CBSE Class 11 Mathematics Sets Competitive Exam MCQs Set 01

Answer: (b) {x : x is a real number and x² + 1 = 0}

Explanation:
1. x² − 1 = 0 gives x = ±1, and x² − 9 = 0 gives x = ±3, so (a) and (c) are non-empty.
2. x² = x + 2 gives (x − 2)(x + 1) = 0, so x = 2, −1 and (d) is non-empty.
3. x² + 1 = 0 means x² = −1, which no real number satisfies, so (b) is the empty set.

Teacher's Note:
1. The square of a real number is never negative, so any condition like x² = −1 over real numbers gives φ.
2. To test a set for emptiness, simply try to solve its defining condition within the stated universal set.

Answer: (c) {0} ⊂ A^c w.r.t. X

Explanation:
1. A^c w.r.t. X = X − A = {0, 1}.
2. The elements of A^c are the numbers 0 and 1, so the sets {0} and φ are not elements of it; (a) and (b) are false.
3. 0 is an element, not a set, so '0 ⊂ A^c' is meaningless; (d) is false.
4. Every element of {0} is in A^c, so {0} ⊂ A^c; the answer is (c).

Teacher's Note:
1. Use ∈ between an element and a set, and ⊂ between two sets.
2. φ is a subset of every set, but it is an element of a set only when it is listed inside the braces.

Answer: (d) not a well defined collection

Explanation:
1. A set must be a collection of well-defined objects: we must be able to decide clearly whether an object belongs to it.
2. Whether a student is 'intelligent' depends on personal opinion, so membership cannot be decided clearly.
3. Hence it is not a well defined collection, i.e. not a set.

Teacher's Note:
1. Words like intelligent, beautiful, tall or good make a collection ill-defined unless a precise rule is given.
2. A collection such as 'students who scored above 90%' would be well defined and hence a set.

Answer: (b) N

Explanation:
1. (A ∪ B) ∩ B^c = (A ∩ B^c) ∪ (B ∩ B^c) = (A − B) ∪ φ = A − B.
2. A and B have no common element, so A − B = A.
3. Hence the expression = A^c ∪ A = N, the universal set.

Teacher's Note:
1. (A ∪ B) ∩ B^c always equals A − B, by the distributive law.
2. For any set, A ∪ A^c is the universal set, whatever the elements of A are.

Answer: (c) {15, 30, 45, …}

Explanation:
1. A ∩ B contains the numbers that are multiples of both 3 and 5.
2. Since 3 and 5 are co-prime, such numbers are exactly the multiples of 15.
3. A ∩ B = {15, 30, 45, …}.

Teacher's Note:
1. The common multiples of two numbers are the multiples of their LCM.
2. LCM(3, 5) = 15 because 3 and 5 have no common factor other than 1.

Answer: (a) The smallest set of Y is {3, 5, 9}

Explanation:
1. Y must contain 3, 5 and 9, because these are in the union but not in {1, 2}.
2. Y may or may not contain 1 and 2, and it cannot contain anything outside {1, 2, 3, 5, 9}.
3. So the smallest possible Y is {3, 5, 9} and the largest possible Y is {1, 2, 3, 5, 9}; only (a) is correct.

Teacher's Note:
1. Elements of the union that are missing from the known set must belong to the unknown set.
2. The unknown set can never contain an element that is absent from the union.

Answer: (c) {1, 2, 3, 4}

Explanation:
1. B ∩ C = {3, 4} ∩ {4, 5, 6} = {4}.
2. A ∪ (B ∩ C) = {1, 2, 3} ∪ {4} = {1, 2, 3, 4}.

Teacher's Note:
1. Always evaluate the bracket first, just as in arithmetic.
2. As a check, (A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4} ∩ {1, 2, 3, 4, 5, 6} = {1, 2, 3, 4}, which is the same answer.

Answer: (d) A

Explanation:
1. A ∩ B ⊆ A, so every element of A ∩ B is already in A.
2. Taking the union of A with a subset of A adds nothing new.
3. Hence A ∪ (A ∩ B) = A.

