JEE Mathematics Set Operations Relations and Mappings MCQs Set 05

Answer: (d) \( \{(8,11),(10,13)\} \)

Explanation:
1. x = 11 gives y = 8 and x = 13 gives y = 10; x = 12 gives y = 9, which is not in the second set.
2. \( R=\{(11,8),(13,10)\} \).
3. \( R^{-1}=\{(8,11),(10,13)\} \).

Teacher's Note:
1. Only pairs with y in the target set belong to R.
2. The inverse simply swaps the two entries of each pair.

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

Explanation:
1. \( x^2-5x+6=0 \) gives \( A=\{2,3\} \).
2. \( A-B=\{3\} \) and \( A-C=\{2,3\} \).
3. \( (A-B)\times(A-C)=\{(3,2),(3,3)\} \).

Teacher's Note:
1. Solve for A first, then take the differences.
2. Keep the order: first factor first.

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

Explanation:
1. Pairs with x < y: (1,2), (1,3), (1,4), (2,3), (2,4), (3,4).
2. \( |x^2-y^2| \): 3, 8, 15, 5, 12, 7.
3. Keep those below 9: (1,2), (1,3), (2,3), (3,4).

Teacher's Note:
1. List the pairs with x < y and test the second condition.
2. Both conditions must hold at the same time.

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

Explanation:
1. \( y=3x\le14 \) gives \( x\le4 \).
2. \( R=\{(1,3),(2,6),(3,9),(4,12)\} \), so the domain is \( \{1,2,3,4\} \) and the range is \( \{3,6,9,12\} \).

Teacher's Note:
1. Do not confuse domain and range.
2. Domain collects the first entries of the pairs.

Answer: (a) \( \{-2,-1,0,1,2\} \)

Explanation:
1. \( x^2\le4-y^2\le4 \) gives \( |x|\le2 \).
2. Each of \( x=-2,-1,0,1,2 \) has a partner (for example y = 0).
3. Domain = \( \{-2,-1,0,1,2\} \).

Teacher's Note:
1. Look at the integers inside the circle of radius 2.
2. Both positive and negative values count.

Answer: (d) 21

Explanation:
1. The remainder on division by 7 can be 0, 1, 2, 3, 4, 5 or 6, and each value occurs.
2. Range = \( \{0,1,2,3,4,5,6\} \).
3. Sum = 21.

Teacher's Note:
1. Remainders on division by n run from 0 to n - 1.
2. Then add the seven values.

Answer: (c) \( \{(2,8),(3,27),(5,125),(7,343)\} \)

Explanation:
1. The primes below 10 are 2, 3, 5, 7.
2. Their cubes are 8, 27, 125, 343.
3. So \( R=\{(2,8),(3,27),(5,125),(7,343)\} \).

Teacher's Note:
1. Roster form lists the pairs explicitly.
2. Compute each cube carefully.

Answer: (d) both reflexive and symmetric

Explanation:
1. Reflexive: \( a+a=2a \) is even.
2. Symmetric: \( a+b=b+a \).
3. (It is also transitive, so it is an equivalence relation; the classes are the odd and the even numbers.)

Teacher's Note:
1. A sum is even when both numbers have the same parity.
2. Same parity is an equivalence relation.

Answer: (c) R is symmetric

Explanation:
1. (1,1), (2,2), (3,3) are in R, so R is reflexive.
2. The only chain is (2,3), (3,3), giving (2,3) in R, so R is transitive.
3. \( (2,3)\in R \) and \( (3,2)\notin R \), so R is antisymmetric but not symmetric.

Teacher's Note:
1. Test each property in turn.
2. Antisymmetric and symmetric are different ideas.

Answer: (c) not reflexive

Explanation:
1. \( (c,c)\notin R \), so R is not reflexive.
2. \( (b,c)\in R \) but \( (c,b)\notin R \), so R is not symmetric.
3. \( (a,b),(b,c)\in R \) but \( (a,c)\notin R \), so R is not transitive.
4. The only true statement is that R is not reflexive.

Teacher's Note:
1. Check the diagonal pairs first.
2. Then look for a missing reverse pair or missing chain.

