JEE Mathematics Set Operations Relations and Mappings MCQs Set 03

Answer: (c) A × B

Explanation:
1. A relation from A to B is a set of ordered pairs (x, y) with x in A and y in B.
2. Hence R ⊆ A × B.

Teacher's Note:
1. A relation from A to B always sits inside A × B, in that order.
2. B × A would have the elements in the wrong order.

Answer: (d) None of these

Explanation:
1. n(A × B) = 3 × 4 = 12.
2. Every subset of A × B is a relation, so there are 2¹² relations (including the empty relation).
3. This equals none of the first three options.

Teacher's Note:
1. The count depends only on n(A) · n(B), not on the common elements.
2. Count the empty relation too.

Answer: (c) \( R \subseteq A \times B \)

Explanation:
1. A relation from A to B is by definition a subset of A × B.

Teacher's Note:
1. Keep the order A first, B second.

Answer: (d) \( R_4 = \{(1, 3), (2, 5), (2, 4), (7, 9)\} \)

Explanation:
1. A relation from X to Y needs every first entry in X and every second entry in Y.
2. R₄ contains (7, 9) with 7 not in X, and (2, 4) with 4 not in Y.
3. The other three sets pass the test.

Teacher's Note:
1. Check both coordinates of each pair.
2. One bad pair is enough to fail.

Answer: (d) Reflexive and symmetric

Explanation:
1. All four pairs (a, a) are present: reflexive.
2. (1, 2) and (2, 1), (3, 1) and (1, 3) both appear: symmetric.
3. (3, 1) and (1, 2) are in R but (3, 2) is not: not transitive.
4. So R is reflexive and symmetric, not an equivalence relation.

Teacher's Note:
1. Test all three properties before choosing.
2. Equivalence needs reflexive, symmetric and transitive together.

Answer: (a) Reflexive but not symmetric

Explanation:
1. (1, 1), (2, 2), (3, 3) present: reflexive.
2. (1, 2) is in R but (2, 1) is not: not symmetric.
3. (1, 2), (2, 3) give (1, 3), which is present; all other chains also work: transitive.
4. So R is reflexive and transitive but not symmetric.

Teacher's Note:
1. Look for one missing mirror pair to disprove symmetry.
2. Check each chain (a, b), (b, c) for transitivity.

Answer: (b) Only transitive

Explanation:
1. a < a is false: not reflexive.
2. a < b does not give b < a: not symmetric.
3. a < b and b < c give a < c: transitive.

Teacher's Note:
1. Strict inequalities are never reflexive.
2. They are always transitive.

Answer: (a) Reflexive but not symmetric

Explanation:
1. a divides a, so R is reflexive.
2. 2 divides 4 but 4 does not divide 2, so R is not symmetric.
3. If a divides b and b divides c then a divides c, so R is transitive as well.
4. Of the options, (a) is the true description.

Teacher's Note:
1. Divisibility is reflexive, transitive and antisymmetric.
2. Use one numerical counter-example to refute symmetry.

Answer: (c) Transitive

Explanation:
1. (1, 1) is not in R: not reflexive.
2. (1, 2) is in R but (2, 1) is not: not symmetric.
3. x < y and y < z imply x < z: transitive.

Teacher's Note:
1. Strict order is transitive.
2. Disprove symmetry with a single pair.

Answer: (a) Reflexive

Explanation:
1. x − x + √2 = √2 is irrational, so R is reflexive.
2. Take x − y = √2: x R y holds, but y − x + √2 = 0 is rational, so y R x fails. Not symmetric.
3. Take x − y = √2 and y − z = 1 − 2√2: both pairs are related, but x − z + √2 = 1 is rational. Not transitive.

Teacher's Note:
1. Choose numbers that make the expression rational to build counter-examples.
2. Only the reflexive property survives.

Answer: (a) x R₁ y ⇔ |x| = |y|

Explanation:
1. |x| = |y| is reflexive, symmetric and transitive.
2. x ≥ y is not symmetric.
3. Divisibility is not symmetric, and is not defined for all reals.
4. x < y is not reflexive.

Teacher's Note:
1. Relations defined by 'same value of some quantity' are equivalences.
2. Inequalities are never symmetric.

Answer: (a) R = {(1, 2), (2, 2), (3, 3), (2, 1), (2, 3), (3, 2)}

Explanation:
1. Pairs: (1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2); (1, 3) fails since |1 − 9| = 8.
2. The set in (a) leaves out (1, 1), so (a) is false.
3. Domain and range are both {1, 2, 3}, and the co-domain is A itself.

