JEE Mathematics Set Operations Relations and Mappings MCQs Set 07

Answer: (b) B

Explanation:
1. A ∪ B = A means every element of B is already in A, so B ⊆ A.
2. When B ⊆ A, the common part A ∩ B is B itself.

Teacher's Note:
1. Translate a set equation into a subset statement first.
2. B ⊆ A is equivalent to A ∩ B = B and to A ∪ B = A.

Answer: (a) A′

Explanation:
1. By De Morgan, (A ∪ B)′ = A′ ∩ B′.
2. So the expression is (A′ ∩ B′) ∪ (A′ ∩ B) = A′ ∩ (B′ ∪ B) = A′ ∩ U = A′.

Teacher's Note:
1. Use De Morgan's laws and then the distributive law to factor out the common set.
2. B′ ∪ B is the universal set.

Answer: (c) A ∩ B ∩ C

Explanation:
1. An element is outside all three differences only if it is in all three sets or in none of them.
2. Since A ∪ B ∪ C = U, no element is in none of them.
3. An element in exactly one set lies in a difference, and an element in exactly two sets also lies in a difference.
4. Hence the complement is A ∩ B ∩ C.

Teacher's Note:
1. Think in terms of membership: how many of the three sets contain a typical element?
2. The condition A ∪ B ∪ C = U removes the case of an element in none.

Answer: (a) A ∪ B

Explanation:
1. A − B, A ∩ B and B − A are three disjoint pieces that together make up A ∪ B.
2. Their union is therefore A ∪ B.

Teacher's Note:
1. Draw a two-circle Venn diagram and shade the three pieces.
2. The only region not included is the outside of both circles.

Answer: (b) 8

Explanation:
1. Only mathematics + both + only physics = 20, and mathematics (including both) = 12.
2. So only physics = 20 − 12 = 8.

Teacher's Note:
1. Everyone teaches at least one subject, so the total is the union.
2. Only physics = total − (all who teach mathematics).

Answer: (a) 43

Explanation:
1. n(C ∪ H ∪ F) = n(C) + n(H) + n(F) − n(C∩H) − n(H∩F) − n(F∩C) + n(C∩H∩F).
2. = 21 + 26 + 29 − 14 − 15 − 12 + 8 = 43.

Teacher's Note:
1. Use the inclusion–exclusion formula for three sets.
2. Add back the triple intersection which was removed three times and added three times.

Answer: (d) an equivalence relation

Explanation:
1. a ≡ a (mod m) since m divides 0, so it is reflexive.
2. If m divides a − b then m divides b − a, so it is symmetric.
3. If m divides a − b and b − c then m divides a − c, so it is transitive.
4. Hence it is an equivalence relation.

Teacher's Note:
1. Congruence modulo m is the standard example of an equivalence relation.
2. Its equivalence classes are the residue classes 0, 1, …, m − 1.

Answer: (b) reflexive and symmetric

Explanation:
1. Reflexive: m·m = m² ≥ 0.
2. Symmetric: mn = nm.
3. Not transitive: 1 R 0 and 0 R −1 (both products are 0), but 1·(−1) = −1 < 0.
4. So R is reflexive and symmetric only.

Teacher's Note:
1. Test each property (reflexive, symmetric, transitive) separately with a concrete counter-example when it fails.
2. An equivalence relation needs all three properties.

Answer: (b) reflexive and symmetric

Explanation:
1. Reflexive: |x − x| = 0 ≤ 1.
2. Symmetric: |x − y| = |y − x|.
3. Not transitive: 0 R 1 and 1 R 2, but |0 − 2| = 2 > 1.
4. So R is reflexive and symmetric.

Teacher's Note:
1. Test each property (reflexive, symmetric, transitive) separately with a concrete counter-example when it fails.
2. An equivalence relation needs all three properties.

Answer: (d) an equivalence relation

Explanation:
1. Reflexive: a + b = b + a.
2. Symmetric: a + d = b + c gives c + b = d + a, so (c, d) r (a, b).
3. Transitive: a + d = b + c and c + f = d + e give a + f = b + e after adding and cancelling c + d.
4. So r is an equivalence relation.

Teacher's Note:
1. The condition a − b = c − d defines the relation; equal differences form an equivalence relation.
2. Rewrite the condition in this difference form to see all three properties.

Answer: (a) 2⁹

Explanation:
1. A × A has 3 × 3 = 9 elements.
2. A relation is any subset of A × A, so there are 2⁹ relations.

Teacher's Note:
1. The number of relations from a set with m elements to one with n elements is 2^{mn}.
2. Here m = n = 3.

Answer: (c) not symmetric

Explanation:
1. Not a function: 2 is related to both 4 and 3.
2. Not transitive: (1, 3) and (3, 1) are in R but (1, 1) is not.
3. Not symmetric: (2, 3) is in R but (3, 2) is not.
4. Not reflexive: (1, 1) is missing.

Teacher's Note:
1. Check each option with one specific pair.
2. A single failing pair is enough to reject a property.

