)
(1) = (2). Verified.
(ii) B intersection C = {3}
A x (B intersection C) = {(0,3), (1,3)} ...(1)
(A x B) intersection (A x C): common pairs
A x B has (0,3) and (1,3); A x C also has (0,3) and (1,3).
(A x B) intersection (A x C) = {(0,3), (1,3)} ...(2)
(1) = (2). Verified.
Q7. Let A = set of all natural numbers less than 8, B = set of all prime numbers less than 8, C = set of even prime number. Verify: (i) (A intersection B) x C = (A x C) intersection (B x C) (ii) A x (B - C) = (A x B) - (A x C)
A = {1, 2, 3, 4, 5, 6, 7}
B = {2, 3, 5, 7} (primes less than 8)
C = {2} (only even prime)
(i) A intersection B = {2, 3, 5, 7}
(A intersection B) x C = {2,3,5,7} x {2}
= {(2,2), (3,2), (5,2), (7,2)} ...(1)
A x C = {(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (7,2)}
B x C = {(2,2), (3,2), (5,2), (7,2)}
(A x C) intersection (B x C) = {(2,2), (3,2), (5,2), (7,2)} ...(2)
(1) = (2). Verified.
(ii) B - C = {3, 5, 7}
A x (B - C) = {1,2,3,4,5,6,7} x {3,5,7}
= {(1,3), (1,5), (1,7), (2,3), (2,5), (2,7), (3,3), (3,5), (3,7),
(4,3), (4,5), (4,7), (5,3), (5,5), (5,7), (6,3), (6,5), (6,7),
(7,3), (7,5), (7,7)} ...(1)
A x B = all pairs from A x B (21 pairs)
A x C = {(1,2), (2,2), (3,2), (4,2), (5,2), (6,2), (7,2)}
(A x B) - (A x C) = remove all pairs with second element 2
= same as A x (B - C) ...(2)
(1) = (2). Verified.
Exercise 1.2 — Relations
Q1. Let A = {1, 2, 3, 4} and B = {a, b, c}. Which of the following are relations from A to B? R1 = {(1, a), (2, b)}, R2 = {(a, 1), (b, 2)}, R3 = {(1, a), (2, d)}, R4 = {(1, a), (2, b), (3, c)}
A relation from A to B must be a subset of A x B.
A x B = {(1,a), (1,b), (1,c), (2,a), (2,b), (2,c), (3,a), (3,b), (3,c), (4,a), (4,b), (4,c)}
R1 = {(1,a), (2,b)}: Both pairs are in A x B. R1 is a relation. YES.
R2 = {(a,1), (b,2)}: First element 'a' is not in A; second element 1 is not in B.
These pairs are from B x A, not A x B. NOT a relation from A to B. NO.
R3 = {(1,a), (2,d)}: (2,d) has 'd' which is not in B. NOT a relation from A to B. NO.
R4 = {(1,a), (2,b), (3,c)}: All three pairs are in A x B. R4 is a relation. YES.
Q2. Let A = {1, 2, 3, 4, ..., 45}. Let R be the relation "is the square of" from A to A. Write R as a set of ordered pairs. Find the domain and range.
R = {(a, b) | a, b in A and a = b squared}
Squares of elements of A that are also in A:
1^2 = 1 -> (1,1)
2^2 = 4 -> (4,2)
3^2 = 9 -> (9,3)
4^2 = 16 -> (16,4)
5^2 = 25 -> (25,5)
6^2 = 36 -> (36,6)
7^2 = 45 (not in A since 45 is in A, but 7^2 = 49 which is not in A)
Wait: 6^2 = 36 (in A), 7^2 = 49 (not in A). So stop at 6.
R = {(1,1), (4,2), (9,3), (16,4), (25,5), (36,6)}
Domain = {1, 4, 9, 16, 25, 36}
Range = {1, 2, 3, 4, 5, 6}
Q3. Let A = {1, 2, 3, 4, 6} and R be the relation on A defined by {(a, b) | a divides b}. Write R as a set. Find domain and range.