Teacher's Note:
1. List all nine possible pairs and test each.
2. Do not forget the pairs (a, a).

Answer: (b) Symmetric

Explanation:
1. x² + x² = 1 does not hold for every x: not reflexive.
2. x² + y² = y² + x², so (x, y) ∈ P gives (y, x) ∈ P: symmetric.
3. (1, 0) and (0, 1) are in P but (1, 1) is not: not transitive.

Teacher's Note:
1. Symmetry follows because the equation is unchanged on swapping x and y.
2. Use specific points on the unit circle for counter-examples.

Answer: (d) None of these

Explanation:
1. R₁ lacks (q, q) and (r, r).
2. R₂ lacks (p, p).
3. R₃ is reflexive but (p, q) has no mirror (q, p).
4. So none of the three is an equivalence relation.

Teacher's Note:
1. First check that all three (a, a) pairs are present.
2. Then check mirror pairs.

Answer: (b) Reflexive, transitive but not symmetric

Explanation:
1. a ≥ a holds: reflexive.
2. 3 ≥ 2 but 2 ≥ 3 is false: not symmetric.
3. a ≥ b and b ≥ c give a ≥ c: transitive.

Teacher's Note:
1. Non-strict inequalities are reflexive and transitive.
2. They fail symmetry.

Answer: (b) Symmetric

Explanation:
1. A line is not perpendicular to itself: not reflexive.
2. If α ⊥ β then β ⊥ α: symmetric.
3. If α ⊥ β and β ⊥ γ then α ∥ γ, not perpendicular: not transitive.

Teacher's Note:
1. Draw three lines to test transitivity.
2. Perpendicularity is a mutual property.

Answer: (a) Reflexive and symmetric but not transitive

Explanation:
1. 1 + a² > 0 for all a: reflexive.
2. 1 + ab = 1 + ba: symmetric.
3. Take a = 1, b = 0, c = −2: 1 + 0 > 0 and 1 + 0 > 0, but 1 + (1)(−2) = −1 < 0. Not transitive.

Teacher's Note:
1. Use a zero value as the middle element to break transitivity.
2. Check that the counter-example satisfies both given relations.

Answer: (d) None of these

Explanation:
1. For (x, x) we need 3x = 41, impossible: not reflexive.
2. (1, 39) ∈ R but 2(39) + 1 = 79 ≠ 41, so (39, 1) ∉ R: not symmetric.
3. (20, 1) and (1, 39) are in R but 2(20) + 39 = 79 ≠ 41: not transitive.
4. So none of the properties holds.

Teacher's Note:
1. Write down a few actual pairs of R.
2. Test them against each property.

Answer: (d) 600

Explanation:
1. Both languages: 750 + 400 − 1000 = 150.
2. Hindi only = 750 − 150 = 600.

Teacher's Note:
1. Everybody speaks at least one language, so the union is 1000.
2. Find the overlap first.

Answer: (b) 250

Explanation:
1. Both languages: 750 + 400 − 1000 = 150.
2. Bengali only = 400 − 150 = 250.

Teacher's Note:
1. Use the same overlap as for Hindi only.
2. Only = total in that language minus overlap.

Answer: (c) 150

Explanation:
1. n(H ∪ B) = 1000 = 750 + 400 − n(H ∩ B).
2. n(H ∩ B) = 1150 − 1000 = 150.

Teacher's Note:
1. Rearrange the union formula for the overlap.
2. Everyone is in the union.

Answer: (b) {(−3, 4), (−2, 3), (−1, 2), (0, 1), (1, 0), (2, 1), (3, 2)}

Explanation:
1. x takes the values −3, −2, −1, 0, 1, 2, 3.
2. y = |x − 1| gives 4, 3, 2, 1, 0, 1, 2 respectively.
3. R = {(−3, 4), (−2, 3), (−1, 2), (0, 1), (1, 0), (2, 1), (3, 2)}.

Teacher's Note:
1. List every allowed integer x and compute y.
2. Each x gives exactly one y.

Answer: (c) {−3, −2, −1, 0, 1, 2, 3}

Explanation:
1. The first entries are all admissible x: −3 to 3.

Teacher's Note:
1. Domain means the set of first components.
2. Use the allowed values of x.

Answer: (a) {0, 1, 2, 3, 4}

Explanation:
1. The values of y are 4, 3, 2, 1, 0, 1, 2.
2. As a set: {0, 1, 2, 3, 4}.

Teacher's Note:
1. Range means the set of second components.
2. Repeated values are written once.