JEE Mathematics Set Operations Relations and Mappings MCQs Set 04

Answer: (b) 12

Explanation:
1. List the pairs by the first element a: a = 1 gives 5 pairs (b = 1, 2, 3, 4, 6); a = 2 gives (2,2), (2,4), (2,6); a = 3 gives (3,3), (3,6); a = 4 gives (4,4); a = 6 gives (6,6).
2. Total = 5 + 3 + 2 + 1 + 1 = 12.
3. (The domain and the range are both the whole set A.)

Teacher's Note:
1. Write the relation in roster form first and then count.
2. Always check each divisor pair, including (a, a).

Answer: (c) reflexive, antisymmetric and transitive (a partial order)

Explanation:
1. Reflexive: every natural number divides itself.
2. Transitive: x divides y and y divides z give x divides z.
3. Antisymmetric: if x divides y and y divides x then \( x=y \) (both are positive).
4. It is not symmetric (2 divides 6 but 6 does not divide 2), so it is a partial order, not an equivalence relation.

Teacher's Note:
1. A partial order is reflexive, antisymmetric and transitive.
2. Divisibility on the natural numbers is the standard example.

Answer: (a) 8

Explanation:
1. The smallest equivalence relation containing R must relate 2, 3 and 4 to each other and 1 only to itself.
2. Its equivalence classes are \( \{1\} \) and \( \{2,3,4\} \), giving \( 1^2+3^2=10 \) ordered pairs.
3. R already has 2 of them, so \( 10-2=8 \) pairs must be added.

Teacher's Note:
1. An equivalence relation is a union of full squares on its classes.
2. Count the pairs in the classes and subtract what you already have.

Answer: (a) symmetric only

Explanation:
1. A line is never perpendicular to itself, so R is not reflexive.
2. If \( l_1\perp l_2 \) then \( l_2\perp l_1 \), so R is symmetric.
3. If \( l_1\perp l_2 \) and \( l_2\perp l_3 \) then \( l_1\parallel l_3 \), so \( l_1 \) is not perpendicular to \( l_3 \): R is not transitive.

Teacher's Note:
1. Think of three lines: two parallel ones and a common perpendicular.
2. That example breaks transitivity immediately.

Answer: (c) transitive only

Explanation:
1. No integer is greater than itself, so it is not reflexive.
2. If \( x>y \) then \( y>x \) is false, so it is not symmetric.
3. If \( x>y \) and \( y>z \) then \( x>z \), so it is transitive.

Teacher's Note:
1. Strict inequalities are transitive but neither reflexive nor symmetric.
2. The same holds for 'less than'.

Answer: (d) an equivalence relation

Explanation:
1. Reflexive: each line is parallel to itself.
2. Symmetric: \( l_1\parallel l_2 \) implies \( l_2\parallel l_1 \).
3. Transitive: \( l_1\parallel l_2 \) and \( l_2\parallel l_3 \) give \( l_1\parallel l_3 \).
4. So R is an equivalence relation; its classes are the sets of lines with a given direction.

Teacher's Note:
1. Parallelism groups lines by direction.
2. Equivalence relations always split a set into classes like this.

Answer: (c) \( P\times Q\ne Q\times P \)

Explanation:
1. \( P\times Q=\{(a,r),(b,r),(c,r)\} \) and \( Q\times P=\{(r,a),(r,b),(r,c)\} \).
2. The ordered pairs are different, so \( P\times Q\ne Q\times P \).
3. Both sets have 3 elements, so the numbers of elements are equal.

Teacher's Note:
1. Ordered pairs (a, r) and (r, a) are different unless a = r.
2. The number of elements n(P x Q) is always n(P) times n(Q), whatever the order.

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

Explanation:
1. \( B\cap C=\{4\} \).
2. \( A\times\{4\}=\{(1,4),(2,4),(3,4)\} \).

Teacher's Note:
1. Do the intersection inside the bracket first.
2. Then pair every element of A with each element of the result.

Answer: (b) 19

Explanation:
1. Compute \( |a^2-b^2| \): the condition fails only when it is 16 or more.
2. Pairs that fail: \( (1,5),(5,1) \) (24), \( (2,5),(5,2) \) (21), \( (3,5),(5,3) \) (16).
3. Pairs that fail: 6, so 25 - 6 = 19 pairs satisfy the condition.

Teacher's Note:
1. Count the complement: it is shorter than counting the pairs that work.
2. Strict inequality means a difference of exactly 16 is excluded.

Answer: (b) \( \phi \)

Explanation:
1. \( (Y\times A)\cap(Y\times B)=Y\times(A\cap B) \).
2. \( A\cap B=\phi \), so \( Y\times\phi=\phi \).

