How to Use This Page
Solve each NCERT exercise question independently first, then use the solutions below to check your approach and final answer. Every question from Chapter 2 (Relations and Functions) is worked out with full steps — read the concept, apply the definition, and verify.
Key Concepts (Quick Recap)
- Ordered pair: (a, b) where (a, b) ≠ (b, a) unless a = b
- Cartesian product: A × B = {(a, b) : a ∈ A, b ∈ B}; n(A × B) = n(A) × n(B)
- Relation R from A to B: R ⊆ A × B; domain = set of first elements, range = set of second elements
- Function f: A → B: Each a ∈ A maps to exactly one b ∈ B; f(a) denotes the image of a
- One-one (injective): f(a₁) = f(a₂) ⟹ a₁ = a₂
- Onto (surjective): For every b ∈ B, there exists a ∈ A such that f(a) = b
- Composition: (g ∘ f)(x) = g(f(x))
Exercise 2.1
Q1. If (x/3 + 1, y − 2/3) = (5/3, 1/3), find the values of x and y.
By equality of ordered pairs: x/3 + 1 = 5/3 and y − 2/3 = 1/3
From first equation: x/3 = 5/3 − 1 = 5/3 − 3/3 = 2/3 ⟹ x = 2
From second equation: y = 1/3 + 2/3 = 3/3 = 1
x = 2, y = 1
Q2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in A × B.
n(A) = 3, n(B) = 3 n(A × B) = n(A) × n(B) = 3 × 3 = 9
Q3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.
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)}
Note: G × H ≠ H × G (order matters in ordered pairs).
Q4. State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly: (i) 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 (iii) If A = {1, 2} and B = {3, 4}, then A × (B ∩ ∅) = ∅
(i) False. P × Q = {(m,n), (m,m), (n,n), (n,m)} — we need all ordered pairs, not just the "cross" ones. Correct: P × Q = {(m, n), (m, m), (n, n), (n, m)}.
(ii) True. If A and B are non-empty, there exist a ∈ A and b ∈ B, so (a, b) ∈ A × B, making A × B non-empty.
(iii) True. B ∩ ∅ = ∅, so A × ∅ = ∅.
Q5. If A = {−1, 1}, find A × A × A.
A × A = {(−1,−1), (−1,1), (1,−1), (1,1)} A × A × A = all ordered triples with each component from {−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)}
Number of elements = 2³ = 8
Q6. If A × B = {(a, x), (a, y), (b, x), (b, y)}, find A and B.
From the ordered pairs, first components are {a, b} and second components are {x, y}. A = {a, b}, B = {x, y}
Q7. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that: (i) A × (B ∩ C) = (A × B) ∩ (A × C) (ii) A × C is a subset of B × D.
(i) B ∩ C = ∅ (no common elements between {1,2,3,4} and {5,6}) A × (B ∩ C) = A × ∅ = ∅ A × B ∩ A × C: A × B has first components from A with second from B; A × C has first from A with second from C. No pair can have second component in both B and C since B ∩ C = ∅. So (A × B) ∩ (A × C) = ∅. Verified: ∅ = ∅ ✓
(ii) A × C = {(1,5), (1,6), (2,5), (2,6)} B × D = {(1,5), (1,6), (1,7), (1,8), (2,5), (2,6), (2,7), (2,8), (3,5), (3,6), (3,7), (3,8), (4,5), (4,6), (4,7), (4,8)} Every element of A × C is in B × D since A ⊂ B and C ⊂ D. Verified ✓
Q8. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have?
A × B = {(1,3), (1,4), (2,3), (2,4)} n(A × B) = 4 Number of subsets = 2⁴ = 16
Q9. 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, where x, y, z are distinct elements.
Second components of the given pairs are 1 and 2, so B = {1, 2}. First components are x, y, z (distinct), so A = {x, y, z} and n(A) = 3 ✓.
Q10. 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.
n(A × A) = 9 ⟹ n(A) = 3 (since n(A)² = 9) From the given pairs, −1, 0 and 1 appear as first/second components. 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 7 elements (excluding (−1,0) and (0,1)): {(−1,−1), (−1,1), (0,−1), (0,0), (1,−1), (1,0), (1,1)}
Exercise 2.2
Q1. Let A = {1, 2, 3, ..., 14} and R be a relation from A to A defined by R = {(x, y) : 3x − y = 0, x, y ∈ A}. Write R in roster form. Find its domain and range.
3x − y = 0 ⟹ y = 3x. Both x and y must be in A = {1, 2, ..., 14}.
