NCERT Solutions to NCERT Class 12 Mathematics Exercise 1.1: Relations and Functions

The document provided is a Class 12 NCERT Mathematics textbook chapter on "Relations and Functions," specifically covering Exercise 1.1. Based on the standard NCERT Class 12 Mathematics textbook (2022-23 edition), Exercise 1.1 consists of 16 questions that focus on relations, their properties (reflexive, symmetric, transitive), and equivalence relations.

Below are the solutions to all 16 questions from Exercise 1.1 of the NCERT Class 12 Mathematics textbook, with explanations based on the definitions and examples.

Exercise 1.1 Questions and Solutions

Question 1

Determine whether each of the following relations are reflexive, symmetric, and transitive:
(i) Relation R in the set A = {1, 2, 3, ..., 13, 14} defined as R = {(x, y) : 3x – y = 0}
(ii) Relation R in the set N of natural numbers defined as R = {(x, y) : y = x + 5 and x < 4}
(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} defined as R = {(x, y) : y is divisible by x}
(iv) Relation R in the set Z of all integers defined as R = {(x, y) : x – y is an integer}
(v) Relation R in the set A of human beings in a town at a particular time defined as
(a) R = {(x, y) : x and y work at the same place}
(b) R = {(x, y) : x and y live in the same locality}
(c) R = {(x, y) : x is exactly 7 cm taller than y}
(d) R = {(x, y) : x is wife of y}
(e) R = {(x, y) : x is father of y}

Solution
(i) R = {(x, y) : 3x – y = 0} on A = {1, 2, 3, ..., 13, 14}

• Reflexive: For (a, a) ∈ R, 3a – a = 2a = 0, so a = 0. Since 0 ∉ A, not reflexive.
• Symmetric: If (a, b) ∈ R, then 3a – b = 0, so b = 3a. For (b, a) ∈ R, 3b – a = 0, so a = 3b. Substituting b = 3a, we get a = 3(3a) = 9a, so 8a = 0, implying a = 0, which is not in A. Not symmetric.
• Transitive: If (a, b) ∈ R and (b, c) ∈ R, then b = 3a and c = 3b = 3(3a) = 9a. For (a, c) ∈ R, c = 3a. Since 9a ≠ 3a (unless a = 0, not in A), not transitive.
• Conclusion: Neither reflexive, symmetric, nor transitive.

(ii) R = {(x, y) : y = x + 5 and x < 4} on N

• Pairs: For x = 1, 2, 3 (since x < 4 and x ∈ N), y = x + 5 gives R = {(1, 6), (2, 7), (3, 8)}.
• Reflexive: For (a, a) ∈ R, a = a + 5, which is impossible. Not reflexive.
• Symmetric: If (a, b) ∈ R, b = a + 5. For (b, a) ∈ R, a = b + 5 and b < 4. Since b = a + 5, a = (a + 5) + 5 = a + 10, impossible. Not symmetric.
• Transitive: If (a, b) ∈ R and (b, c) ∈ R, b = a + 5, c = b + 5 = a + 10, and b < 4. For (a, c) ∈ R, c = a + 5. Since a + 10 ≠ a + 5, not transitive.
• Conclusion: Neither reflexive, symmetric, nor transitive.

(iii) R = {(x, y) : y is divisible by x} on A = {1, 2, 3, 4, 5, 6}

• Reflexive: For (a, a) ∈ R, a ÷ a = 1 (integer), so reflexive.
• Symmetric: If (a, b) ∈ R, b = ka. For (b, a) ∈ R, a = mb. Then b = ka and a = mb imply b = k(mb) = mkb, so mk = 1. Thus, m = ±1, k = ±1. If k = 1, b = a; if k = –1, b = –a, not in A. Not symmetric (e.g., (2, 4) ∈ R, but (4, 2) ∉ R).
• Transitive: If (a, b) ∈ R and (b, c) ∈ R, b = ka, c = mb. Then c = m(ka) = (mk)a, so (a, c) ∈ R. Transitive.
• Conclusion: Reflexive and transitive, not symmetric.

(iv) R = {(x, y) : x – y is an integer} on Z

• Reflexive: For (a, a) ∈ R, a – a = 0, an integer. Reflexive.
• Symmetric: If (a, b) ∈ R, a – b is an integer. Then b – a = –(a – b), also an integer. Symmetric.
• Transitive: If (a, b) ∈ R and (b, c) ∈ R, a – b and b – c are integers. Then a – c = (a – b) + (b – c), an integer. Transitive.
• Conclusion: Reflexive, symmetric, transitive (equivalence relation).