Teacher's Note:
1. Use the identity (X x A) intersection (X x B) = X x (A intersection B).
2. A product with the empty set is empty.

Answer: (d) \( \{1,2,3,4,5\} \)

Explanation:
1. y = x + 1 must also lie in A, so \( x+1\le6 \), i.e. \( x\le5 \).
2. \( R=\{(1,2),(2,3),(3,4),(4,5),(5,6)\} \), so the domain is \( \{1,2,3,4,5\} \).

Teacher's Note:
1. Domain means the set of first elements that really occur.
2. Check that the second element stays inside A.

Answer: (d) \( [-\sqrt3,\sqrt3] \)

Explanation:
1. For a real y to exist we need \( 2x^2\le6-3y^2\le6 \), and the largest allowed \( 2x^2 \) occurs at \( y=0 \).
2. So \( 2x^2\le6 \), i.e. \( x^2\le3 \).
3. Domain: \( [-\sqrt3,\sqrt3] \).

Teacher's Note:
1. Look for the x values that have at least one real partner y.
2. The region is an ellipse, so the domain is its horizontal extent.

Answer: (a) \( m\ge n \)

Explanation:
1. Reflexivity forces the n pairs \( (a,a) \), \( a\in A \), to be in R.
2. R may contain further pairs, so \( m\ge n \).

Teacher's Note:
1. Reflexive means every element is related to itself.
2. The identity relation is the smallest reflexive relation.

Answer: (a) \( I\subseteq R \)

Explanation:
1. Reflexive means \( (a,a)\in R \) for every a in A.
2. The identity relation is exactly \( \{(a,a): a\in A\} \), so every pair of I is in R, i.e. \( I\subseteq R \).
3. R may have extra pairs, so \( R=I \) need not hold.

Teacher's Note:
1. Compare the pairs of R with the diagonal pairs.
2. The identity relation is the smallest reflexive relation.

Answer: (d) reflexive but neither symmetric nor transitive

Explanation:
1. Reflexive: (1,1), (2,2), (3,3) are all in R.
2. Not symmetric: \( (2,1)\in R \) but \( (1,2)\notin R \).
3. Not transitive: \( (2,1)\in R \) and \( (1,3)\in R \) but \( (2,3)\notin R \).

Teacher's Note:
1. Check reflexivity first because it is quick.
2. Then search for a pair whose reverse is missing.

Answer: (b) transitive only

Explanation:
1. (1,1) is missing, so R is not reflexive.
2. \( (1,2)\in R \) but \( (2,1)\notin R \), so R is not symmetric.
3. The only pair that could chain is \( (1,2),(2,2)\Rightarrow(1,2)\in R \), so R is transitive.

Teacher's Note:
1. For transitivity, list every chain (a, b), (b, c).
2. Here the only chain returns a pair already in R.

Answer: (a) an equivalence relation

Explanation:
1. Reflexive: \( a+b=b+a \).
2. Symmetric: \( a+d=b+c\Rightarrow c+b=d+a \), so \( (c,d)R(a,b) \).
3. Transitive: \( a+d=b+c \) and \( c+f=d+e \) add to \( a+f=b+e \), so \( (a,b)R(e,f) \).

Teacher's Note:
1. The relation says a - b = c - d, i.e. equal differences.
2. Equality of a quantity is always an equivalence relation.

Answer: (a) 64

Explanation:
1. A reflexive relation must contain the 3 identity pairs.
2. The other \( 3^2-3=6 \) pairs may be included or not.
3. Number \( =2^6=64 \).

Teacher's Note:
1. General rule: a set with n elements has 2^(n^2 - n) reflexive relations.
2. Here n = 3 gives 2^6.

Answer: (c) an equivalence relation

Explanation:
1. Reflexive: OA = OA.
2. Symmetric: OA = OB gives OB = OA.
3. Transitive: OA = OB and OB = OC give OA = OC.
4. So the relation is an equivalence relation; its classes are circles centred at O.

Teacher's Note:
1. Equal distance from a fixed point is an equality of a quantity.
2. The classes are concentric circles.

Answer: (d) \( \{(3,3),(3,5),(5,3),(5,5)\} \)

Explanation:
1. \( R=\{(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)\} \) and \( R^{-1}=\{(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)\} \).
2. \( R\,o\,R^{-1} \) means: first apply \( R^{-1} \), then R.
3. From 3: \( 3\to1\to\{3,5\} \), and \( 3\to2\to\{3,5\} \). From 5: \( 5\to1,2,3,4 \) leads to \( \{3,5\} \) and \( 5 \).
4. Result: \( \{(3,3),(3,5),(5,3),(5,5)\} \).

Teacher's Note:
1. For a composition, start with the relation on the right of the symbol.
2. Follow each element through the intermediate set.