NCERT Exemplar - Relations and Functions

Q.1. Let A = {-1, 2, 3} and B = {1, 3}. Determine
(i) A × B
(ii) B × A
(iii) B × B
(iv) A × A
Ans.
Given that: A = {-1, 2, 3} and B = {1, 3}
(i) A × B = {-1, 2, 3} × {1, 3} = {(-1, 1), (-1, 3), (2, 1), (2, 3), (3, 1), (3, 3)}.
Explanation: Each element of A is paired with every element of B to form ordered pairs.
(ii) B × A = {1, 3} × {-1, 2, 3} = {(1, -1), (1, 2), (1, 3), (3, -1), (3, 2), (3, 3)}.
Explanation: Order matters in Cartesian product; here elements of B appear first in each ordered pair.
(iii) B × B = {1, 3} × {1, 3} = {(1, 1), (1, 3), (3, 1), (3, 3)}.
(iv) A × A = {-1, 2, 3} × {-1, 2, 3} = {(-1, -1), (-1, 2), (-1, 3), (2, -1), (2, 2), (2, 3), (3, -1), (3, 2), (3, 3)}.
Q.2. If P = {x : x < 3, x ∈ N}, Q = {x : x ≤ 2, x ∈ W}. Find (P ∪ Q) × (P ∩ Q), where W is the set of whole numbers.
Ans.
Given that:
P = {x : x < 3, x ∈ N} ⇒ P = {1, 2}.
Q = {x : x ≤ 2, x ∈ W} ⇒ Q = {0, 1, 2}.
Now (P ∪ Q) = {0, 1, 2} and (P ∩ Q) = {1, 2}.
Therefore (P ∪ Q) × (P ∩ Q) = {0, 1, 2} × {1, 2} = {(0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)}.
Q.3. If A = {x : x ∈ W, x < 2} B = {x : x ∈ N, 1 < x < 5} C = {3, 5} find
(i) A × (B ∩ C)
(ii) A × (B ∪ C)
Ans.
Given that:
A = {x : x ∈ W, x < 2} ⇒ A = {0, 1}.
B = {x : x ∈ N, 1 < x < 5} ⇒ B = {2, 3, 4}.
C = {3, 5}.
Now (B ∩ C) = {3} and (B ∪ C) = {2, 3, 4, 5}.
(i) A × (B ∩ C) = {0, 1} × {3} = {(0, 3), (1, 3)}.
(ii) A × (B ∪ C) = {0, 1} × {2, 3, 4, 5} = {(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)}.
Q.4. In each of the following cases, find a and b.
(i) (2a + b, a - b) = (8, 3)
(ii) (a/4, a- 2b) = (0, 6 + b)
Ans.
(i) Given (2a + b, a - b) = (8, 3). Equate components:
2a + b = 8 ...(1)
a - b = 3 ...(2)
Add (1) and (2): 3a = 11 ⇒ a = 11/3.
Then b = a - 3 = 11/3 - 9/3 = 2/3.
Hence a = 11/3, b = 2/3.
(ii) Given (a/4, a - 2b) = (0, 6 + b). Equate components:
a/4 = 0 ⇒ a = 0.
Then a - 2b = 6 + b ⇒ 0 - 2b = 6 + b ⇒ -3b = 6 ⇒ b = -2.
Hence a = 0, b = -2.
Q.5. Given A = {1, 2, 3, 4, 5}, S = {(x, y) : x ∈ A, y ∈ A}. Find the ordered pairs which satisfy the conditions given below:
(i) x + y = 5
(ii) x + y < 5
(iii) x + y > 8
Ans.
Given A = {1, 2, 3, 4, 5} and S = {(x, y) : x ∈ A, y ∈ A}.
(i) x + y = 5 ⇒ ordered pairs: (1, 4), (2, 3), (3, 2), (4, 1).
(ii) x + y < 5 ⇒ ordered pairs: (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1).
(iii) x + y > 8 ⇒ ordered pairs: (4, 5), (5, 4), (5, 5).
Q.6. Given R = {(x, y) : x, y ∈ W, x² + y² = 25}. Find the domain and Range of R.
Ans.
We seek non-negative integer solutions (x, y) with x² + y² = 25.
Possible ordered pairs (with x, y ∈ W = {0,1,2,...}) are: (0, 5), (3, 4), (4, 3), (5, 0).
Therefore domain = {0, 3, 4, 5} and range = {0, 3, 4, 5}.
Q.7. If R₁ = {(x, y) | y = 2x + 7, where x ∈ R and - 5 ≤ x ≤ 5} is a relation. Then find the domain and Range of R₁.
