Ncert Solutions Class 12 Maths Chapter 1 Relations And Functions Ex 1 1

Concept used. A relation R on a non-empty set A is a subset of A × A. It is called reflexive when (a,a) ∈ R for every a ∈ A. It is called symmetric when (a,b) ∈ R forces (b,a) ∈ R for every pair. It is called transitive when (a,b) ∈ R together with (b,c) ∈ R forces (a,c) ∈ R. To prove a property we argue for an arbitrary element; to disprove a property a single counter-example is enough.

1. (i) On A = 1,2,…,14 with R = (x,y) : 3x - y = 0. Rewriting the rule as y = 3x and restricting both coordinates to A gives R = (1,3),(2,6),(3,9),(4,12). Reflexive? Take a = 1. Then (1,1) requires 3(1) - 1 = 0, but 3 - 1 = 2 ≠ 0. So (1,1) ∉ R and R is not reflexive.
   Symmetric? (1,3) ∈ R, but (3,1) requires 3(3) - 1 = 0, i.e. 8 = 0, which is false. So R is not symmetric.
   Transitive? The only chain available is (1,3) and (3,9); for transitivity we would need (1,9) ∈ R, but (1,9) requires 3(1) - 9 = -6 = 0, false. So R is not transitive.
2. (ii) On N with R = (x,y) : y = x + 5, x < 4. Only x = 1, 2, 3 are allowed, so R = (1,6),(2,7),(3,8).
   Reflexive? (1,1) ∈ R would require 1 = 1 + 5, false. Not reflexive.
   Symmetric? (1,6) ∈ R but (6,1) fails because the rule forces 1 = 6 + 5. Not symmetric.
   Transitive? For a chain (a,b) and (b,c) in R, b must occur as a second coordinate and also as a first coordinate. The first coordinates of R are 1,2,3 and the second coordinates are 6,7,8; they share nothing. So the premise of transitivity is never met, making the implication vacuously true. Hence R is transitive.
3. (iii) On A = 1,2,3,4,5,6 with R = (x,y) : x y. Reflexive? For every a ∈ A, a divides a (with quotient 1). So (a,a) ∈ R for all a. Reflexive.
   Symmetric? (2,4) ∈ R since 2 4, but (4,2) ∉ R since 4 2. Not symmetric.
   Transitive? If x y and y z, write y = xk and z = ym for integers k, m. Then z = x(km), so x z. Hence (x,y), (y,z) ∈ R ⇒ (x,z) ∈ R. Transitive.
4. (iv) On Z with R = (x,y) : x - y ∈ Z. Every difference of two integers is itself an integer, so R = Z × Z.
   Reflexive? a - a = 0 ∈ Z, so (a,a) ∈ R for all a. Reflexive.
   Symmetric? If x - y ∈ Z, then y - x = -(x - y) ∈ Z. So (y,x) ∈ R. Symmetric.
   Transitive? If x - y ∈ Z and y - z ∈ Z, then x - z = (x - y) + (y - z) ∈ Z. Transitive.
5. (v) Five relations on the set A of humans in a town.
   • [(a)] R = (x,y) : x,y work at the same place. Every person works at the same place as themselves: reflexive. If x works where y works, then y works where x works: symmetric. If x and y share a workplace and y and z share a workplace, all three share that workplace: transitive. [(b)] R = (x,y) : x,y live in the same locality. By the same argument as (a): reflexive, symmetric, and transitive. [(c)] R = (x,y) : x is exactly 7 cm taller than y. Reflexive? A person is not 7 cm taller than themselves (the difference is 0 cm). Not reflexive. Symmetric? If x is 7 cm taller than y, then y is 7 cm shorter than x, not taller. Not symmetric. Transitive? If x is 7 cm taller than y and y is 7 cm taller than z, then x is 14 cm taller than z, not 7 cm. Not transitive. [(d)] R = (x,y) : x is wife of y. Reflexive? A person is not their own wife. Not reflexive. Symmetric? If x is wife of y, then y is husband of x, not wife of x. Not symmetric. Transitive? The premise is rarely met, but more importantly if x is wife of y and y is wife of z the chain cannot exist (no person is wife of two distinct people in this sense), so the implication is vacuously true; the relation is technically transitive. [(e)] R = (x,y) : x is father of y. Not reflexive, since nobody fathers themselves. Not symmetric, since the reverse pair would mean the child fathers the parent. Not transitive: if x fathers y and y fathers z, then x is grandfather (not father) of z.