R = {(a,b) | a divides b, a and b in A = {1,2,3,4,6}}
1 divides everything: (1,1), (1,2), (1,3), (1,4), (1,6)
2 divides 2, 4, 6: (2,2), (2,4), (2,6)
3 divides 3, 6: (3,3), (3,6)
4 divides 4: (4,4)
6 divides 6: (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)}
Domain = {1, 2, 3, 4, 6}
Range = {1, 2, 3, 4, 6}
Q4. Represent the following relations by (i) arrow diagram, (ii) graph, (iii) roster form: (a) R = {(x, 2x) | x in {1, 2, 3, 4}} (b) R = {(x, x + 3) | x in {1, 2, 3}}
(a) R = {(x, 2x) | x in {1,2,3,4}}
Roster form: R = {(1,2), (2,4), (3,6), (4,8)}
Domain = {1, 2, 3, 4}, Range = {2, 4, 6, 8}
Arrow diagram: 1->2, 2->4, 3->6, 4->8
Graph: Plot points (1,2), (2,4), (3,6), (4,8)
(b) R = {(x, x+3) | x in {1,2,3}}
Roster form: R = {(1,4), (2,5), (3,6)}
Domain = {1, 2, 3}, Range = {4, 5, 6}
Arrow diagram: 1->4, 2->5, 3->6
Graph: Plot points (1,4), (2,5), (3,6)
Q5. A company has four employees E1, E2, E3, E4 with salaries Rs 12,000, Rs 18,000, Rs 12,000, Rs 25,000 respectively. Express the relation R (employee-salary) as ordered pairs. Is R a function?
R = {(E1, 12000), (E2, 18000), (E3, 12000), (E4, 25000)}
Domain = {E1, E2, E3, E4}
Range = {12000, 18000, 25000}
Each employee (input) maps to exactly one salary (output).
Even though E1 and E3 have the same salary, that is allowed.
A function requires each INPUT to have one output, not the reverse.
R is a function.
Exercise 1.3 — Functions
Q1. Let f = {(x, y) | x, y in N and y = 2x} be a relation on N. Find the domain, co-domain and range. Is this relation a function?
f = {(1,2), (2,4), (3,6), (4,8), ...}
Each natural number x maps to exactly one value y = 2x.
So this is a function.
Domain = N = {1, 2, 3, 4, ...}
Co-domain = N = {1, 2, 3, 4, ...}
Range = {2, 4, 6, 8, ...} = set of even natural numbers
Q2. Let X = {3, 4, 6, 8}. Determine whether the relation R = {(x, f(x)) | x in X, f(x) = x squared + 1} is a function from X to N.
f(3) = 3^2 + 1 = 9 + 1 = 10
f(4) = 4^2 + 1 = 16 + 1 = 17
f(6) = 6^2 + 1 = 36 + 1 = 37
f(8) = 8^2 + 1 = 64 + 1 = 65
R = {(3,10), (4,17), (6,37), (8,65)}
Each element of X maps to exactly one element in N.
R is a function from X to N.
Q3. Given the function f: x maps to x squared - 5x + 6, evaluate: (i) f(-1) (ii) f(2a) (iii) f(2) (iv) f(x - 1)
f(x) = x^2 - 5x + 6
(i) f(-1) = (-1)^2 - 5(-1) + 6
= 1 + 5 + 6 = 12
(ii) f(2a) = (2a)^2 - 5(2a) + 6
= 4a^2 - 10a + 6
(iii) f(2) = (2)^2 - 5(2) + 6
= 4 - 10 + 6 = 0
(iv) f(x-1) = (x-1)^2 - 5(x-1) + 6
= x^2 - 2x + 1 - 5x + 5 + 6
= x^2 - 7x + 12
Q4. A function f(x) = 2x + 5 is given. If x is not 0, find [f(x+2) - f(2)] / x.