Ans.
Given R₁ = {(x, y) | y = 2x + 7, x ∈ R, -5 ≤ x ≤ 5}.
Domain: All real x with -5 ≤ x ≤ 5, so domain = [-5, 5].
Range: Compute y at endpoints: y(-5) = 2(-5) + 7 = -10 + 7 = -3; y(5) = 2·5 + 7 = 17.
Since y = 2x + 7 is continuous and linear on [-5, 5], range = [-3, 17].
Q.8. If R₂ = {(x, y) | x and y are integers and x² + y² = 64} is a relation. Then find R₂.
Ans.
We seek integer solutions of x² + y² = 64.
Integer pairs: (0, ±8), (±8, 0). (Also (±6, ±√28) are not integers.)
Hence R₂ = {(0, 8), (0, -8), (8, 0), (-8, 0)}.
Q.9. If R₃ = {(x, x ) | x is a real number} is a relation. Then find domain and range of R₃.
Ans.
R₃ = {(x, x) | x ∈ R} is the identity relation on R.
Therefore domain = R and range = R.
Q.10. Is the given relation a function? Give reasons for your answer.
(i) h = {(4, 6), (3, 9), (- 11, 6), (3, 11)}
(ii) f = {(x, x) | x is a real number}
(iii) g = { (n, 1/n) | n is a positive integer }
(iv) s = {(n, n²) | n is a positive integer}
(v) t = {(x, 3) | x is a real number.}
Ans.
(i) h is not a function because the element 3 in the domain has two different images 9 and 11. A function must assign a unique image to each domain element.
(ii) f = {(x, x) | x ∈ R} is a function (the identity function) because each real x has exactly one image x.
(iii) g = {(n, 1/n) | n ∈ N} is a function from positive integers to real numbers: each n has the unique image 1/n.
(iv) s = {(n, n²) | n ∈ N} is a function because each positive integer n maps to the unique value n².
(v) t = {(x, 3) | x ∈ R} is a function (a constant function) since every domain element x maps to the single value 3.
Q.11. If f and g are real functions defined by f (x) = x² + 7 and g (x) = 3x + 5, find each of the following
(a) f (3) + g (- 5)
(b) f(1/2) × g(14)
(c) f (- 2) + g (- 1)
(d) f (t) - f (- 2)
(e) [f(t) - f(5)]/(t-5), if t ≠ 5
Ans.
Given f(x) = x² + 7 and g(x) = 3x + 5.
(a) f(3) + g(-5) = (3² + 7) + (3(-5) + 5) = (9 + 7) + (-15 + 5) = 16 - 10 = 6.
(b) f(1/2) = (1/2)² + 7 = 1/4 + 7 = 29/4.
g(14) = 3·14 + 5 = 42 + 5 = 47.
So f(1/2) × g(14) = (29/4) × 47 = 1363/4 = 340.75.
(c) f(-2) + g(-1) = ((-2)² + 7) + (3(-1) + 5) = (4 + 7) + (-3 + 5) = 11 + 2 = 13.
(d) f(t) - f(-2) = (t² + 7) - ((-2)² + 7) = t² + 7 - 11 = t² - 4.
(e) [f(t) - f(5)]/(t - 5), t ≠ 5:
f(5) = 5² + 7 = 25 + 7 = 32.
So numerator = t² + 7 - 32 = t² - 25 = (t - 5)(t + 5).
Therefore [f(t) - f(5)]/(t - 5) = t + 5, for t ≠ 5.
Q.12. Let f and g be real functions defined by f (x) = 2x + 1 and g (x) = 4x - 7. 
(a) For what real numbers x, f (x) = g (x)?
(b) For what real numbers x, f (x) < g (x)?
Ans.
Given f(x) = 2x + 1 and g(x) = 4x - 7.
(a) Solve 2x + 1 = 4x - 7 ⇒ 1 + 7 = 4x - 2x ⇒ 8 = 2x ⇒ x = 4.
(b) Solve 2x + 1 < 4x - 7 ⇒ 1 + 7 < 4x - 2x ⇒ 8 < 2x ⇒ x > 4.
Q.13. If f and g are two real valued functions defined as f (x) = 2x + 1, g (x) = x² + 1, then find.
(i) f + g
(ii) f - g
(iii) f·g
(iv) f/g
Ans.
Given f(x) = 2x + 1, g(x) = x² + 1.
(i) (f + g)(x) = f(x) + g(x) = x² + 2x + 2.
(ii) (f - g)(x) = f(x) - g(x) = (2x + 1) - (x² + 1) = -x² + 2x.