Teacher's Note:
1. This is called the absorption law; similarly A ∩ (A ∪ B) = A.
2. If X ⊆ A, then A ∪ X = A and A ∩ X = X.

Answer: (a) 6

Explanation:
1. A ∪ B contains every element of B, so it has at least 6 elements.
2. If A ⊆ B, then A ∪ B = B, which has exactly 6 elements.
3. Hence the minimum number of elements in A ∪ B is 6.

Teacher's Note:
1. The union is smallest when the smaller set lies completely inside the larger set.
2. The union is largest (3 + 6 = 9 elements) when the two sets are disjoint.

Answer: (a) φ

Explanation:
1. 2x = 6 gives x = 3.
2. But x² = 16 gives x = ±4, and 3² = 9 ≠ 16.
3. No real number satisfies both conditions, so A = φ.

Teacher's Note:
1. When a set is defined by two conditions joined by 'and', an element must satisfy both conditions at the same time.
2. Solve the simpler condition first and then test the result in the other condition.

Answer: (d) {x : x ≠ x}

Explanation:
1. In rule (set-builder) method a set is written as {x : P(x)} where P(x) is a property of its elements.
2. No object x can satisfy x ≠ x, so {x : x ≠ x} has no element.
3. { } and φ are the null set but not in rule form, and {x : x = x} contains every object. Hence the answer is {x : x ≠ x}.

Teacher's Note:
1. The rule form of the empty set needs a condition that no object can satisfy, such as x ≠ x or x² + 1 = 0 with x real.
2. Read the question carefully: { } and φ also denote the null set, but only option (d) is written in rule method.

Answer: (b) 1 ∈ Q

Explanation:
1. Q = {1, \( \frac{1}{2} \), \( \frac{1}{3} \), \( \frac{1}{4} \), …}, obtained by putting y = 1, 2, 3, … .
2. For y = 1, x = 1, so 1 ∈ Q.
3. 0, 2 and \( \frac{2}{3} \) are not of the form \( \frac{1}{y} \) with y a natural number. Hence 1 ∈ Q.

Teacher's Note:
1. Converting a set from set-builder form to roster form by substituting y = 1, 2, 3, … makes membership questions easy.
2. Every element of Q is a unit fraction lying in (0, 1], so 0 and 2 can be rejected at once.

Answer: (d) { }

Explanation:
1. The empty set has no element, so the condition ‘every element of φ is in A’ is true for every set A.
2. Hence the empty set { } is a subset of every set.
3. {1}, {0} and {1, 2, 3, …} are not subsets of each other. So the answer is { }.

Teacher's Note:
1. The empty set is a subset of every set; this is true because there is no element in φ that could fail the test.
2. Also remember that every set is a subset of itself, but only φ is a subset of all sets.

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

Explanation:
1. A ∩ B contains the elements common to A and B.
2. Common elements of {1, 2, 3, 4} and {2, 3, 4, 5, 6} are 2, 3, 4.
3. ∴ A ∩ B = {2, 3, 4}.

Teacher's Note:
1. For intersection pick only the common elements; for union take all elements, writing common ones once.
2. A ∩ B is always a subset of both A and B, which quickly rules out {5, 6} and {1}.

Answer: (b) {3, 5, 9}

Explanation:
1. A ∪ {1, 2} must contain 3, 5 and 9, and these can come only from A.
2. So A must contain 3, 5, 9; it may or may not contain 1 and 2.
3. The smallest such set is A = {3, 5, 9}.

Teacher's Note:
1. ‘Smallest’ means we put in A only those elements that the other set cannot supply.
2. Check: {3, 5, 9} ∪ {1, 2} = {1, 2, 3, 5, 9}, as required.

Answer: (b) B ⊆ A

Explanation:
1. A ∩ B ⊆ A always.
2. Given A ∩ B = B, so B ⊆ A.
3. (Conversely, if B ⊆ A then every element of B is common to A and B, so A ∩ B = B.)

Teacher's Note:
1. A ∩ B = B and A ∪ B = A both mean the same thing: B ⊆ A.
2. Test with a quick example such as A = {1, 2}, B = {1}: A ∩ B = {1} = B and B ⊆ A.