f(x+2) = 2(x+2) + 5 = 2x + 4 + 5 = 2x + 9
f(2) = 2(2) + 5 = 4 + 5 = 9
[f(x+2) - f(2)] / x = (2x + 9 - 9) / x = 2x / x = 2
Q5. A function f is defined by f(x) = 2x - 3. Find: (i) [f(0) + f(1)] / 2 (ii) x such that f(x) = 0 (iii) x such that f(x) = x (iv) x such that f(x) = f(1 - x)
(i) f(0) = 2(0) - 3 = -3
f(1) = 2(1) - 3 = -1
[f(0) + f(1)] / 2 = [-3 + (-1)] / 2 = -4/2 = -2
(ii) f(x) = 0
2x - 3 = 0
2x = 3
x = 3/2
(iii) f(x) = x
2x - 3 = x
x = 3
(iv) f(x) = f(1 - x)
2x - 3 = 2(1 - x) - 3
2x - 3 = 2 - 2x - 3
2x - 3 = -1 - 2x
4x = 2
x = 1/2
Q6. A function f is defined by f(x) = 3 - 2x. Find x such that f(x squared) = [f(x)] squared.
f(x^2) = 3 - 2x^2
[f(x)]^2 = (3 - 2x)^2 = 9 - 12x + 4x^2
Setting them equal:
3 - 2x^2 = 9 - 12x + 4x^2
Rearranging:
0 = 9 - 12x + 4x^2 - 3 + 2x^2
0 = 6x^2 - 12x + 6
0 = x^2 - 2x + 1
0 = (x - 1)^2
x = 1
Q7. A square piece of material 24 cm on a side has equal squares cut from the corners and the sides turned up to form an open box. Express the volume V as a function of x (side of cut square).
After cutting squares of side x from each corner:
Length = 24 - 2x
Breadth = 24 - 2x
Height = x
V(x) = (24 - 2x)(24 - 2x)(x)
= (24 - 2x)^2 * x
= (576 - 96x + 4x^2)(x)
= 4x^3 - 96x^2 + 576x
Q8. A plane is flying at 500 km per hour. Express the distance d as a function of time t in hours.
Distance = Speed x Time
d = 500t
Domain: t >= 0 (time cannot be negative)
Q9. A forehand length x (in cm) and height y (in inches) are related by y = ax + b. Given data: (35, 56), (45, 65), (50, 69.5), (55, 74). Find a and b. Is this relation a function?
Using two data points: (35, 56) and (55, 74)
56 = 35a + b ...(1)
74 = 55a + b ...(2)
Subtract (1) from (2): 18 = 20a
a = 18/20 = 0.9
Substituting in (1): 56 = 35(0.9) + b
56 = 31.5 + b
b = 24.5
y = 0.9x + 24.5
Each forehand length gives exactly one height, so this is a function.
Verification with other points:
x = 45: y = 0.9(45) + 24.5 = 40.5 + 24.5 = 65 matches
x = 50: y = 0.9(50) + 24.5 = 45 + 24.5 = 69.5 matches
Q10. Using y = 0.9x + 24.5 from Q9, find: (i) height when forehand length is 40 cm (ii) forehand length when height is 53.3 inches.
(i) y = 0.9(40) + 24.5 = 36 + 24.5 = 60.5 inches
(ii) 53.3 = 0.9x + 24.5
28.8 = 0.9x
x = 32 cm
Exercise 1.4 — Types of Functions
Q1. Represent f(x) = x/2 - 1 by (i) a set of ordered pairs (ii) a table (iii) an arrow diagram (iv) a graph, for the domain {2, 4, 6, 10, 12}.
f(x) = x/2 - 1
f(2) = 2/2 - 1 = 0
f(4) = 4/2 - 1 = 1
f(6) = 6/2 - 1 = 2
f(10) = 10/2 - 1 = 4
f(12) = 12/2 - 1 = 5
(i) Ordered pairs: f = {(2,0), (4,1), (6,2), (10,4), (12,5)}
(ii) Table:
x | 2 | 4 | 6 | 10 | 12
f(x)| 0 | 1 | 2 | 4 | 5
(iii) Arrow diagram: 2->0, 4->1, 6->2, 10->4, 12->5
(iv) Graph: Plot points (2,0), (4,1), (6,2), (10,4), (12,5)
Q2. Show that f: N to N defined by f(x) = 2x - 1 is one-one but not onto.
