JEE Mathematics Sets MCQs Set 03

Answer: (a) 31

Explanation:
1. A set with n elements has 2ⁿ subsets, of which 2ⁿ − 1 are proper.
2. Here n = 5, so the number of proper subsets = 2⁵ − 1 = 31.

Teacher's Note:
1. Proper subsets exclude the set itself, so subtract 1 from 2ⁿ.
2. The empty set is counted as a proper subset.

Answer: (d) None of these

Explanation:
1. The elements of {{a}, b} are {a} and b, so a ∉ {{a}, b}: (a) is false.
2. The elements of {a, {b, c}} are a and {b, c}. The set {b, c} is an element, not a subset, because b and c are not elements: (b) is false.
3. {a, b} ⊂ {a, {b, c}} would need b to be an element, which it is not: (c) is false.
4. Hence none of (a), (b), (c) is true.

Teacher's Note:
1. Keep ∈ (is an element of) and ⊂ (is a subset of) apart.
2. List the elements of the bigger set carefully before testing each statement.

Answer: (b) 23

Explanation:
1. There are 6 × 6 = 36 pairs (p, q), but equal fractions must be counted once.
2. Count pairs with gcd(p, q) = 1 (these give distinct fractions in lowest terms): for q = 1,...,6 there are 6, 3, 4, 3, 5, 2 such values of p, a total of 23.
3. Every other pair reduces to one of these, so |S| = 23.

Teacher's Note:
1. Distinct fractions correspond to pairs in lowest terms.
2. Counting all 36 pairs ignores repeated values like 1/2 = 2/4 = 3/6.

Answer: (a) {x : |x| = 5, x ∈ N}

Explanation:
1. (a) |x| = 5 with x ∈ N gives x = 5 only: one element.
2. (b) x = 6 or −6: two elements.
3. (c) x = −1 is not a natural number: empty set.
4. (d) x = ±√7 is not natural: empty set.

Teacher's Note:
1. A singleton set has exactly one element.
2. Always apply the condition on the domain (N, Z) after solving the equation.

Answer: (d) 30

Explanation:
1. B is a 3-element subset of {1, ..., 9}. The residues mod 3 of the elements are three 1s, three 2s and three 0s (1,4,7; 2,5,8; 3,6,9).
2. The sum is divisible by 3 when all three residues are equal (3 × 1 = 3 ways each, so 3) or when they are all different (3 × 3 × 3 = 27 ways).
3. Total = 3 + 27 = 30.

Teacher's Note:
1. Group the numbers by their remainder on division by 3.
2. Sum is a multiple of 3 when remainders are all equal or all different.

Answer: (b) m = 6, n = 3

Explanation:
1. 2^m − 2ⁿ = 56.
2. 56 = 64 − 8 = 2⁶ − 2³, so m = 6 and n = 3.

Teacher's Note:
1. Write 56 as a difference of two powers of 2.
2. The first set has more subsets, so it has more elements.

Answer: (d) 4

Explanation:
1. A = ∅ has 0 elements, so P(A) has 2⁰ = 1 element.
2. P{P(A)} has 2¹ = 2 elements.
3. P{P{P(A)}} has 2² = 4 elements.

Teacher's Note:
1. Each application of P turns n elements into 2ⁿ elements.
2. P(∅) = {∅} is not empty; it has one element.

Answer: (a) A = {x : x ∈ Z and x² − 5x + 6 = 0}

Explanation:
1. (a) x² − 5x + 6 = 0 gives x = 2, 3: A = {2, 3} is finite.
2. (b) B contains every even integer: infinite.
3. (c) D = {−9, −8, −7, ...} is infinite.
4. So only (a) is finite.

Teacher's Note:
1. Solve the defining condition and list the set to judge whether it is finite.
2. Inequalities over Z without an upper bound give infinite sets.

Answer: (c) (A − B) ∪ (A ∩ C)

Explanation:
1. B − C = B ∩ C′, so A − (B − C) = A ∩ (B ∩ C′)′ = A ∩ (B′ ∪ C).
2. Distribute: (A ∩ B′) ∪ (A ∩ C) = (A − B) ∪ (A ∩ C).

Teacher's Note:
1. Rewrite differences with complements: X − Y = X ∩ Y′.
2. Then apply De Morgan and the distributive law.

Answer: (b) A ∪ B

Explanation:
1. The three parts are: only A, only B and both A and B.
2. Together they cover every element of A or B, so the union is A ∪ B.

Teacher's Note:
1. Think of the Venn diagram of two circles split into three regions.
2. The three regions together make up A ∪ B.

Answer: (b) 5

Explanation:
1. Let x = n(E ∩ M). Then 40 = 16 + 26 + 22 − x − 5 − 14 + 2 = 47 − x, so x = 7.
2. Students studying E and M but not Economics = 7 − 2 = 5.

Teacher's Note:
1. Use n(E ∪ M ∪ C) with the three-set formula first.
2. Subtract the triple overlap to get 'but not' regions.

Answer: (a) A ∪ B

Explanation:
1. X − B = B′, so the expression is (A ∩ B′) ∪ B.
2. By the distributive law this equals (A ∪ B) ∩ (B′ ∪ B) = (A ∪ B) ∩ X = A ∪ B.

Teacher's Note:
1. Replace X − B by B′ and distribute the union.
2. B ∪ B′ is the whole universal set.