Answer: (b) reflexive and transitive only

Explanation:
1. Reflexive: all four pairs (a, a) are present.
2. Not symmetric: (6, 12) is present but (12, 6) is not.
3. Transitive: every chain such as (3, 6), (6, 12) gives (3, 12), which is in R; the other chains also stay inside R.
4. So R is reflexive and transitive only.

Teacher's Note:
1. Test each property (reflexive, symmetric, transitive) separately with a concrete counter-example when it fails.
2. An equivalence relation needs all three properties.

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

Explanation:
1. Both 7 − x and x − 3 must be non-negative integers with x − 3 ≤ 7 − x, which gives x = 3, 4, 5.
2. x = 3: ⁴P₀ = 1; x = 4: ³P₁ = 3; x = 5: ²P₂ = 2.
3. Range = {1, 2, 3}.

Teacher's Note:
1. The domain of a permutation function is restricted by n ≥ r ≥ 0 with integers n, r.
2. Find the domain first, then evaluate.

Answer: (d) many-one and into

Explanation:
1. sin x repeats its values (for example sin 0 = sin π), so f is many-one.
2. The range is [−1, 1], which is smaller than R, so f is into.

Teacher's Note:
1. Periodic functions are always many-one on the whole real line.
2. The range of sine is [−1, 1].

Answer: (d) none of these

Explanation:
1. f(1) = f(−1) = 1/2, so f is not injective.
2. The range is (0, 1], so f is not surjective.
3. Hence f is neither injective, surjective nor bijective.

Teacher's Note:
1. An even function cannot be injective on R.
2. Find the range by observing 1 + x² ≥ 1.

Answer: (a) many-one and into

Explanation:
1. Many points on a circle x² + y² = r² have the same image r, so f is many-one.
2. The values are non-negative only, so negative reals are not images and f is into.

Teacher's Note:
1. f is the modulus of the complex number x + iy.
2. Equal modulus means the same circle, so different points share a value.

Answer: (b) x|x|

Explanation:
1. |x| is not one-one since |1| = |−1|.
2. sin πx gives sin 0 = sin π·1 = 0 for x = 0 and x = 1, so it is not one-one.
3. x|x| is strictly increasing from −1 to 1 and continuous, so it is one-one and onto [−1, 1].

Teacher's Note:
1. A continuous strictly monotonic function from [−1, 1] onto itself is bijective.
2. x|x| equals x² for x ≥ 0 and −x² for x < 0.

Answer: (b) {(1, b), (2, a), (3, a)}

Explanation:
1. A mapping must pair every element of A with exactly one element of B.
2. Option (a) pairs 2 with both a and b; option (c) leaves 3 unpaired.
3. Option (b) pairs each of 1, 2, 3 exactly once.

Teacher's Note:
1. Check two things: every element of the domain appears, and none appears twice.
2. Violating either rule breaks the definition of a function.

Answer: (a) 1 only

Explanation:
1. Symmetric: if l₁ ⊥ l₂ then l₂ ⊥ l₁.
2. Not reflexive: a line is not perpendicular to itself.
3. Not transitive: if l₁ ⊥ l₂ and l₂ ⊥ l₃ then l₁ ∥ l₃, not perpendicular.
4. So only statement 1 is correct.

Teacher's Note:
1. Test each property (reflexive, symmetric, transitive) separately with a concrete counter-example when it fails.
2. An equivalence relation needs all three properties.

Answer: (b) 1 and 3

Explanation:
1. Reflexive: m = 1·m.
2. Not symmetric: 6 is a multiple of 3, but 3 is not a multiple of 6.
3. Transitive: if m = kn and n = lp then m = klp.
4. So statements 1 and 3 are correct.

Teacher's Note:
1. Test each property (reflexive, symmetric, transitive) separately with a concrete counter-example when it fails.
2. An equivalence relation needs all three properties.

Answer: (b) 1 and 2

Explanation:
1. Reflexive: all (a, a) are present.
2. Symmetric: (1,2)↔(2,1) and (3,1)↔(1,3) are both present.
3. Not transitive: (2,1) and (1,3) are in R but (2,3) is not.
4. So statements 1 and 2 are correct.

Teacher's Note:
1. Test each property (reflexive, symmetric, transitive) separately with a concrete counter-example when it fails.
2. An equivalence relation needs all three properties.

Answer: (b) 1 and 3

Explanation:
1. f(1) = f(−1) = 1/2, so f is many-one.
2. Writing f(x) = 1 − 1/(1 + x²), the range is [0, 1), which is smaller than R, so f is into.
3. So statements 1 and 3 are correct.

Teacher's Note:
1. Rewriting the fraction as 1 minus a simpler fraction makes the range clear.
2. Even functions are never one-one on R.

Answer: (b) f: 27 and g: 6

Explanation:
1. Each of the 3 elements of A can map to any of the 3 elements of B, so there are 3³ = 27 functions f.
2. A one-one map needs distinct images: 3 × 2 × 1 = 6 choices for g.

Teacher's Note:
1. Number of functions from an m-set to an n-set is n^m.
2. Number of one-one functions is nPm.