(i) None. (ii) Transitive only. (iii) Reflexive, transitive. (iv) All three (equivalence). (v) (a),(b) all three; (c) none; (d) transitive only; (e) none.

Concept used. For a single number a the pair (a,a) belongs to R exactly when a ≤ a². The inequality a ≤ a² rearranges to a(a-1) ≥ 0, which holds when a ≤ 0 or a ≥ 1 but fails when 0 < a < 1. We exploit this gap by choosing a counter-example in (0,1), typically a = 12.

1. Not reflexive. Pick a = 12. Compute the right-hand side first: a² = (12)² = 14. Now check the inequality a ≤ a²: 12 ≤ 14 ? ⇒ 0.5 ≤ 0.25 ? . Hence (12, 12) ∉ R, so R is not reflexive.
2. Not symmetric. Choose a = 1 and b = 2. Check (1,2): is 1 ≤ 2²? 1 ≤ 4, true. So (1,2) ∈ R. Check (2,1): is 2 ≤ 1²? 2 ≤ 1, false. So (2,1) ∉ R. Symmetry is broken.
3. Not transitive. Choose a = 10, b = -2, c = 1. Check (10,-2): is 10 ≤ (-2)² = 4? 10 ≤ 4, false; this pair is NOT in R, so it does not produce a chain. Try instead a = 5, b = -3, c = 1. Check (5,-3): is 5 ≤ (-3)² = 9? Yes. So (5,-3) ∈ R. Check (-3,1): is -3 ≤ 1² = 1? Yes. So (-3,1) ∈ R. Check (5,1): is 5 ≤ 1² = 1? No. So (5,1) ∉ R. We have a chain (5,-3), (-3,1) in R but no shortcut (5,1). Transitivity fails.

R is neither reflexive, nor symmetric, nor transitive.

Concept used. The rule b = a + 1 lists out every ``successor'' pair on the set A = 1,2,3,4,5,6. Writing the relation explicitly makes property checks routine.

1. Enumerate R. Allowed first coordinates are 1, 2, 3, 4, 5 (since a = 6 would force b = 7 ∉ A): R = (1,2), (2,3), (3,4), (4,5), (5,6).
2. Reflexive? Pick any a, say a = 1. The pair (1,1) requires 1 = 1 + 1 = 2, false. So (1,1) ∉ R and reflexivity fails.
3. Symmetric? (1,2) ∈ R since 2 = 1 + 1. For symmetry we need (2,1) ∈ R, which requires 1 = 2 + 1 = 3, false. So (2,1) ∉ R. Not symmetric.
4. Transitive? Take the chain (1,2) and (2,3). Both are in R since 2 = 1+1 and 3 = 2+1. For transitivity we need (1,3) ∈ R. That requires 3 = 1 + 1 = 2, false. So (1,3) ∉ R. Not transitive.

R is neither reflexive, nor symmetric, nor transitive.

Concept used. The standard order ≤ on the real line satisfies three axioms: a ≤ a (reflexivity), a ≤ b and b ≤ c ⇒ a ≤ c (transitivity), and a ≤ b together with b ≤ a ⇒ a = b (antisymmetry, which prevents two-way comparison unless a = b).

1. Reflexive. For every real number a, a ≤ a holds (any number equals itself, hence is ≤ itself). So (a, a) ∈ R for every a ∈ R.
2. Transitive. Suppose (a, b) ∈ R and (b, c) ∈ R. Then a ≤ b and b ≤ c. By the transitivity axiom of the real order, a ≤ c. So (a, c) ∈ R.
3. Not symmetric. Counter-example: choose a = 1, b = 2. Check (1, 2): is 1 ≤ 2? Yes. So (1, 2) ∈ R. Check (2, 1): is 2 ≤ 1? No. So (2, 1) ∉ R. Symmetry fails.

R is reflexive and transitive but not symmetric.

Concept used. Cubing preserves sign and order on R: b³ > 0 iff b > 0, b³ < 0 iff b < 0, and b³ = 0 iff b = 0. Unlike squaring, cubing does not turn negatives into positives, so the relation a ≤ b³ treats negative numbers and positive numbers asymmetrically.