One-one test:
Let f(a) = f(b)
2a - 1 = 2b - 1
2a = 2b
a = b
So f is one-one.
Onto test:
Co-domain = N = {1, 2, 3, 4, ...}
For f(x) = 2x - 1, the range is {1, 3, 5, 7, ...} (odd natural numbers).
Even numbers like 2, 4, 6, ... have no pre-image.
Range is not equal to Co-domain.
So f is NOT onto.
Q3. Determine whether f: R to R is bijective for: (i) f(x) = 2x + 1 (ii) f(x) = 3 - 4x squared
(i) f(x) = 2x + 1
One-one: f(a) = f(b) => 2a+1 = 2b+1 => a = b. Yes, one-one.
Onto: For any y in R, set y = 2x+1 => x = (y-1)/2, which is in R. Yes, onto.
Bijective: Yes.
(ii) f(x) = 3 - 4x^2
One-one: f(1) = 3-4 = -1 and f(-1) = 3-4 = -1.
Two different inputs give same output. NOT one-one.
(No need to check onto since already not bijective.)
But for completeness: f(x) = 3-4x^2 <= 3 for all x.
Values greater than 3 have no pre-image. NOT onto.
Not bijective.
Q4. Find a and b such that f(x) = ax + b is an onto function for A = {-1, 1} and B = {0, 2}.
f(-1) = a(-1) + b = -a + b
f(1) = a(1) + b = a + b
For f to be onto, both elements of B = {0, 2} must be images.
Case 1: f(-1) = 0 and f(1) = 2
-a + b = 0 => b = a
a + b = 2 => a + a = 2 => a = 1, b = 1
Case 2: f(-1) = 2 and f(1) = 0
-a + b = 2
a + b = 0 => b = -a
-a + (-a) = 2 => -2a = 2 => a = -1, b = 1
Both solutions work. Taking the first: a = 1, b = 1.
Q5. Evaluate the piecewise function: f(x) = x^2 if x <= 2; f(x) = 2x + 1 if x > 2. Find f(3), f(0), f(-1.5), f(2) + f(-2).
f(3): 3 > 2, so f(3) = 2(3) + 1 = 7
f(0): 0 <= 2, so f(0) = 0^2 = 0
f(-1.5): -1.5 <= 2, so f(-1.5) = (-1.5)^2 = 2.25
f(2): 2 <= 2, so f(2) = 2^2 = 4
f(-2): -2 <= 2, so f(-2) = (-2)^2 = 4
f(2) + f(-2) = 4 + 4 = 8
Q6. The function t(C) = (9/5)C + 32 converts Celsius to Fahrenheit. Find: (i) t(0) (ii) t(28) (iii) t(-10) (iv) C when t(C) = 212 (v) temperature where C = F.
(i) t(0) = (9/5)(0) + 32 = 32 degrees F
(ii) t(28) = (9/5)(28) + 32 = 252/5 + 32 = 50.4 + 32 = 82.4 degrees F
(iii) t(-10) = (9/5)(-10) + 32 = -18 + 32 = 14 degrees F
(iv) t(C) = 212
(9/5)C + 32 = 212
(9/5)C = 180
C = 180 x 5/9 = 100 degrees C
(v) C = F means C = (9/5)C + 32
C - (9/5)C = 32
-(4/5)C = 32
C = -40
At -40 degrees, Celsius equals Fahrenheit.
