JEE Mathematics Set Operations Relations and Mappings MCQs Set 01

Answer: (d) 4

Explanation:
1. \( P(\phi)=\{\phi\} \) has 1 element.
2. \( P[P(\phi)]=\{\phi,\{\phi\}\} \) has 2 elements.
3. \( P[P[P(\phi)]] \) has \( 2^2=4 \) elements.

Teacher's Note:
1. The power set of a set with k elements has 2^k elements.
2. Apply this three times starting from the empty set.

Answer: (c) the set of integers

Explanation:
1. A is the set of odd integers and B the set of even integers.
2. Together they make up all integers.

Teacher's Note:
1. Every integer is either odd or even.
2. Both sets contain negative numbers too.

Answer: (d) \( \{x:x=12n,\ n\in\mathbb Z\} \)

Explanation:
1. A common element is a multiple of both 3 and 4.
2. Hence it is a multiple of lcm(3, 4) = 12: \( \{x:x=12n\} \).

Teacher's Note:
1. Elements of the intersection are common multiples.
2. The LCM of 3 and 4 is 12.

Answer: (b) 6

Explanation:
1. \( n(A\cup B)=n(A)+n(B)-n(A\cap B)=9-n(A\cap B) \).
2. The intersection has at most 3 elements (when A ⊂ B), so the minimum is 9 − 3 = 6.

Teacher's Note:
1. The union is smallest when the intersection is largest.
2. A ∩ B can have at most 3 elements.

Answer: (c) \( 2^{mn} \)

Explanation:
1. A × B has mn elements.
2. A set with mn elements has \( 2^{mn} \) subsets.

Teacher's Note:
1. Count the elements of A × B first.
2. The power set of a k-element set has 2^k elements.

Answer: (c) \( n^2 \)

Explanation:
1. An ordered pair (x, y) lies in both A × B and B × A iff x ∈ A ∩ B and y ∈ A ∩ B.
2. There are n choices for x and n for y, so \( n^2 \) common elements.

Teacher's Note:
1. Both coordinates must belong to A ∩ B.
2. Count ordered pairs.

Answer: (c) \( \{9,6,3\} \)

Explanation:
1. Solutions of \( x+3y=12 \) in natural numbers: y = 1, x = 9; y = 2, x = 6; y = 3, x = 3 (y = 4 gives x = 0, not natural).
2. Domain \( =\{9,6,3\} \).

Teacher's Note:
1. List the pairs (x, y) with x, y natural.
2. The domain is the set of first coordinates.

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

Explanation:
1. Pairs of R: (8, 1), (6, 2), (4, 3), (2, 4) (y = 5 gives x = 0, which is not in A).
2. The domain of R⁻¹ is the range of R: \( \{1,2,3,4\} \).

Teacher's Note:
1. The domain of the inverse relation is the range of the original relation.
2. List all pairs first.

Answer: (b) reflexive and symmetric

Explanation:
1. Reflexive: (2,2), (3,3), (4,4), (5,5) are all present.
2. Symmetric: (2,3)/(3,2) and (3,5)/(5,3) both occur.
3. Not transitive: (2,3) and (3,5) are in R but (2,5) is not. Not antisymmetric either, since (2,3) and (3,2) both belong with 2 ≠ 3.

Teacher's Note:
1. Test the three properties one at a time.
2. Look for one counter-example to transitivity.