Answer: (a) 325

Explanation:
1. Viewers watching at least one game = 500 − 50 = 450.
2. 450 = 285 + 195 + 115 − 45 − 70 − 50 + x, so x (all three) = 20.
3. Exactly one = (285 + 195 + 115) − 2(45 + 70 + 50) + 3 × 20 = 595 − 330 + 60 = 325.

Teacher's Note:
1. First find the triple intersection from the union formula.
2. Exactly one = sum of sets − 2 × (sum of pairs) + 3 × (triple).

Answer: (d) {12n − 6 : n ∈ N}

Explanation:
1. A = {6, 10, 14, 18, ...} and B = {3, 6, 9, 12, ...}.
2. Common elements: 6, 18, 30, 42, ... an AP with first term 6 and common difference 12.
3. The general term is 6 + 12(n − 1) = 12n − 6.

Teacher's Note:
1. List a few terms of each set and spot the common ones.
2. The common difference is the LCM of the two differences, 4 and 3.

Answer: (a) 20

Explanation:
1. Fish lovers = 100 − 50 = 50. Persons liking fish or chicken = 100 − 10 = 90.
2. Both = 50 + 60 − 90 = 20.

Teacher's Note:
1. Convert 'do not prefer' statements into the sets themselves.
2. Use n(F ∩ C) = n(F) + n(C) − n(F ∪ C).

Answer: (c) C

Explanation:
1. A′ ∩ B′ ∩ C is the part of C outside A and B; B ∩ C and A ∩ C together are the part of C inside A or B.
2. Together they make up the whole of C.

Teacher's Note:
1. Draw the Venn diagram and look only at the circle C.
2. Split C into the part outside A ∪ B and the part inside it.

Answer: (b) A − (B ∪ C) = (A ∩ B′) ∩ C′

Explanation:
1. A − (B ∪ C) = A ∩ (B ∪ C)′ = A ∩ B′ ∩ C′, so (b) is true.
2. (a) is false (left side contains all of A), (c) is false since A − (B ∩ C) = A ∩ (B′ ∪ C′), and (d) is false since A ∩ (B − C) = A ∩ B ∩ C′.

Teacher's Note:
1. Use X − Y = X ∩ Y′ and De Morgan's laws.
2. Test each option on a small Venn diagram if unsure.

Answer: (c) 4

Explanation:
1. F(20) = {2, 4, 5, 10, 20} and F(16) = {2, 4, 8, 16}, so F(20) ∩ F(16) = {2, 4}.
2. F(y) must contain both 2 and 4, so 4 must divide y.
3. The least such y is 4, with F(4) = {2, 4}.

Teacher's Note:
1. The condition means both 2 and 4 divide y.
2. Check the candidate: F(2) = {2} misses 4, so y = 2 fails.

Answer: (a) φ

Explanation:
1. P − μ = φ and μ − P = P′.
2. So the bracket equals P′, and P ∩ P′ = φ.

Teacher's Note:
1. Everything in P is also in μ, so P − μ is empty.
2. A set and its complement never overlap.

Answer: (b) 2 only

Explanation:
1. A − (B − C) = (A − B) ∪ (A ∩ C), which differs from (A − B) ∪ C because C may have elements outside A. So 1 is false.
2. A − (B ∪ C) = A ∩ B′ ∩ C′ = (A − B) − C, so 2 is true.

Teacher's Note:
1. Rewrite both sides using complements.
2. A single counter-example with C outside A kills statement 1.

Answer: (b) 300

Explanation:
1. Everyone speaks at least one language: 500 = 475 + 200 − n(H ∩ B), so n(H ∩ B) = 175.
2. Hindi only = 475 − 175 = 300.

Teacher's Note:
1. The total of 500 is the union of the two sets.
2. Hindi only = n(H) − n(H ∩ B).

Answer: (d) Neither 1 nor 2

Explanation:
1. The correct distributive laws are A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) and A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).
2. Both given statements swap ∪ and ∩ on the right side, so both are wrong in general.

Teacher's Note:
1. Union distributes over intersection and intersection over union.
2. Both statements here have the operations interchanged.

Answer: (c) 60 percent

Explanation:
1. n(Car ∪ Bus) = 20 + 50 − 10 = 60 percent.

Teacher's Note:
1. Subtract the overlap once when adding the two percentages.

Answer: (b) Only II is true

Explanation:
1. At least one juice = 100 + 150 − 75 = 175, so statement I is false.
2. Neither = 400 − 175 = 225, so statement II is true.

Teacher's Note:
1. Compute the union first, then the complement.
2. Statement I quotes 150, which is only the orange-juice count.

Answer: (c) (A′ ∩ B′) ∪ C′

Explanation:
1. By De Morgan, [(A ∪ B) ∩ C]′ = (A ∪ B)′ ∪ C′ = (A′ ∩ B′) ∪ C′.

Teacher's Note:
1. Apply De Morgan to the outer intersection first.
2. Then apply it to the inner union.

Answer: (c) 1170

Explanation:
1. n(P₁ ∪ P₂) = 1720 + 1450 − n(both) ≤ 2000.
2. So n(both) ≥ 3170 − 2000 = 1170.

Teacher's Note:
1. The union cannot exceed the total number surveyed.
2. The minimum overlap occurs when the union is the whole group.