JEE Mathematics Set Operations Relations and Mappings MCQs Set 06

Answer: (d) not reflexive

Explanation:
1. \( x+x=100 \) holds only for \( x=50 \), so \( (1,1)\notin R \) and R is not reflexive.
2. It is symmetric, but not transitive (1R99 and 99R1 would need 1R1).

Teacher's Note:
1. One failed diagonal pair is enough to break reflexivity.
2. Symmetry here is automatic because addition is commutative.

Answer: (d) 1

Explanation:
1. Chains: (1,2), (2,3) require (1,3).
2. (1,1), (1,2) give (1,2), already present.
3. No other chains exist, so only (1,3) must be added.

Teacher's Note:
1. List all chains (a, b), (b, c) and check the pair (a, c).
2. Add only what is missing.

Answer: (a) \( \{(4,4)\} \)

Explanation:
1. Pairs (x, y) with \( x\ge y \), x in A and y in B: only \( x=4,\ y=4 \).
2. \( R=\{(4,4)\} \), so \( R^{-1}=\{(4,4)\} \).

Teacher's Note:
1. Maximum of A is 4 and minimum of B is 4.
2. The relation can therefore hold only for (4, 4).

Answer: (b) 8

Explanation:
1. \( P=\{1,2\} \) and \( Q=\{0,1,2,3\} \).
2. \( P\cup Q=\{0,1,2,3\} \) and \( P\cap Q=\{1,2\} \).
3. \( n=4\times2=8 \).

Teacher's Note:
1. Natural numbers start from 1, whole numbers from 0.
2. Then multiply the sizes of the union and the intersection.

Answer: (c) \( \mathbb Z \)

Explanation:
1. For every integer y, \( x=12-3y \) is an integer.
2. So every integer occurs as a second entry, and the range is \( \mathbb Z \).

Teacher's Note:
1. Solve for the other variable and check it is allowed.
2. Here no restriction applies.

Answer: (b) \( \{0,1,2,3,4\} \)

Explanation:
1. a runs over \( -3,-2,-1,0,1,2,3 \).
2. \( b=|a+1| \) takes the values 2, 1, 0, 1, 2, 3, 4.
3. Range = \( \{0,1,2,3,4\} \).

Teacher's Note:
1. Substitute each integer a in the allowed interval.
2. The absolute value makes every b non-negative.

Answer: (d) 6

Explanation:
1. An equivalence relation is reflexive, so it must contain the 6 pairs (a, a).
2. The identity relation with exactly these 6 pairs is itself an equivalence relation.
3. So the minimum is 6.

Teacher's Note:
1. The identity relation is the smallest equivalence relation.
2. The universal relation (36 pairs) is the largest.

Answer: (b) both reflexive and symmetric (indeed an equivalence relation)

Explanation:
1. \( x-x=0 \) is divisible by 5, so R is reflexive.
2. If 5 divides \( x-y \) then 5 divides \( y-x \), so R is symmetric.
3. It is also transitive, so R is an equivalence relation (congruence modulo 5).

Teacher's Note:
1. Divisibility of the difference by n is congruence modulo n.
2. Congruence modulo n is always an equivalence relation.

Answer: (a) symmetric

Explanation:
1. \( 1+a\cdot a=1+a^2>0 \), so \( R_1 \) is reflexive.
2. \( 1+ab>0\Rightarrow1+ba>0 \), so it is symmetric.
3. \( 1R_1\tfrac12 \) and \( \tfrac12R_1(-1) \) hold (values 3/2 and 1/2) but \( 1+1\cdot(-1)=0 \), so \( 1R_1(-1) \) fails and \( R_1 \) is not transitive.

Teacher's Note:
1. Reflexive and symmetric are easy here; transitivity needs a counter-example.
2. Pick numbers of opposite sign.

Answer: (a) an equivalence relation

Explanation:
1. Reflexive: \( (x,x)\in R \) gives \( (x,x)\in R^{-1} \).
2. Symmetric: \( (x,y)\in R^{-1} \) means \( (y,x)\in R \), hence \( (x,y)\in R \) by symmetry of R, hence \( (y,x)\in R^{-1} \).
3. Transitive: similar. So \( R^{-1} \) is an equivalence relation (in fact \( R^{-1}=R \)).

Teacher's Note:
1. For a symmetric relation, the inverse equals the relation itself.
2. So the inverse of an equivalence relation is the same equivalence relation.

Answer: (d) All of the above

Explanation:
1. \( a+d=b+c \) means \( a-b=c-d \): an equality of differences, so an equivalence relation.
2. \( ad=bc \) means \( \tfrac ab=\tfrac cd \): equality of ratios, an equivalence relation.
3. Dividing the first condition by abcd gives \( \tfrac1b+\tfrac1c=\tfrac1a+\tfrac1d \), i.e. \( \tfrac1a-\tfrac1b=\tfrac1c-\tfrac1d \): equality of a quantity, again an equivalence relation.
4. So all three are equivalence relations.