Checking: x=1 → y=3 ✓; x=2 → y=6 ✓; x=3 → y=9 ✓; x=4 → y=12 ✓; x=5 → y=15 > 14 ✗
R = {(1, 3), (2, 6), (3, 9), (4, 12)} Domain = {1, 2, 3, 4} Range = {3, 6, 9, 12}
Q2. Define a relation R on the set N of natural numbers by R = {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈ N}. Depict this relationship using (i) roster form (ii) an arrow diagram.
x ∈ {1, 2, 3} and y = x + 5. x=1 → y=6; x=2 → y=7; x=3 → y=8
(i) R = {(1, 6), (2, 7), (3, 8)} (ii) Arrow diagram: 1 → 6, 2 → 7, 3 → 8
Domain = {1, 2, 3}, Range = {6, 7, 8}
Q3. A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y) : the difference x − y is odd, x ∈ A, y ∈ B}. Write R in roster form.
x − y is odd when one of x, y is even and the other is odd.
Checking all pairs:
| (1,4): 1−4=−3 (odd) ✓ | (1,6): 1−6=−5 (odd) ✓ | (1,9): 1−9=−8 (even) ✗ |
|---|---|---|
| (2,4): 2−4=−2 (even) ✗ | (2,6): 2−6=−4 (even) ✗ | (2,9): 2−9=−7 (odd) ✓ |
| (3,4): 3−4=−1 (odd) ✓ | (3,6): 3−6=−3 (odd) ✓ | (3,9): 3−9=−6 (even) ✗ |
| (5,4): 5−4=1 (odd) ✓ | (5,6): 5−6=−1 (odd) ✓ | (5,9): 5−9=−4 (even) ✗ |
R = {(1,4), (1,6), (2,9), (3,4), (3,6), (5,4), (5,6)}
Q4. The given arrow diagram shows a relation R from A to B. Write R in roster form. (Assume A = {1, 2, 3, 4}, B = {a, b, c} with arrows: 1→a, 2→b, 2→c, 3→c, 4→a)
R = {(1, a), (2, b), (2, c), (3, c), (4, a)} Domain = {1, 2, 3, 4} Range = {a, b, c}
Q5. 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}. (i) Write R in roster form (ii) Find the domain and range of R.
a=1: all elements → (1,1), (1,2), (1,3), (1,4), (1,6) a=2: b = 2,4,6 → (2,2), (2,4), (2,6) a=3: b = 3,6 → (3,3), (3,6) a=4: b = 4 → (4,4) a=6: b = 6 → (6,6)
| (i) For each a ∈ A, find all b ∈ A such that a | b: |
|---|
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 = A = {1, 2, 3, 4, 6}, Range = {1, 2, 3, 4, 6}
Q6. Determine the domain and range of the relation R defined by R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}.
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}
Q7. Write the relation R = {(x, x³) : x is a prime number less than 10} in roster form.
Primes less than 10: 2, 3, 5, 7 R = {(2, 8), (3, 27), (5, 125), (7, 343)}
Q8. Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.
n(A × B) = 3 × 2 = 6 Number of relations = number of subsets of A × B = 2⁶ = 64
Exercise 2.3
Q1. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range. (i) {(1, 2), (1, 3), (2, 4), (3, 5)} (ii) {(1, 2), (2, 3), (3, 4), (4, 5)} (iii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5)}
(i) Not a function. Element 1 maps to both 2 and 3 (one input, two outputs). (ii) Function. Each element in the domain maps to exactly one element. Domain = {1, 2, 3, 4}, Range = {2, 3, 4, 5}. (iii) Function. Each input has exactly one output. Domain = {2, 4, 6, 8, 10}, Range = {1, 2, 3, 4, 5}.
| Q2. Find the domain and range of the following real functions: (i) f(x) = − | x | (ii) f(x) = √(9 − x²) |
|---|
| (i) f(x) = − | x | |||
|---|---|---|---|---|
| Domain: All real numbers ( | x | is defined for all x). Domain = R | ||
| Range: | x | ≥ 0, so − | x | ≤ 0. Range = (−∞, 0] |
(ii) f(x) = √(9 − x²) For the square root to be real: 9 − x² ≥ 0 ⟹ x² ≤ 9 ⟹ −3 ≤ x ≤ 3 Domain = [−3, 3] When x = 0, f(0) = 3 (maximum); when x = ±3, f(±3) = 0 (minimum). Range = [0, 3]
Q3. A function f is defined by f(x) = 2x − 5. Write down the values of: (i) f(0) (ii) f(7) (iii) f(−3)
(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
Q4. The function 't' which maps temperature in Celsius to temperature in Fahrenheit is defined by t(C) = (9C/5) + 32. Find: (i) t(0) (ii) t(28) (iii) t(−10) (iv) The value of C when t(C) = 212
(i) t(0) = 9(0)/5 + 32 = 32 (ii) t(28) = 9(28)/5 + 32 = 252/5 + 32 = 50.4 + 32 = 82.4 (iii) t(−10) = 9(−10)/5 + 32 = −18 + 32 = 14 (iv) 212 = 9C/5 + 32 ⟹ 180 = 9C/5 ⟹ C = 100
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
(i) When x > 0, 3x > 0, so −3x < 0, and 2 − 3x < 2. As x → 0⁺, f(x) → 2⁻; as x → ∞, f(x) → −∞. Range = (−∞, 2).