(iii) (f·g)(x) = (2x + 1)(x² + 1) = 2x³ + x² + 2x + 1.
(iv) (f/g)(x) = (2x + 1)/(x² + 1), provided x² + 1 ≠ 0 (always true for real x).
Q.14. Express the following functions as set of ordered pairs and determine their range. f : X → R, f (x) = x³ + 1, where X = {-1, 0, 3, 9, 7}
Ans.
Compute f(x) for each x in X:
f(-1) = (-1)³ + 1 = 0.
f(0) = 0 + 1 = 1.
f(3) = 27 + 1 = 28.
f(7) = 343 + 1 = 344.
f(9) = 729 + 1 = 730.
Ordered pairs: (-1, 0), (0, 1), (3, 28), (7, 344), (9, 730).
Range = {0, 1, 28, 344, 730}.
Q.15. Find the values of x for which the functions f (x) = 3x² - 1 and g (x) = 3 + x are equal?
Ans.
Set f(x) = g(x): 3x² - 1 = 3 + x ⇒ 3x² - x - 4 = 0.
Factorise: (3x - 4)(x + 1) = 0 ⇒ x = 4/3 or x = -1.
Hence x = -1 and x = 4/3.
Long Answer Type
Q.16. Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} a function? Justify. If this is described by the relation, g (x) = αx + β, then what values should be assigned to α and β?
Ans.
g is a function because every element of the domain {1, 2, 3, 4} has a unique image.
Assume g(x) = αx + β. Use two ordered pairs:
g(1) = α + β = 1 ...(i)
g(2) = 2α + β = 3 ...(ii)
Subtract (i) from (ii): α = 2. Then β = 1 - α = 1 - 2 = -1.
Therefore α = 2 and β = -1. (Check with other points: g(3) = 2·3 - 1 = 5, g(4) = 8 - 1 = 7.)
Q.17. Find the domain of each of the following functions given by

(iii) f(x) = x|x|

Ans.
(i) From the working it is clear the function is of the form f(x) = 1/(1 - cos x). The denominator 1 - cos x = 0 when cos x = 1, i.e. x = 2nπ for n ∈ Z. So domain = R \ {2nπ : n ∈ Z}.
(ii) The expression involves x + |x| in denominator. Note x + |x| = 2x for x ≥ 0, and = 0 for x < 0. Hence denominator zero for x < 0, so function is defined only for x ≥ 0. Thus domain = [0, ∞).
(iii) f(x) = x|x| is defined for all real x, so domain = R.
(iv) The function has denominator x² - 1, so it is undefined where x = ±1. Hence domain = R \ {-1, 1}.
(v) The function has denominator 28 - x, so undefined at x = 28. Hence domain = R \ {28}.
Q.18. Find the range of the following functions given by
(ii) f(x) = 1 - |x - 2|
(iii) f(x) = |x - 3|
(iv) f(x) = 1 + 3 cos 2x
Ans.
(ii) Since |x - 2| ≥ 0, we have 1 - |x - 2| ≤ 1. As |x - 2| can be arbitrarily large, 1 - |x - 2| has no lower bound. Therefore range = (-∞, 1].
(iii) |x - 3| ≥ 0 for all real x, so range = [0, ∞).
(iv) cos 2x ∈ [-1, 1] ⇒ 3 cos 2x ∈ [-3, 3]. Hence 1 + 3 cos 2x ∈ [-2, 4]. So range = [-2, 4].
Q.19. Redefine the function f(x) = |x - 2| + |2 + x|, -3 ≤ x ≤ 3.
Ans.
We split the interval into regions where the absolute values change sign: critical points at x = -2 and x = 2, so consider intervals [-3, -2), [-2, 2), [2, 3].
- For -3 ≤ x < -2: |x - 2| = 2 - x, |2 + x| = -(2 + x) = -2 - x, so f(x) = (2 - x) + (-2 - x) = -2x. Thus f(x) = -2x for -3 ≤ x < -2.
- For -2 ≤ x < 2: |x - 2| = 2 - x, |2 + x| = 2 + x, so f(x) = (2 - x) + (2 + x) = 4. Thus f(x) = 4 for -2 ≤ x < 2.
- For 2 ≤ x ≤ 3: |x - 2| = x - 2, |2 + x| = 2 + x, so f(x) = (x - 2) + (2 + x) = 2x. Thus f(x) = 2x for 2 ≤ x ≤ 3.
Therefore f(x) is given piecewise by:
f(x) = { -2x, -3 ≤ x < -2; 4, -2 ≤ x < 2; 2x, 2 ≤ x ≤ 3 }.
Q.20. If f(x) = x - 1/x + 1, then show that ...