Answer: (a) B ⊆ A

Explanation:
1. B ⊆ A ∪ B always.
2. If A ∪ B = A, then B ⊆ A.
3. Conversely, if B ⊆ A, B adds no new element to A, so A ∪ B = A.

Teacher's Note:
1. Union with a subset adds nothing: if B ⊆ A then A ∪ B = A and A ∩ B = B.
2. A = B is only a special case of B ⊆ A, so it is not the ‘iff’ condition.

Answer: (c) A = B

Explanation:
1. A ⊆ A ∪ B = A ∩ B ⊆ B, so A ⊆ B.
2. Similarly B ⊆ A ∪ B = A ∩ B ⊆ A, so B ⊆ A.
3. Hence A = B; conversely, if A = B both sides equal A.

Teacher's Note:
1. Use the chain A ∩ B ⊆ A ⊆ A ∪ B: if the two ends are equal, everything in the chain is equal.
2. To prove two sets equal, show each is a subset of the other.

Answer: (b) A ∩ B ⊆ A ∪ B

Explanation:
1. Let x ∈ A ∩ B. Then x ∈ A, so x ∈ A ∪ B.
2. ∴ A ∩ B ⊆ A ∪ B for all sets A and B.
3. Options (a) and (c) hold only when A = B, so they are not true in general.

Teacher's Note:
1. A ∩ B ⊆ A ⊆ A ∪ B holds for every pair of sets.
2. A quick counter-example such as A = {1}, B = {2} rules out options (a) and (c).

Answer: (a) {3, 4, 10}

Explanation:
1. A ∩ B = {3, 4, 10}
2. A ∩ C = {4}
3. (A ∩ B) ∪ (A ∩ C) = {3, 4, 10}.

Teacher's Note:
1. You may also use the distributive law: (A ∩ B) ∪ (A ∩ C) = A ∩ (B ∪ C).
2. B ∪ C = {3, 4, 5, 6, 10, 12, 14} and A ∩ (B ∪ C) = {3, 4, 10}, which confirms the answer.

Answer: (b) B = C

Explanation:
1. Let x ∈ B. If x ∈ A, then x ∈ A ∩ B = A ∩ C, so x ∈ C.
2. If x ∉ A, then x ∈ A ∪ B = A ∪ C, so x ∈ C. Thus B ⊆ C.
3. Similarly C ⊆ B. Hence B = C.

Teacher's Note:
1. Neither condition alone is enough: A ∪ B = A ∪ C alone does not force B = C (take B ⊆ A, C = A).
2. Splitting into the cases x ∈ A and x ∉ A is a standard way to prove such set identities.

Answer: (a) {a, b, c}

Explanation:
1. B ∪ C = {b, c, d} ∪ {a, b, d, e} = {a, b, c, d, e}
2. A ∩ (B ∪ C) = {a, b, c} ∩ {a, b, c, d, e}
3. = {a, b, c} = A.

Teacher's Note:
1. Always work out the bracket first, then take the outer operation.
2. When B ∪ C contains the whole of A, the intersection A ∩ (B ∪ C) is simply A.

Answer: (a) φ

Explanation:
1. B − A = {x : x ∈ B and x ∉ A}.
2. So no element of B − A lies in A.
3. ∴ A ∩ (B − A) = φ.

Teacher's Note:
1. A set and the part of another set lying outside it are always disjoint.
2. Similarly (A − B) ∩ B = φ for any sets A and B.

Answer: (c) A ∩ B = φ

Explanation:
1. By definition, A and B are disjoint if they have no element in common.
2. That is, A ∩ B = φ.

Teacher's Note:
1. Overlapping (joint) sets have A ∩ B ≠ φ; disjoint sets have A ∩ B = φ.
2. A − B = A is an equivalent result, but the defining condition of disjoint sets is A ∩ B = φ.

Answer: (d) A and the complement of B are always non-disjoint

Explanation:
1. Since A ⊄ B, there is an element x ∈ A with x ∉ B.
2. Then x ∈ B′ (complement of B in X).
3. So x ∈ A ∩ B′, i.e. A and B′ are non-disjoint.

