JEE Mathematics Set Operations Relations and Mappings MCQs Set 02

Answer: (d) not a well defined collection

Explanation:
1. A set must be a well defined collection: for any object we can say definitely whether it belongs or not.
2. Whether a student is 'intelligent' depends on opinion, so membership is not definite.
3. Hence it is not a well defined collection.

Teacher's Note:
1. Always test a 'set' for a clear membership rule first.
2. Words like intelligent, beautiful or tall have no sharp boundary.

Answer: (b) \( \{x : x \in \mathbb{R},\ x^2 + 1 = 0\} \)

Explanation:
1. (a) x² = 1 gives x = ±1, two elements.
2. (b) x² = −1 has no real solution, so the set is empty.
3. (c) x² = 9 gives x = ±3.
4. (d) x² − x − 2 = 0 gives x = 2 or −1.
5. Only (b) is empty.

Teacher's Note:
1. Solve each defining equation inside the stated universe.
2. Real numbers cannot square to a negative value.

Answer: (a) Null set

Explanation:
1. x² = 16 gives x = 4 or x = −4.
2. 2x = 6 gives x = 3.
3. Both conditions must hold together, but 3 is not 4 or −4.
4. So A has no element: it is the null set.

Teacher's Note:
1. 'And' means intersection of the solution sets.
2. Solve each condition separately, then compare.

Answer: (d) 32

Explanation:
1. |x| < 3 with x an integer gives A = {−2, −1, 0, 1, 2}.
2. n(A) = 5.
3. Number of subsets = 2⁵ = 32.

Teacher's Note:
1. A set with n elements has 2ⁿ subsets, including the empty set and the set itself.
2. Do not forget that x must be an integer, so only five values qualify.

Answer: (a) [3, 7] ⊆ (2, 10)

Explanation:
1. (a) Every number from 3 to 7 lies between 2 and 10, so it is true.
2. (b) The number 1 is in (0, ∞) but not in (4, ∞): false.
3. (c) The number 7 is in (5, 7] but not in [5, 7): false.
4. (d) The number 2.5 is in [2, 7] but not in (2.9, 8): false.

Teacher's Note:
1. To disprove a subset claim, find one element of the left set missing from the right set.
2. Check the end points of intervals carefully.

Answer: (d) 256

Explanation:
1. n(A) = 3, so n(P(A)) = 2³ = 8.
2. The number of subsets of P(A) is 2⁸ = 256.

Teacher's Note:
1. The power set of an n-element set has 2ⁿ elements.
2. Apply the 2ⁿ rule twice.

Answer: (b) 6

Explanation:
1. n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 9 − n(A ∩ B).
2. n(A ∩ B) is largest when A ⊆ B, namely 3.
3. Minimum n(A ∪ B) = 9 − 3 = 6.

Teacher's Note:
1. The union is smallest when the overlap is largest.
2. The overlap can never exceed the smaller set.

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

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

Teacher's Note:
1. Work the bracket first.
2. List elements without repetition.

Answer: (b) \( \{x : 1 \le x < 2\} \)

Explanation:
1. A = (−1, 1).
2. |x − 1| ≥ 1 gives x − 1 ≥ 1 or x − 1 ≤ −1, so B = (−∞, 0] ∪ [2, ∞).
3. A ∪ B = (−∞, 1) ∪ [2, ∞).
4. The real numbers missing are 1 ≤ x < 2, so D = {x : 1 ≤ x < 2}.

Teacher's Note:
1. Draw both sets on a number line to see the gap.
2. Check whether each end point is included.

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

Explanation:
1. A must supply the elements 3, 5, 9 that {1, 2} lacks.
2. Elements 1 and 2 are already present, so A need not contain them.
3. The smallest such set is {3, 5, 9}.

Teacher's Note:
1. 'Smallest' means no unnecessary elements.
2. Compare the target set with the known part.

Answer: (b) \( A \cap B \subseteq A \cup B \)

Explanation:
1. Every element of A ∩ B is in both sets, hence in A ∪ B.
2. So A ∩ B ⊆ A ∪ B always holds.
3. (a) holds only if A = B, and (c) only if A = B as well, so they are not true in general.

Teacher's Note:
1. Intersection is the smaller region and union the larger one.
2. Statements true only for special sets are not 'always' true.