1. Not reflexive. A pair (a,a) ∈ R requires a ≤ a³, i.e. a³ - a ≥ 0, i.e. a(a-1)(a+1) ≥ 0. This holds for a ∈ [-1, 0] ∪ [1, ∞) and fails otherwise. Pick a counter-example in the gap (0,1): take a = 12. a³ = (12)³ = 18 = 0.125. Now check a ≤ a³: 0.5 ≤ 0.125? False. So (12, 12) ∉ R. Not reflexive.
2. Not symmetric. Take a = 1, b = 2. Check (1, 2): 1 ≤ 2³ = 8? True. So (1,2) ∈ R. Check (2, 1): 2 ≤ 1³ = 1? False. So (2,1) ∉ R. Not symmetric.
3. Not transitive. Take a = 10, b = -2, c = 1? First test whether the pairs are in R: (10, -2) needs 10 ≤ (-2)³ = -8, false. So this chain does not start in R. Retry. Take a = 3, b = 32, c = 1110. Compute: b³ = (32)³ = 278 = 3.375, c³ = (1110)³ = 1.331. Check (3, 32): 3 ≤ 3.375? True. In R. Check (32, 1110): 1.5 ≤ 1.331? False. So this pair is NOT in R. Retry once more. Take a = 9, b = 3, c = 2. b³ = 27, c³ = 8. Check (9, 3): 9 ≤ 27? True. In R. Check (3, 2): 3 ≤ 8? True. In R. Check (9, 2): 9 ≤ 8? False. Not in R. Chain (9,3),(3,2) ∈ R but (9,2) ∉ R, so not transitive.

R is neither reflexive, nor symmetric, nor transitive.

Concept used. A relation with only two pairs is small enough to check each property by listing.

1. Symmetric. (1,2) ∈ R and its reverse (2,1) ∈ R. (2,1) ∈ R and its reverse (1,2) ∈ R. Every pair has its mirror in R, so R is symmetric.
2. Not reflexive. Reflexivity demands (a,a) ∈ R for every a ∈ 1,2,3. We need (1,1), (2,2), (3,3), none of which appears in R. Hence not reflexive.
3. Not transitive. Chain: (1,2) ∈ R and (2,1) ∈ R. For transitivity we need (1,1) ∈ R. But (1,1) ∉ R. So R is not transitive.

R is symmetric, but neither reflexive nor transitive.

Concept used. A relation defined by ``x and y share the same value of some function g'' is always an equivalence relation, because equality of values is itself reflexive, symmetric, and transitive. Here g(x) is the page count of book x.

1. Reflexive. For every book x, x has the same number of pages as itself (g(x) = g(x)). So (x, x) ∈ R for every x ∈ A.
2. Symmetric. Suppose (x, y) ∈ R. Then g(x) = g(y). Equality is symmetric, so g(y) = g(x), meaning (y, x) ∈ R.
3. Transitive. Suppose (x, y) ∈ R and (y, z) ∈ R. Then g(x) = g(y) and g(y) = g(z). Equality is transitive, so g(x) = g(z), i.e. (x, z) ∈ R.

R is an equivalence relation.

Concept used. The integer difference a - b is even iff a and b have the same parity (both odd or both even). So R is exactly the ``same-parity'' relation on 1, …, 5, which is a kernel relation of the parity function g(n) = n 2.

1. Reflexive. For any a, |a - a| = 0 is even. So (a, a) ∈ R.
2. Symmetric. If (a, b) ∈ R then |a - b| is even. But |b - a| = |a - b|, so |b - a| is even too. Hence (b, a) ∈ R.
3. Transitive. If (a, b), (b, c) ∈ R then |a - b| and |b - c| are both even. Write a - b = 2m, b - c = 2n for integers m, n. Then a - c = (a - b) + (b - c) = 2m + 2n = 2(m + n), which is even, so |a - c| is even and (a, c) ∈ R.
4. Class 1, 3, 5. All three are odd. Differences: |1-3|=2, |1-5|=4, |3-5|=2 are all even. So every pair among 1,3,5 is in R.
5. Class 2, 4. Both even. |2-4| = 2 is even, so (2,4),(4,2) ∈ R.
6. No cross pairs. Pick any x ∈ 1,3,5 (odd) and y ∈ 2,4 (even). |x - y| is odd (odd minus even is odd), so (x, y) ∉ R.

R is an equivalence relation; classes are 1,3,5 and 2,4.

Concept used. A = 0, 1, 2, …, 12 has 13 elements. We use the same template as Q8: ``|a - b| is a multiple of 4'' is the kernel relation of the map g(n) = n 4, and ``a = b'' is the identity relation, which is the smallest equivalence relation on any set.