Teacher's Note:
1. Rewrite each condition as 'the same value of some expression of (a, b) and (c, d)'.
2. Such a relation is automatically an equivalence relation.

Answer: (b) 4

Explanation:
1. \( R=\{(8,1),(6,2),(4,3),(2,4)\} \).
2. The domain of \( R^{-1} \) equals the range of R, which is \( \{1,2,3,4\} \).
3. So its number of elements is 4.

Teacher's Note:
1. Domain of the inverse relation is the range of the original.
2. Find the pairs of R first.

Answer: (a) \( 3^5 \)

Explanation:
1. Each element of X has three choices: it is in Y only, in Z only, or in neither (it cannot be in both).
2. For 5 elements, the number of choices is \( 3^5 \).

Teacher's Note:
1. Decide the fate of each element independently.
2. Multiply the number of options over all elements.

Answer: (b) R is symmetric and reflexive but not transitive

Explanation:
1. \( C_iC_i=0\le2\sqrt2 \), so R is reflexive.
2. The distance \( C_iC_j \) equals \( C_jC_i \), so R is symmetric.
3. Neighbouring centres are 2 apart and diagonal ones are \( 2\sqrt2 \) apart, but \( C_1C_3=4>2\sqrt2 \).
4. \( (A_1,A_2),(A_2,A_3)\in R \) but \( (A_1,A_3)\notin R \), so R is not transitive.

Teacher's Note:
1. Work out the possible distances between centres: 2, 2 root 2, 4, 2 root 5.
2. A chain of two close pairs can give a far pair.

Answer: (b) reflexive but not symmetric and not transitive

Explanation:
1. \( x^2-4xy+3y^2=(x-y)(x-3y)=0 \), so \( xRy \) iff \( x=y \) or \( x=3y \).
2. Reflexive: \( x=x \).
3. Not symmetric: \( 3R1 \) (3 = 3 x 1) but \( 1R3 \) would need 1 = 3 or 1 = 9.
4. Not transitive: \( 9R3 \) and \( 3R1 \) but \( 9R1 \) is false.

Teacher's Note:
1. Factorise the quadratic expression first.
2. Then the relation is 'equal or one is thrice the other'.

Answer: (a) \( (x,y)\in R_3\iff |x|+|y|=1,\ x,y\in\mathbb R \)

Explanation:
1. 'a < = b', divisibility and parallelism are transitive.
2. For \( R_3 \): \( (0.6,0.4)\in R_3 \) and \( (0.4,-0.6)\in R_3 \) because \( 0.6+0.4=1 \) and \( 0.4+0.6=1 \).
3. But \( (0.6,-0.6) \) has \( |x|+|y|=1.2\ne1 \), so \( R_3 \) is not transitive.

Teacher's Note:
1. Look for a chain with values that add up.
2. Equation-type relations such as |x| + |y| = 1 are rarely transitive.

Answer: (d) an equivalence relation (the universal relation)

Explanation:
1. The dot product of two vectors is always a scalar, so every pair \( (\vec a,\vec b) \) belongs to R. R is the universal relation.
2. The universal relation on a non-empty set is reflexive, symmetric and transitive, so R is an equivalence relation.

Teacher's Note:
1. Check whether any pair of vectors can fail the condition.
2. The universal relation is always an equivalence relation.

Answer: (a) Both \( \operatorname{dom}(R^{-1})=\operatorname{range}(R) \) and \( \operatorname{range}(R^{-1})=\operatorname{dom}(R) \) hold

Explanation:
1. \( R^{-1}=\{(b,a):(a,b)\in R\} \) reverses every ordered pair.
2. The first elements of \( R^{-1} \) are the second elements of R, so \( \operatorname{dom}(R^{-1})=\operatorname{range}(R) \), and the second elements of \( R^{-1} \) are the first elements of R, so \( \operatorname{range}(R^{-1})=\operatorname{dom}(R) \).

Teacher's Note:
1. Reversing the pairs swaps the roles of the first and second elements.
2. Both facts follow from the same swap.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Reflexive: take \( P=I \), so \( A=I^{-1}AI \). Symmetric: \( A=P^{-1}BP\Rightarrow B=PAP^{-1}=\left(P^{-1}\right)^{-1}AP^{-1} \), which has the required form. Transitive: \( A=P^{-1}BP \) and \( B=Q^{-1}CQ \) give \( A=(QP)^{-1}C(QP) \). So the assertion is true.
2. The reason is the standard rule for the inverse of a product, so it is also true.
3. The proof of symmetry and transitivity uses \( (QP)^{-1}=P^{-1}Q^{-1} \), but the reason alone does not establish the three properties, so it is not the full explanation.

Teacher's Note:
1. Prove reflexive, symmetric and transitive separately.
2. The product rule for inverses is a tool used in the proof, not the whole argument.