Answer: (a) \( aRb \) if \( a < b \)

Explanation:
1. a + b is even and a - b is even both mean a and b have the same parity, which is an equivalence relation.
2. a = b is the identity relation, also an equivalence.
3. a < b is not reflexive and not symmetric, so it is not an equivalence relation.

Teacher's Note:
1. Recognise equality and same-remainder relations as equivalences.
2. Strict inequality never is one.

Answer: (c) 5

Explanation:
1. Equivalence relations on a set correspond to partitions of the set.
2. Partitions of {a, b, c}: one block {a,b,c}; {a},{b,c}; {b},{a,c}; {c},{a,b}; and {a},{b},{c}.
3. That is 5 partitions, hence 5 equivalence relations.

Teacher's Note:
1. Count partitions of the set rather than relations.
2. The number of partitions of a 3-element set is the third Bell number, 5.

Answer: (a) \( x-y \) is divisible by 5

Explanation:
1. x - y divisible by 5 means x and y leave the same remainder on division by 5: reflexive, symmetric and transitive.
2. x < y and x > y are not reflexive or symmetric.
3. 'x divides y' is not symmetric (and not reflexive for 0).

Teacher's Note:
1. Congruence modulo n is the model equivalence relation on the integers.
2. The relations in the other options are orders.

Answer: (d) \( \{(3,3),(2,6),(1,9)\} \)

Explanation:
1. y = 1 gives x = 9; y = 2 gives x = 6; y = 3 gives x = 3; y = 4 gives x = 0, which is not natural.
2. \( R=\{(9,1),(6,2),(3,3)\} \).
3. \( R^{-1}=\{(1,9),(2,6),(3,3)\} \).

Teacher's Note:
1. List the natural-number solutions first.
2. Then swap the entries of each pair.

Answer: (a) \( \phi \)

Explanation:
1. \( B\times\phi \) has no elements, so \( B\times\phi=\phi \).
2. \( A\times\phi=\phi \).

Teacher's Note:
1. A Cartesian product with the empty set is empty.
2. Evaluate the bracket first.

Answer: (b) \( \{(4,5),(5,4),(5,5)\} \)

Explanation:
1. The largest possible sum is 10.
2. Sums above 8 are 9 and 10: \( 4+5=9 \), \( 5+4=9 \), \( 5+5=10 \).
3. So the pairs are (4,5), (5,4), (5,5).

Teacher's Note:
1. Only large values can work, so start with the biggest elements.
2. Both orders of an unequal pair count.

Answer: (c) Domain = {1, 2, 3}, Range = {7, 5}

Explanation:
1. For y to be a natural number, x must divide 6, and \( x < 6 \) leaves \( x=1,2,3 \).
2. \( x=1: y=7;\ x=2: y=5;\ x=3: y=5 \).
3. Domain = \( \{1,2,3\} \), range = \( \{5,7\} \).

Teacher's Note:
1. The fraction 6/x must be an integer.
2. Remember that two different x values can give the same y.

Answer: (a) 4

Explanation:
1. The number of relations on A is \( 2^{n^2} \).
2. \( n^2=16 \) gives \( n=4 \).

Teacher's Note:
1. A relation on A is a subset of A x A, which has n^2 elements.
2. Equate the exponents.

Answer: (b) reflexive and symmetric but not transitive

Explanation:
1. \( |a-a|=0\le1 \), so R is reflexive.
2. \( |a-b|=|b-a| \), so R is symmetric.
3. \( 0R1 \) and \( 1R2 \) but \( |0-2|=2>1 \), so R is not transitive.

Teacher's Note:
1. A 'closeness' relation is reflexive and symmetric.
2. Transitivity fails because small steps add up.

Answer: (a) R is not reflexive, symmetric and not transitive

Explanation:
1. \( R=\{(1,4),(4,1),(2,3),(3,2)\} \).
2. No pair (a, a) is in R, so R is not reflexive.
3. Each pair has its reverse, so R is symmetric.
4. \( (1,4),(4,1)\in R \) but \( (1,1)\notin R \), so R is not transitive.

Teacher's Note:
1. List the pairs explicitly.
2. The sum condition x + y = 5 is symmetric in x and y.