1. (i) Reflexive. |a - a| = 0 = 4 · 0 is a multiple of 4. So (a, a) ∈ R.
2. (i) Symmetric. |a - b| = |b - a|. If one is a multiple of 4, so is the other.
3. (i) Transitive. If |a - b| and |b - c| are multiples of 4, write a - b = 4m, b - c = 4n. Then a - c = (a - b) + (b - c) = 4(m + n), a multiple of 4. So |a - c| is a multiple of 4.
4. (i) Set related to 1. We need x ∈ A with |x - 1| a multiple of 4. So x - 1 ∈ 0, ± 4, ± 8, ± 12. Restricted to A: x ∈ 1, 5, 9 (taking x - 1 = 0, 4, 8; x - 1 = -4 gives x = -3 ∉ A; x - 1 = 12 gives x = 13 ∉ A).
5. (ii) Reflexive, symmetric, transitive. For ``a = b'': a = a is true (reflexive); a = b ⇒ b = a (symmetric); a = b and b = c ⇒ a = c (transitive). All three hold by the axioms of equality.
6. (ii) Set related to 1. We need x ∈ A with x = 1, i.e. x = 1. So the related set is 1.

(i) 1, 5, 9;   (ii) 1.

Concept used. Different combinations of the three properties yield different qualitative ``shapes'' for a relation. Constructing examples is a creativity exercise: pick a small set, draw an arrow diagram, and tune until the desired properties hold and the unwanted ones fail.

1. (i) Symmetric only. On A = 1, 2, 3 take R = (1, 2), (2, 1). Symmetric: each pair has its reverse. Not reflexive: (1,1), (2,2), (3,3) are all absent. Not transitive: (1,2) ∈ R and (2,1) ∈ R but (1,1) ∉ R.
2. (ii) Transitive only. On R take R = (a, b) : a < b, the strict less-than relation. Transitive: a < b and b < c ⇒ a < c. Not reflexive: a < a is false. Not symmetric: 1 < 2 but 2 < 1 is false.
3. (iii) Reflexive and symmetric but not transitive. On A = 1, 2, 3 take R = (1,1), (2,2), (3,3), (1,2), (2,1), (2,3), (3,2). Reflexive: every self-loop present. Symmetric: every cross-pair has its reverse. Not transitive: (1,2),(2,3) ∈ R but (1,3) ∉ R.
4. (iv) Reflexive and transitive but not symmetric. On N take R = (a, b) : a ≤ b. Reflexive: a ≤ a. Transitive: a ≤ b and b ≤ c ⇒ a ≤ c. Not symmetric: 1 ≤ 2 but 2 ≤ 1 is false.
5. (v) Symmetric and transitive but not reflexive. On A = 1, 2, 3 take R = (1,2), (2,1), (1,1), (2,2). Symmetric: yes. Transitive: every chain in R stays inside 1, 2². Check (1,2),(2,1) → (1,1) ∈ R; (2,1),(1,2) → (2,2) ∈ R; (1,1),(1,2) → (1,2) ∈ R, etc. All chains close inside R. Not reflexive: (3, 3) ∉ R.

See the five examples above; each satisfies exactly the requested combination of properties.

Concept used. R is the kernel relation of the function d : A → [0, ∞) defined by d(P) = distance of P from the origin O = (0, 0). Kernel relations are automatically equivalence relations (Q7), so we just need to identify the equivalence class.

1. Reflexive. d(P) = d(P) for every point P. So (P, P) ∈ R.
2. Symmetric. If d(P) = d(Q), then d(Q) = d(P) (equality is symmetric).
3. Transitive. If d(P) = d(Q) and d(Q) = d(R), then d(P) = d(R).
4. Class of P ≠ O. Set r = d(P) > 0. A point Q is related to P iff d(Q) = r. The set of all points at distance r from O is precisely the circle of radius r centred at O. Since d(P) = r, this circle passes through P. Hence [P] = Q ∈ A : d(Q) = r = Q : Q lies on the circle through P centred at O.

R is an equivalence; class of P is the circle through P centred at the origin.

Concept used. Two triangles are similar when their corresponding angles are equal, equivalently, when their corresponding sides are in the same ratio. Similarity is reflexive, symmetric, and transitive, making it an equivalence relation on the set of all triangles.

1. Reflexive. Every triangle is similar to itself (angles equal to themselves, sides in ratio 1:1:1). So (T, T) ∈ R.
2. Symmetric. If T₁ ∼ T₂ with ratio k, then T₂ ∼ T₁ with ratio 1/k. So (T₂, T₁) ∈ R.
3. Transitive. If T₁ ∼ T₂ with ratio k and T₂ ∼ T₃ with ratio m, then T₁ ∼ T₃ with ratio km. So (T₁, T₃) ∈ R.
4. Which of T₁, T₂, T₃ are related? Two triangles are similar iff their sides are proportional. Compute ratios for each pair.