Teacher's Note:
1. ‘A is not a subset of B’ means at least one element of A lies outside B; it says nothing about the other elements.
2. Options (a), (b) and (c) can all be disproved with A = {1, 2}, B = {2, 3} in X = {1, 2, 3}.

Answer: (a) A

Explanation:
1. Every element of A is in B.
2. So all elements of A are common to A and B.
3. ∴ A ∩ B = A.

Teacher's Note:
1. If A ⊆ B, then A ∩ B = A and A ∪ B = B.
2. Drawing the circle of A inside the circle of B shows this at once.

Answer: (c) φ

Explanation:
1. By De Morgan’s law, (A ∪ B)′ = A′ ∩ B′.
2. A ∩ (A ∪ B)′ = A ∩ A′ ∩ B′.
3. Since A ∩ A′ = φ, the result is φ.

Teacher's Note:
1. (A ∪ B)′ is the region outside both circles, so it cannot meet A.
2. Remember A ∩ A′ = φ and A ∪ A′ = U.

Answer: (b) A

Explanation:
1. B′ = U − B = {1, 2, 3, 4, 5, 8, 9, 10}.
2. A ∩ B′ = {1, 2, 5} ∩ {1, 2, 3, 4, 5, 8, 9, 10} = {1, 2, 5}.
3. ∴ A ∩ B′ = A.

Teacher's Note:
1. A ∩ B′ is the same as A − B; since A and B are disjoint here, A − B = A.
2. Whenever A ∩ B = φ, we get A ⊆ B′ and hence A ∩ B′ = A.

Answer: (b) A ∪ A′ = U

Explanation:
1. Every element of U is either in A or not in A (then it is in A′).
2. So A ∪ A′ = U.
3. Also A ∩ A′ = φ, so options (a) and (c) are false.

Teacher's Note:
1. The complement laws are A ∪ A′ = U and A ∩ A′ = φ.
2. Do not mix them up: union with the complement fills U, intersection with it is empty.

Answer: (c) N₂₄

Explanation:
1. N₆ = {6, 12, 18, 24, …} and N₈ = {8, 16, 24, 32, …}.
2. Common elements are the common multiples of 6 and 8, i.e. multiples of LCM(6, 8) = 24.
3. ∴ N₆ ∩ N₈ = N₂₄.

Teacher's Note:
1. N_a ∩ N_b = N_l, where l is the LCM of a and b; it equals N_ab only when a and b are coprime.
2. Choosing N₄₈ is a common slip, since 6 × 8 = 48 is not the LCM.

Answer: (a) 21N

Explanation:
1. 3N is the set of multiples of 3 and 7N is the set of multiples of 7.
2. Common multiples of 3 and 7 are multiples of LCM(3, 7) = 21.
3. ∴ 3N ∩ 7N = 21N.

Teacher's Note:
1. Since 3 and 7 are coprime, their LCM is just the product 21.
2. The intersection of multiples of a and of b is always the set of multiples of LCM(a, b).

Answer: (d) A − (B ∪ C)

Explanation:
1. The shaded part lies inside A but outside both B and C.
2. Points outside both B and C form (B ∪ C)′.
3. So the region is A ∩ (B ∪ C)′ = A − (B ∪ C).

Teacher's Note:
1. To name a region, list which circles it is inside and which it is outside.
2. By De Morgan’s law, A − (B ∪ C) = (A − B) ∩ (A − C).

Answer: (a) [f(x)]² + [g(x)]² = 0

Explanation:
1. x ∈ A ∩ B ⇔ f(x) = 0 and g(x) = 0.
2. For real values, [f(x)]² + [g(x)]² = 0 ⇔ f(x) = 0 and g(x) = 0, since a sum of squares is zero only when each square is zero.
3. ∴ A ∩ B = {x : [f(x)]² + [g(x)]² = 0}.

Teacher's Note:
1. A sum of squares of real numbers is zero only when every term is zero; this turns ‘and’ conditions into one equation.
2. \( \frac{f(x)}{g(x)} = 0 \) needs f(x) = 0 with g(x) ≠ 0, which is the opposite of a common root.

