Class 12 Maths Chapter 1: Relations and Functions worksheet with answer key

Answer: (c) 64

A × A has 9 pairs. A reflexive relation must contain the 3 pairs (a, a); each of the other 9 − 3 = 6 pairs may be in or out, giving 2⁶ = 64. 512 = 2⁹ counts all relations.

Answer: (b) 64

A relation from A to B is any subset of A × B. A × B has 2 × 3 = 6 elements, so it has 2⁶ = 64 subsets. 6 counts the pairs, not the subsets; 32 and 8 are 2⁵ and 2³.

Answer: (b) the universal relation

The squares of the elements are 1, 4, 9, 16, so the largest value of |a² − b²| is 16 − 1 = 15 < 16. Every pair of A × A is in R, so R = A × A, the universal relation (which is also reflexive and symmetric, so (c) and (d) are false).

Answer: (c) 8

The three pairs (1, 1), (2, 2), (3, 3) must be present. Symmetry makes the other six pairs come in three couples {(1, 2), (2, 1)}, {(1, 3), (3, 1)}, {(2, 3), (3, 2)}, each wholly in or wholly out: 2³ = 8. 64 counts reflexive relations without the symmetry condition.

Answer: (c) 9

P(X) = {φ, {1}, {2}, {1, 2}}. Pairs with A ⊂ B: φ with all four sets (4), {1} with {1} and {1, 2} (2), {2} with {2} and {1, 2} (2), {1, 2} with itself (1): 4 + 2 + 2 + 1 = 9. Equivalently, each of 1 and 2 is in neither set, in B only, or in both, giving 3² = 9. 16 counts all pairs; 5 leaves out the four pairs (A, A).

x₁³ + 2 = x₂³ + 2 gives x₁³ = x₂³, so x₁ = x₂ (x³ is one-one on R). For any real y, x = ∛(y − 2) is real and f(x) = y, so f is onto. Being one-one and onto, f is invertible, with f⁻¹(y) = ∛(y − 2). f⁻¹(10) = ∛8 = 2.

• one-one and onto shown, hence invertible (1)
• f⁻¹(10) = 2 (1)

f(x) = 1 + 3/x. If f(x₁) = f(x₂), then 3/x₁ = 3/x₂, so x₁ = x₂: f is one-one. For y ∈ R, y = 1 + 3/x gives x = 3/(y − 1), which exists when y ≠ 1, and 3/(y − 1) ≠ 0. For y = 1 we would need 3/x = 0, impossible. So 1 is the only real number not in the range, and f is not onto.

• one-one: 3/x₁ = 3/x₂ gives x₁ = x₂ (1)
• x = 3/(y − 1); 1 is not an image (1)

[1] = {b ∈ R : f(b) = f(1)} and f(1) = 1 − 4 = −3. So b² − 4b = −3, b² − 4b + 3 = 0, (b − 1)(b − 3) = 0, b = 1 or 3. Hence [1] = {1, 3}.

• sets up f(b) = f(1) = −3 (1)
• [1] = {1, 3} (1)

f(x) = 3x for x ≥ 0 and f(x) = x for x < 0; f(x) ≥ 0 when x ≥ 0 and f(x) < 0 when x < 0. One-one: if f(x₁) = f(x₂), the common value has one sign, so x₁, x₂ are on the same side of 0; then 3x₁ = 3x₂ or x₁ = x₂, giving x₁ = x₂. Onto: for y ≥ 0, x = y/3 ≥ 0 and f(x) = y; for y < 0, x = y < 0 and f(x) = y. So f is one-one and onto.

• f written piecewise, with the sign of f(x) noted (1)
• one-one by cases (1)
• onto: preimages y/3 and y (1)

a R b exactly when a and b lie in the same subset. From {1, 4}: (1, 1), (1, 4), (4, 1), (4, 4). From {2, 3, 5}: all 9 pairs (2, 2), (2, 3), (2, 5), (3, 2), (3, 3), (3, 5), (5, 2), (5, 3), (5, 5). From {6}: (6, 6). R is the union of these; it is reflexive (every element is in some subset), symmetric and transitive (pairs never cross subsets). Number of pairs = 2² + 3² + 1² = 14.

• rule: related iff in the same subset (1)
• roster form of R (1)
• 14 ordered pairs (1)

(i) Each of 1, 2, 3 has 2 choices of image: 2³ = 8. (ii) Three distinct images are needed but B has only two elements, so there are 0 one-one functions. (iii) A function fails to be onto only when all three images are p or all are q: 2 such functions, so 8 − 2 = 6 are onto.

• 8 functions (1)
• 0 one-one, with reason (1)
• 6 onto (1)