   T₁ (3,4,5) vs T₃ (6,8,10): 36 = 12, 48 = 12, 510 = 12. All three ratios equal 12. So T₁ ∼ T₃.

   T₁ (3,4,5) vs T₂ (5,12,13): 35 = 0.6, 412 = 13 ≈ 0.333, 513 ≈ 0.385. Ratios differ, so T₁ ∼ T₂.

   T₂ (5,12,13) vs T₃ (6,8,10): 56 ≈ 0.833, 128 = 1.5, 1310 = 1.3. Ratios differ, so T₂ ∼ T₃.

R is an equivalence relation; among the three triangles, only T₁ and T₃ are related.

Concept used. R is the kernel relation of the function s : A → N that returns the number of sides of a polygon. Kernel relations are equivalences (Q7 again).

1. Reflexive. s(P) = s(P) trivially. So (P, P) ∈ R.
2. Symmetric. s(P₁) = s(P₂) ⇒ s(P₂) = s(P₁).
3. Transitive. s(P₁) = s(P₂) and s(P₂) = s(P₃) ⇒ s(P₁) = s(P₃).
4. Class of T. T is a triangle, so s(T) = 3. The class of T is the set of all polygons in A with exactly 3 sides, i.e. all triangles.

R is an equivalence; the class of T is the set of all triangles in A.

Concept used. Two non-vertical lines y = m₁ x + c₁ and y = m₂ x + c₂ in the XY plane are parallel iff they have the same slope m₁ = m₂. (Two vertical lines x = a₁ and x = a₂ are parallel; a vertical and a non-vertical line are never parallel.) The convention in NCERT is that a line is parallel to itself.

1. Reflexive. Every line is parallel to itself (slope equal to itself, or both vertical). So (L, L) ∈ R.
2. Symmetric. If L₁ ∥ L₂, both have the same slope, so L₂ ∥ L₁. Hence (L₂, L₁) ∈ R.
3. Transitive. If L₁ ∥ L₂ and L₂ ∥ L₃, then L₁, L₂, L₃ all share the same slope (or all are vertical). So L₁ ∥ L₃.
4. Class of y = 2x + 4. The slope of this line is m = 2. Every line with slope 2 has the equation y = 2x + c for some real c. So [ y = 2x + 4 ] = y = 2x + c : c ∈ R .

R is an equivalence; the class of y = 2x + 4 is y = 2x + c : c ∈ R.

Concept used. Check the three properties on the explicit list of pairs.

1. Reflexive? Need (a, a) ∈ R for every a ∈ 1, 2, 3, 4. From the list: (1,1), (2,2), (3,3), (4,4) all present. Reflexive.
2. Symmetric? Check the non-diagonal pairs. (1, 2) ∈ R; is (2, 1) ∈ R? Not in the list. So symmetry fails. Not symmetric.
3. Transitive? Check chains. From the off-diagonal pairs (1,2),(1,3),(3,2):
   • (1,3),(3,2) ⇒ (1,2)? Yes, (1,2) ∈ R. (ok)
   • (3,2),(2,2) ⇒ (3,2)? Yes, (3,2) ∈ R. (ok)
   • (1,2),(2,2) ⇒ (1,2)? Yes. (ok)
   • (1,3),(3,3) ⇒ (1,3)? Yes. (ok)
   • (1,1),(1,2) ⇒ (1,2)? Yes. (ok)
   • (1,1),(1,3) ⇒ (1,3)? Yes. (ok)
   All chains close inside R. Transitive.
4. Reflexive + transitive but not symmetric. Matches option (B).

(B)

Concept used. A pair (a, b) is in R iff both conditions hold: a = b - 2 and b > 6.

1. (A) (2, 4)? b = 4 > 6? No. Fails. Not in R.
2. (B) (3, 8)? b = 8 > 6? Yes. a = b - 2 = 6? But a = 3 ≠ 6. Fails. Not in R.
3. (C) (6, 8)? b = 8 > 6? Yes. a = b - 2 = 8 - 2 = 6? Yes, a = 6. Both conditions hold. In R.
4. (D) (8, 7)? b = 7 > 6? Yes. a = b - 2 = 5? But a = 8 ≠ 5. Fails. Not in R.

(C) (6, 8) ∈ R.