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

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

Teacher's Note:
1. Find each intersection separately and then join them.
2. Here A ∩ C lies inside A ∩ B.

Answer: (d) \( A - (B \cup C) \)

Explanation:
1. The shaded part lies inside A.
2. It excludes every point of B and every point of C.
3. So it is A minus (B ∪ C).

Teacher's Note:
1. Identify the circle that contains the shading, then say what is cut out of it.
2. Points in neither B nor C but in A form A − (B ∪ C).

Answer: (c) 300

Explanation:
1. n(A ∪ B) = 200 + 300 − 100 = 400.
2. A′ ∩ B′ = (A ∪ B)′ by De Morgan's law.
3. n = 700 − 400 = 300.

Teacher's Note:
1. A′ ∩ B′ means 'in neither set'.
2. Always compute the union first.

Answer: (b) A

Explanation:
1. A and B have no common element.
2. A ∩ B′ = A − B = A.

Teacher's Note:
1. A − B removes from A whatever is also in B.
2. Disjoint sets leave A unchanged.

Answer: (b) Y

Explanation:
1. 4ⁿ = (1 + 3)ⁿ = 1 + 3n + 9(...), so 4ⁿ − 3n − 1 is a multiple of 9.
2. Hence every element of X is a multiple of 9 and X ⊆ Y.
3. Y is the set of all non-negative multiples of 9, which includes 0 (for n = 1).
4. So X ∪ Y = Y.

Teacher's Note:
1. Expand (1 + 3)ⁿ with the binomial theorem to see divisibility by 9.
2. If X ⊆ Y then X ∪ Y = Y.

Answer: (d) Diagram (D)

Explanation:
1. Every mother is a female, so the mothers circle lies inside the females circle.
2. Some doctors are females (and some of those are mothers), and some doctors are not.
3. So the doctors circle overlaps the females circle and also part of the mothers circle.
4. Only diagram (D) shows a small circle inside a large one with a third circle cutting both.

Teacher's Note:
1. Translate each statement into inside, overlapping or outside.
2. Mothers are a subset of females, not a separate group.

Answer: (c) exactly 25

Explanation:
1. Total student-newspaper readings = 300 × 5 = 1500.
2. Each newspaper accounts for 60 readings.
3. Number of newspapers = 1500 ÷ 60 = 25.

Teacher's Note:
1. Count the same total in two ways.
2. Total readings equals students × papers per student.

Answer: (b) 3300

Explanation:
1. A only = n(A) − n(A∩B) − n(A∩C) + n(A∩B∩C).
2. = 40% − 5% − 4% + 2% = 33%.
3. 33% of 10,000 = 3300.

Teacher's Note:
1. Subtracting both overlaps removes the triple overlap twice, so add it back once.
2. Work in percentages first.

Answer: (c) 60 percent

Explanation:
1. n(car ∪ bus) = 20 + 50 − 10 = 60.

Teacher's Note:
1. Use n(A ∪ B) = n(A) + n(B) − n(A ∩ B).
2. The people using both are counted twice and must be subtracted once.

Answer: (c) 60

Explanation:
1. Maths alone = 100 − 30 − 28 + 18 = 60.

Teacher's Note:
1. Maths alone = n(M) − n(M∩P) − n(M∩C) + n(M∩P∩C).
2. Draw the three-circle Venn diagram if unsure.

Answer: (c) \( A \times (B \cup C) \)

Explanation:
1. B ∪ C = {c, d, e}.
2. A × (B ∪ C) pairs each of a, b with each of c, d, e: six pairs, exactly the given set.
3. (d) would give only (a, d), (b, d).

Teacher's Note:
1. Ordered pairs come only from a Cartesian product.
2. Count: 2 × 3 = 6 pairs.

Answer: (b) 9

Explanation:
1. n(A × B) = n(A) · n(B) = 3 × 3 = 9.

Teacher's Note:
1. Each element of A pairs with every element of B.
2. Multiply the sizes.

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

Explanation:
1. x² − 5x + 6 = (x − 2)(x − 3), so A = {2, 3}.
2. B ∩ C = {4}.
3. A × {4} = {(2, 4), (3, 4)}.

Teacher's Note:
1. The first coordinate comes from A and the second from B ∩ C.
2. Factorise to get A.