Ans.
Solution retained as per original image placeholders. (No change to image-based derivation.)
Q.21. Let f (x) = √x and g (x) = x be two functions defined in the domain R⁺ ∪ {0}. Find
(i) (f + g) (x)
(ii) (f - g) (x)
(iii) (fg) (x)
(iv) (f/g) (x)
Ans.
Given f(x) = √x, g(x) = x, domain x ∈ [0, ∞).
(i) (f + g)(x) = √x + x.
(ii) (f - g)(x) = √x - x.
(iii) (fg)(x) = √x · x = x · √x = x^3/2.
(iv) (f/g)(x) = √x / x. For x > 0 this simplifies to 1/√x. At x = 0 the quotient is not defined. Hence domain of f/g is (0, ∞) and (f/g)(x) = 1/√x for x > 0.
Q.22. Find the domain and Range of the function f (x) = √(x - 5)
Ans.
For √(x - 5) to be real we require x - 5 ≥ 0 ⇒ x ≥ 5. Thus domain = [5, ∞).
Range: For x ≥ 5, √(x - 5) takes all values y ≥ 0, so range = [0, ∞).
Q.23. If f (x) = (ax - b)/(cx - a), then prove that f (f(x)) = x.
Ans.
Solution retained as per original image placeholders. (Mobius transformation computation shown in images.)
Objective Type Questions
Q.24. Let n (A) = m, and n (B) = n. Then the total number of non-empty relations that can be defined from A to B is
(a) mⁿ
(b) n^m - 1
(c) mn - 1
(d) 2^mn - 1
Ans. (d)
Explanation: A × B has mn ordered pairs. Any relation from A to B is a subset of A × B. There are 2^mn subsets in total; excluding the empty relation gives 2^mn - 1 non-empty relations.
Q.25. If [x]² - 5 [x] + 6 = 0, where [ . ] denote the greatest integer function, then
(a) x ∈ [3, 4]
(b) x ∈ (2, 3]
(c) x ∈ [2, 3]
(d) x ∈ [2, 4)
Ans. (d)
Explanation: Let m = [x]. Then m² - 5m + 6 = 0 ⇒ (m - 2)(m - 3) = 0 ⇒ m = 2 or m = 3.
If [x] = 2 then x ∈ [2, 3). If [x] = 3 then x ∈ [3, 4). Union gives x ∈ [2, 4). Hence option (d) is correct.
Q.26. Range of f (x) = 1/(1 - 2 cos x) is
Ans.
Explanation: cos x ∈ [-1, 1] so denominator d = 1 - 2 cos x ranges over [-1, 3]. The denominator is zero when cos x = 1/2 (so those x are excluded). As d varies in (-1, 0) we get y = 1/d ∈ (-∞, -1). At d = -1 we have y = -1. As d varies in (0, 3] we get y ∈ [1/3, ∞). Hence the full range is (-∞, -1] ∪ [1/3, ∞).
Q.27. Let f be defined by ... then
Ans.
Solution retained as per original image placeholders and hint. The correct relation is f(xy) ≤ f(x)·f(y) (option (c) in original key).
Q.28. Domain of √(a² - x²) (a > 0) is
(a) (-a, a)
(b) [-a, a]
(c) [0, a]
(d) (-a, 0]
Ans. (b)
Explanation: For a² - x² ≥ 0 we need x² ≤ a² ⇒ -a ≤ x ≤ a. Hence domain = [-a, a].
Q.29. If f (x) = ax + b, where a and b are integers, f (-1) = -5 and f (3) = 3, then a and b are equal to
(a) a = -3, b = -1
(b) a = 2, b = -3
(c) a = 0, b = 2
(d) a = 2, b = 3
Ans. (b)
Explanation: From f(-1) = -a + b = -5 and f(3) = 3a + b = 3. Subtract: 4a = 8 ⇒ a = 2. Then b = -5 + a = -5 + 2 = -3. So (a, b) = (2, -3).
Q.30. The domain and range of the real function f defined by f (x) = ... is given by
Ans. (A) Domain = R, Range = {-1, 1} (as per original key)
Solution. Retained as per image reasoning in the original content.
Q.31. The domain and range of the real function f defined by f (x) = ... is given by
Ans. (c) (see original keyed solution)
Solution. Retained as per image reasoning in the original content.
Q.32. The domain and range of real function f given by f (x) = √(x - 1) is
(a) Domain = (1, ∞), Range = (0, ∞)
(b) Domain = [1, ∞), Range = (0, ∞)
(c) Domain = [1, ∞), Range = [0, ∞)
(d) Domain = [1, ∞), Range = [0, ∞)
Ans. (d)
Explanation: x - 1 ≥ 0 ⇒ x ≥ 1 so domain = [1, ∞). For x ≥ 1, √(x - 1) ≥ 0, so range = [0, ∞).