Exercise 1.5 — Composition of Functions
Q1. Find f o g and g o f: (i) f(x) = x - 6, g(x) = x squared (ii) f(x) = 2/x, g(x) = 2x squared - 1 (iii) f(x) = (x+6)/3, g(x) = 3 - x
(i) f(x) = x - 6, g(x) = x^2
(f o g)(x) = f(g(x)) = f(x^2) = x^2 - 6
(g o f)(x) = g(f(x)) = g(x-6) = (x-6)^2 = x^2 - 12x + 36
(ii) f(x) = 2/x, g(x) = 2x^2 - 1
(f o g)(x) = f(g(x)) = f(2x^2 - 1) = 2/(2x^2 - 1)
(g o f)(x) = g(f(x)) = g(2/x) = 2(2/x)^2 - 1 = 2(4/x^2) - 1 = 8/x^2 - 1
(iii) f(x) = (x+6)/3, g(x) = 3 - x
(f o g)(x) = f(g(x)) = f(3-x) = ((3-x)+6)/3 = (9-x)/3 = 3 - x/3
(g o f)(x) = g(f(x)) = g((x+6)/3) = 3 - (x+6)/3 = (9-x-6)/3 = (3-x)/3
Q2. Find k so that f o g = g o f: (i) f(x) = 3x + 2, g(x) = 6x - k (ii) f(x) = 2x - k, g(x) = 4x + 5
(i) (f o g)(x) = f(6x - k) = 3(6x - k) + 2 = 18x - 3k + 2
(g o f)(x) = g(3x + 2) = 6(3x + 2) - k = 18x + 12 - k
Setting equal: 18x - 3k + 2 = 18x + 12 - k
-3k + 2 = 12 - k
-2k = 10
k = -5
(ii) (f o g)(x) = f(4x + 5) = 2(4x + 5) - k = 8x + 10 - k
(g o f)(x) = g(2x - k) = 4(2x - k) + 5 = 8x - 4k + 5
Setting equal: 8x + 10 - k = 8x - 4k + 5
10 - k = -4k + 5
3k = -5
k = -5/3
Q3. Show that f o g = g o f = x for f(x) = 2x - 1 and g(x) = (x + 1)/2.
(f o g)(x) = f(g(x)) = f((x+1)/2) = 2((x+1)/2) - 1 = (x+1) - 1 = x
(g o f)(x) = g(f(x)) = g(2x-1) = ((2x-1)+1)/2 = 2x/2 = x
Both f o g and g o f equal x.
f and g are inverses of each other.
Q4. If f(x) = 2x + 1 and g(x) = x squared, find the range of f o g and g o f for the domain {1, 2, 3}.
(f o g)(x) = f(g(x)) = f(x^2) = 2x^2 + 1
x = 1: f(g(1)) = 2(1) + 1 = 3
x = 2: f(g(2)) = 2(4) + 1 = 9
x = 3: f(g(3)) = 2(9) + 1 = 19
Range of f o g = {3, 9, 19}
(g o f)(x) = g(f(x)) = g(2x+1) = (2x+1)^2
x = 1: g(f(1)) = 3^2 = 9
x = 2: g(f(2)) = 5^2 = 25
x = 3: g(f(3)) = 7^2 = 49
Range of g o f = {9, 25, 49}
Q5. If f(x) = x squared - 1, find f o f and f o f o f.
f o f = f(f(x)) = f(x^2 - 1) = (x^2 - 1)^2 - 1
= x^4 - 2x^2 + 1 - 1
= x^4 - 2x^2
f o f o f = f(f(f(x))) = f(x^4 - 2x^2)
= (x^4 - 2x^2)^2 - 1
= x^8 - 4x^6 + 4x^4 - 1
Exercise 1.6 — Multiple Choice Questions
Q1. If n(A x B) = 6 and A = {1, 3}, then n(B) is:
n(A x B) = n(A) x n(B)
6 = 2 x n(B)
n(B) = 3
Q2. If A = {1, 2, 3} and B = {a, b}, find n[(A union B) x B].