Answer: (a) A ∪ B

Explanation:
1. A − B, A ∩ B and B − A are the three disjoint pieces of the Venn diagram of A and B.
2. Together they make up both circles completely.
3. ∴ (A − B) ∪ (B − A) ∪ (A ∩ B) = A ∪ B.

Teacher's Note:
1. n(A ∪ B) = n(A − B) + n(A ∩ B) + n(B − A) follows from the same splitting.
2. Draw two overlapping circles and label the three regions; the result is then obvious.

Answer: (a) A′

Explanation:
1. (A ∪ B)′ = A′ ∩ B′ (De Morgan’s law).
2. (A′ ∩ B′) ∪ (A′ ∩ B) = A′ ∩ (B′ ∪ B)
3. = A′ ∩ U = A′.

Teacher's Note:
1. After De Morgan’s law, take A′ common using the distributive law.
2. In a Venn diagram, both pieces lie outside A and together cover all of the outside of A.

Answer: (c) A ∩ B ∩ C

Explanation:
1. x ∉ (A − B) ∪ (B − C) ∪ (C − A) means: (x ∈ A ⇒ x ∈ B), (x ∈ B ⇒ x ∈ C) and (x ∈ C ⇒ x ∈ A).
2. So x is either in all three sets or in none of them.
3. Since A ∪ B ∪ C = U, every x is in at least one set; hence x ∈ A ∩ B ∩ C. So the required set is A ∩ B ∩ C.

Teacher's Note:
1. The union (A − B) ∪ (B − C) ∪ (C − A) contains every point that is in some but not all of the three sets.
2. The condition A ∪ B ∪ C = U removes the possibility of a point lying in none of the sets.

Answer: (c) 6

Explanation:
1. Since A ⊆ B, A ∪ B = B.
2. ∴ n(A ∪ B) = n(B) = 6.

Teacher's Note:
1. When one set is inside the other, the union is the bigger set and the intersection is the smaller set.
2. Adding 3 + 6 = 9 is wrong here because the elements of A are already in B.

Answer: (a) 3

Explanation:
1. Since A ⊆ B, A ∩ B = A.
2. ∴ n(A ∩ B) = n(A) = 3.

Teacher's Note:
1. A ⊆ B gives A ∩ B = A and A ∪ B = B.
2. The intersection can never have more elements than the smaller set.

Answer: (b) A′ ∩ B′

Explanation:
1. x ∈ (A ∪ B)′ ⇔ x ∉ A ∪ B ⇔ x ∉ A and x ∉ B.
2. ⇔ x ∈ A′ and x ∈ B′ ⇔ x ∈ A′ ∩ B′.
3. ∴ (A ∪ B)′ = A′ ∩ B′ (De Morgan’s law).

Teacher's Note:
1. De Morgan’s laws: the complement of a union is the intersection of complements, and vice versa.
2. Shade the outside of both circles to see that it equals A′ ∩ B′.

Answer: (b) A′ ∪ B′

Explanation:
1. x ∈ (A ∩ B)′ ⇔ x ∉ A ∩ B ⇔ x ∉ A or x ∉ B.
2. ⇔ x ∈ A′ or x ∈ B′ ⇔ x ∈ A′ ∪ B′.
3. ∴ (A ∩ B)′ = A′ ∪ B′ (De Morgan’s law).

Teacher's Note:
1. Taking a complement changes ∩ into ∪ and ∪ into ∩.
2. Check with U = {1, 2, 3}, A = {1, 2}, B = {2, 3}: (A ∩ B)′ = {1, 3} = A′ ∪ B′.

Answer: (d) A − B = A ∩ B^c

Explanation:
1. A − B = {x : x ∈ A and x ∉ B}.
2. x ∉ B ⇔ x ∈ B^c.
3. ∴ A − B = {x : x ∈ A and x ∈ B^c} = A ∩ B^c.