Q.33. The domain of the function f given by f(x) = x² + 2x + 1 / (x² - x - 6) is
(a) R - {3, -2}
(b) R - {-3, 2}
(c) R - [3, -2]
(d) R - (3, -2)
Ans. (a)
Explanation: Denominator x² - x - 6 = (x - 3)(x + 2) must not be 0, so x ≠ 3 and x ≠ -2. Hence domain = R \ {-2, 3}.
Q.34. The domain and range of the function f given by f (x) = 2 - |x - 5| is
(a) Domain = R⁺, Range = (-∞, 1]
(b) Domain = R, Range = (-∞, 2]
(c) Domain = R, Range = (-∞, 2)
(d) Domain = R⁺, Range = (-∞, 2]
Ans. (b)
Explanation: |x - 5| ≥ 0 for all real x so domain = R and f(x) ≤ 2 with no lower bound, so range = (-∞, 2].
Q.35. The domain for which the functions f (x) = 3x² - 1 and g (x) = 3 + x are equal is
Ans.
Solving 3x² - 1 = 3 + x ⇒ 3x² - x - 4 = 0 ⇒ (3x - 4)(x + 1) = 0 ⇒ x = 4/3 or x = -1.
Hence domain (values) = {-1, 4/3}.
Fill in the blanks:
Q.36. Let f and g be two real functions given by
f = {(0, 1), (2, 0), (3, - 4), (4, 2), (5, 1)}
g = {(1, 0), (2, 2), (3, - 1), (4, 4), (5, 3)}
then the domain of f . g is given by _______.
Ans.
Domain of f = {0, 2, 3, 4, 5}. Domain of g = {1, 2, 3, 4, 5}.
Domain of f·g = domain(f) ∩ domain(g) = {2, 3, 4, 5}.
Q.37. Let f = {(2, 4), (5, 6), (8, - 1), (10, - 3)}
g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)}
be two real functions. Then Match the following:
Ans.
Compute on domain {2, 8, 10} = domain(f) ∩ domain(g):
(f - g) = {(2, 4 - 5) = (2, -1), (8, -1 - 4) = (8, -5), (10, -3 - 13) = (10, -16)}.
(f + g) = {(2, 4 + 5) = (2, 9), (8, -1 + 4) = (8, 3), (10, -3 + 13) = (10, 10)}.
(f·g) = {(2, 4·5 = 20), (8, (-1)·4 = -4), (10, (-3)·13 = -39)}.
(f/g) = {(2, 4/5), (8, -1/4), (10, -3/13)} (defined where denominator ≠ 0).
Hence the correct matching is as given in the original key: (a) ↔ (iii), (b) ↔ (iv), (c) ↔ (ii), (d) ↔ (i).
State True or False for the following statements
Q.38. The ordered pair (5, 2) belongs to the relation R = {(x, y) : y = x - 5, x, y ∈ Z}
Ans.
Check: For (5, 2) we require y = x - 5 ⇒ 2 = 5 - 5 = 0, which is false. So (5, 2) ∉ R. Statement is False.
Q.39. If P = {1, 2}, then P × P × P = {(1, 1, 1), (2, 2, 2), (1, 2, 2), (2, 1, 1)}
Ans.
P × P × P actually equals the set of all 3-tuples with entries 1 or 2, namely eight tuples: (1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1), (2,2,2). The given list omits several tuples. Statement is False.
Q.40. If A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}, then (A × B) ∪ (A × C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}.
Ans.
Compute A × B = {(1,3),(1,4),(2,3),(2,4),(3,3),(3,4)} and A × C = {(1,4),(1,5),(1,6),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)}. Their union is exactly the set listed. Statement is True.
Q.41. If (x - 2, y + 5) = (-2, 1/3) are two equal ordered pairs, then x = 4, y = -14/3
Ans.
From equality of ordered pairs: x - 2 = -2 ⇒ x = 0 (not 4). Also y + 5 = 1/3 ⇒ y = 1/3 - 5 = 1/3 - 15/3 = -14/3. So x = 0 and y = -14/3. The statement x = 4, y = -14/3 is False.
Q.42. If A × B = {(a, x), (a, y), (b, x), (b, y)}, then A = {a, b}, B = {x, y}
Ans.
Given A = {a, b} and B = {x, y} yields A × B = {(a, x), (a, y), (b, x), (b, y)}. The statement is True.