A union B = {1, 2, 3, a, b}
n(A union B) = 5
n(B) = 2
n[(A union B) x B] = 5 x 2 = 10
Q3. If (a + 2, 4) = (5, 2a + b), find a and b.
Two ordered pairs are equal if corresponding components are equal.
a + 2 = 5 => a = 3
4 = 2a + b => 4 = 2(3) + b => 4 = 6 + b => b = -2
Q4. Find the range of R = {(x, x squared) | x is a prime number less than 13}.
Primes less than 13: {2, 3, 5, 7, 11}
R = {(2,4), (3,9), (5,25), (7,49), (11,121)}
Range = {4, 9, 25, 49, 121}
Q5. If f is bijective and n(B) = 7, find n(A).
A bijective function is both one-one and onto.
For a bijection from A to B, every element of A maps to a unique element of B,
and every element of B has exactly one pre-image.
So n(A) = n(B) = 7.
Common Mistakes
1. Reversing ordered pairs. (a, b) is NOT the same as (b, a) unless a = b. When writing A x B, the first component always comes from A. Writing (b, a) instead of (a, b) is the most common error in cartesian product questions.
2. Confusing domain and range. Domain = set of first elements (inputs). Range = set of second elements (outputs). Students often swap these, especially when the relation is given as a formula rather than ordered pairs.
3. Calling every relation a function. A relation is a function only when every element of the domain maps to exactly ONE element in the co-domain. If any input has two or more outputs, it is not a function. Do not confuse "two inputs giving the same output" (which is allowed) with "one input giving two outputs" (which is not).
4. Assuming f o g = g o f. Composition is NOT commutative in general. f(g(x)) and g(f(x)) give different results most of the time. Always compute both separately.
5. Confusing one-one and onto. One-one means different inputs give different outputs. Onto means every element of the co-domain is covered. A function can be one-one but not onto, onto but not one-one, both, or neither.
6. Forgetting the vertical line test for graphs. When asked whether a graph represents a function, use the vertical line test: if any vertical line cuts the graph more than once, it is NOT a function.
7. Missing the empty set in cartesian product. A x (empty set) = empty set. If B is empty, there are no ordered pairs to form. Students sometimes write A x empty set = A, which is wrong.
8. Counting repeated domain values twice. Domain is a set, so repeated values are counted once. If R = {(1,2), (1,3)}, domain = {1}, not {1, 1}.
Exam Checklist
- [ ] I can form cartesian products A x B, A x A, B x A and count their elements.
- [ ] I can find sets A and B given B x A.
- [ ] I can verify distributive properties of cartesian products over union and intersection.
- [ ] I can identify whether a given set of ordered pairs is a relation from A to B.
- [ ] I can find domain, co-domain, and range of a relation.
- [ ] I can determine whether a relation is a function.
- [ ] I can evaluate functions at specific values, including piecewise functions.
- [ ] I can represent functions by ordered pairs, tables, arrow diagrams, and graphs.
- [ ] I can check if a function is one-one (injective) using definition or horizontal line test.
- [ ] I can check if a function is onto (surjective) by comparing range with co-domain.
- [ ] I can identify bijective functions.
- [ ] I can find f o g and g o f (composition of functions).
- [ ] I can find the value of unknown constants in composition problems.
- [ ] I can identify inverse functions (where f o g = g o f = x).
- [ ] I can apply functions to real-world problems (temperature, distance, volume).
Internal Links
- [Class 10 Maths — Relations and Functions Chapter Guide](/school/tamil-nadu-state-board/class/10/maths/relations-and-functions)
- [All Class 10 Maths Solutions](/school/tamil-nadu-state-board/class/10/maths/solutions)
- [Class 10 Maths Subject Guide](/school/tamil-nadu-state-board/class/10/maths)
Frequently Asked Questions
Every function is a relation, but not every relation is a function. A relation from A to B is any subset of A x B. A function requires that each element of the domain maps to exactly one element in the co-domain. If one input has two different outputs, it is a relation but not a function.