Teacher's Note:
1. Writing a difference as an intersection with a complement, A − B = A ∩ B′, lets you use all the laws of ∩ and ∪.
2. Note that A^c ∩ B is B − A, not A − B.

Answer: (a) (A − B) ∪ (A − C)

Explanation:
1. A − (B ∩ C) = A ∩ (B ∩ C)′ = A ∩ (B′ ∪ C′).
2. = (A ∩ B′) ∪ (A ∩ C′)
3. = (A − B) ∪ (A − C).

Teacher's Note:
1. This is De Morgan’s law for differences: A − (B ∩ C) = (A − B) ∪ (A − C) and A − (B ∪ C) = (A − B) ∩ (A − C).
2. Write each difference as an intersection with a complement and the result follows from the ordinary De Morgan’s law.

Answer: (b) (A ∩ B) ∪ (A ∩ C)

Explanation:
1. By the distributive law, intersection distributes over union.
2. ∴ A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).

Teacher's Note:
1. The two distributive laws are A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) and A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
2. Option (a) equals A ∪ (B ∩ C), the other distributive law, not A ∩ (B ∪ C).

Answer: (b) B = C

Explanation:
1. Let x ∈ B. If x ∈ A, then x ∈ A ∩ B = A ∩ C, so x ∈ C.
2. If x ∉ A, then x ∈ A ∪ B = A ∪ C, and since x ∉ A, x ∈ C. So B ⊆ C.
3. Interchanging the roles of B and C gives C ⊆ B. Hence B = C.

Teacher's Note:
1. Splitting the argument into the cases x ∈ A and x ∉ A is a standard technique for set proofs.
2. Either condition alone is not enough: e.g. A = {1}, B = {1, 2}, C = {2} have equal unions but different B and C.

Answer: (a) Y

Explanation:
1. Y = {0, 9, 18, 27, …}, i.e. all non-negative multiples of 9.
2. \( 4^n = (1 + 3)^n = 1 + 3n + {}^nC_2\, 3^2 + {}^nC_3\, 3^3 + \ldots \), so \( 4^n - 3n - 1 = 9\left[ {}^nC_2 + {}^nC_3\, 3 + \ldots \right] \), a multiple of 9.
3. So X = {0, 9, 54, 243, …} contains only non-negative multiples of 9, i.e. X ⊆ Y.
4. Hence X ∪ Y = Y.

Teacher's Note:
1. Listing the first few elements (0, 9, 54, …) quickly suggests that X is made of multiples of 9.
2. Whenever X ⊆ Y, we have X ∪ Y = Y and X ∩ Y = X.

Answer: (d) P = Q

Explanation:
1. For P: \( \sin\theta = (\sqrt{2} + 1)\cos\theta \), so \( \tan\theta = \sqrt{2} + 1 \) (cos θ = 0 is impossible, as then sin θ = 0 too).
2. For Q: \( \cos\theta = (\sqrt{2} - 1)\sin\theta \), so \( \tan\theta = \frac{1}{\sqrt{2} - 1} = \sqrt{2} + 1 \) (sin θ = 0 is impossible, as then cos θ = 0 too).
3. Both sets are exactly the angles with \( \tan\theta = \sqrt{2} + 1 \), so P = Q.

Teacher's Note:
1. Rationalise: \( \frac{1}{\sqrt{2} - 1} = \frac{\sqrt{2} + 1}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \sqrt{2} + 1 \).
2. To compare two sets defined by equations, reduce each condition to the simplest equivalent form.

Answer: (c) A ⊂ B

Explanation:
1. A is the inside of the square 4 < a < 6, 4 < b < 6.
2. For such points, (a − 6)² < 4 and (b − 5)² < 1, so 4(a − 6)² + 9(b − 5)² < 16 + 9 = 25 ≤ 36. Hence every point of A lies in B.
3. B contains points such as (8, 5) [4 × 4 + 0 = 16 ≤ 36] which are not in A, so B ⊄ A.
4. Hence A ⊂ B.

Teacher's Note:
1. To show A ⊂ B, take a general element of A and show it satisfies the condition defining B.
2. One counter-example, such as (8, 5), is enough to prove that B is not a subset of A.