(ii) x² ≥ 0, so x² + 2 ≥ 2. When x = 0, f(0) = 2 (minimum). Range = [2, ∞).
(iii) f(x) = x for x ∈ R. Range = R (all real numbers).
Miscellaneous Exercise
Q1. The relation f is defined by f(x) = {x², 0 ≤ x ≤ 3; 3x, 3 ≤ x ≤ 10}. Show that f is a function. Also find f(2), f(5), f(7).
Check at x = 3: Using first definition, f(3) = 3² = 9. Using second definition, f(3) = 3(3) = 9. Both give the same value at the overlapping point, so f is well-defined (a function).
f(2) = 2² = 4 (since 0 ≤ 2 ≤ 3) f(5) = 3(5) = 15 (since 3 ≤ 5 ≤ 10) f(7) = 3(7) = 21 (since 3 ≤ 7 ≤ 10)
Q2. The relation g is defined by g(x) = {x², 0 ≤ x ≤ 2; 3x, 2 ≤ x ≤ 10}. Is g a function? Justify.
Check at x = 2: Using first definition, g(2) = 2² = 4. Using second definition, g(2) = 3(2) = 6. Since g(2) = 4 ≠ 6, the relation gives two different values for the same input. g is not a function.
Q3. 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.
Using f(1) = 1: a(1) + b = 1 ⟹ a + b = 1 ... (i) Using f(2) = 3: a(2) + b = 3 ⟹ 2a + b = 3 ... (ii)
Subtracting (i) from (ii): a = 2 From (i): 2 + b = 1 ⟹ b = −1
Verify: f(0) = 2(0) + (−1) = −1 ✓ and f(−1) = 2(−1) + (−1) = −3 ✓
a = 2, b = −1
Q4. Let f, g : R → R be defined by f(x) = x + 1 and g(x) = 2x − 3. Find: (i) f + g (ii) f − g (iii) f · g (iv) f/g
(i) (f + g)(x) = f(x) + g(x) = (x + 1) + (2x − 3) = 3x − 2 (ii) (f − g)(x) = (x + 1) − (2x − 3) = −x + 4 (iii) (f · g)(x) = (x + 1)(2x − 3) = 2x² − x − 3 (iv) (f/g)(x) = (x + 1)/(2x − 3), x ≠ 3/2
Q5. Let f = {(1, 1), (2, 3), (0, −1), (−1, −3)} be a linear function. Find f(3).
From the previous problem, f(x) = 2x − 1. f(3) = 2(3) − 1 = 6 − 1 = 5
Q6. Find the domain of the function f(x) = x² + 2x + 1 / x² − 8x + 12.
The function is undefined when the denominator = 0. x² − 8x + 12 = 0 ⟹ (x − 2)(x − 6) = 0 ⟹ x = 2 or x = 6
Domain = R − {2, 6} (all real numbers except 2 and 6)
| Q7. Let f(x) = | x | and g(x) = | x | . Find: (i) f + g (ii) f − g (iii) f · g |
|---|
| (i) (f + g)(x) = | x | + | x | = **2 | x | ** |
|---|---|---|---|---|---|---|
| (ii) (f − g)(x) = | x | − | x | = 0 | ||
| (iii) (f · g)(x) = | x | · | x | = x² |
Q8. Let f = {(1, 2), (2, 3), (3, 1)} and g = {(2, 1), (3, 2), (1, 3)}. Find gof and fog.
gof: Apply f first, then g. gof(1) = g(f(1)) = g(2) = 1 gof(2) = g(f(2)) = g(3) = 2 gof(3) = g(f(3)) = g(1) = 3 gof = {(1, 1), (2, 2), (3, 3)}
fog: Apply g first, then f. fog(1) = f(g(1)) = f(3) = 1 fog(2) = f(g(2)) = f(1) = 2 fog(3) = f(g(3)) = f(2) = 3 fog = {(1, 1), (2, 2), (3, 3)}
Internal Links
- [Chapter 2 Guide — Relations and Functions](/school/cbse/class/11/maths/cbse-relations-and-functions)
- [All Class 11 Maths Solutions](/school/cbse/class/11/maths/solutions)
- [Class 11 Maths Subject Guide](/school/cbse/class/11/maths)