(v) Relations on A (human beings in a town)
(a) R = {(x, y) : x and y work at the same place}

• Reflexive: x works at the same place as x. Reflexive.
• Symmetric: If x and y work at the same place, y and x do too. Symmetric.
• Transitive: If x and y work at the same place, and y and z do, then x and z do. Transitive.
• Conclusion: Equivalence relation.

(b) R = {(x, y) : x and y live in the same locality}

• Reflexive: x lives in the same locality as x. Reflexive.
• Symmetric: If x and y live in the same locality, y and x do. Symmetric.
• Transitive: If x and y live in the same locality, and y and z do, then x and z do. Transitive.
• Conclusion: Equivalence relation.

(c) R = {(x, y) : x is exactly 7 cm taller than y}

• Reflexive: x cannot be 7 cm taller than x. Not reflexive.
• Symmetric: If x is 7 cm taller than y, y cannot be 7 cm taller than x. Not symmetric.
• Transitive: If x is 7 cm taller than y, and y is 7 cm taller than z, x is 14 cm taller than z. Not transitive.
• Conclusion: Neither reflexive, symmetric, nor transitive.

(d) R = {(x, y) : x is wife of y}

• Reflexive: x cannot be the wife of x. Not reflexive.
• Symmetric: If x is the wife of y, y (male) cannot be the wife of x (female). Not symmetric.
• Transitive: If x is the wife of y, and y is the wife of z, contradiction (y cannot be both male and female). Not transitive.
• Conclusion: Neither reflexive, symmetric, nor transitive.

(e) R = {(x, y) : x is father of y}

• Reflexive: x cannot be the father of x. Not reflexive.
• Symmetric: If x is the father of y, y cannot be the father of x. Not symmetric.
• Transitive: If x is the father of y, and y is the father of z, x is the father of z. Transitive.
• Conclusion: Transitive, not reflexive or symmetric.

Question 2

Show that the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b}, is reflexive and transitive but not symmetric.

Solution

• Reflexive: For (a, a) ∈ R, a ≤ a is true. Reflexive.
• Symmetric: If (a, b) ∈ R, a ≤ b. For (b, a) ∈ R, b ≤ a. This is not always true (e.g., if a = 2, b = 3, 2 ≤ 3, but 3 ≰ 2). Not symmetric.
• Transitive: If (a, b) ∈ R and (b, c) ∈ R, then a ≤ b and b ≤ c, so a ≤ c. Thus, (a, c) ∈ R. Transitive.
• Conclusion: Reflexive and transitive, not symmetric.

Question 3

Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric, or transitive.

Solution

• Pairs: R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}.
• Reflexive: For (a, a) ∈ R, a = a + 1, impossible. Not reflexive.
• Symmetric: If (a, b) ∈ R, b = a + 1. For (b, a) ∈ R, a = b + 1, so a = (a + 1) + 1, impossible. Not symmetric.
• Transitive: If (a, b) ∈ R and (b, c) ∈ R, b = a + 1, c = b + 1 = (a + 1) + 1 = a + 2. For (a, c) ∈ R, c = a + 1, but c = a + 2. Not transitive (e.g., (1, 2) ∈ R, (2, 3) ∈ R, but (1, 3) ∉ R).
• Conclusion: Neither reflexive, symmetric, nor transitive.

Question 4

Show that the relation R in the set A of positive integers given by R = {(a, b) : a divides b} is reflexive and transitive but not symmetric.

Solution

• Reflexive: For (a, a) ∈ R, a divides a (a = 1·a). Reflexive.
• Symmetric: If (a, b) ∈ R, a divides b, so b = ka. For (b, a) ∈ R, b divides a, so a = mb. Then b = ka and a = mb imply b = k(mb) = mkb, so mk = 1. Thus, m = ±1, k = ±1, and b = a (since a, b are positive). Not symmetric (e.g., 2 divides 4, but 4 does not divide 2).
• Transitive: If (a, b) ∈ R and (b, c) ∈ R, a divides b (b = ka), and b divides c (c = mb). Then c = m(ka) = (mk)a, so a divides c. Transitive.
• Conclusion: Reflexive and transitive, not symmetric.

Question 5

Check whether the relation R in the set R of real numbers, defined as R = {(a, b) : a ≤ b²}, is reflexive, symmetric, or transitive.

Solution

• Reflexive: For (a, a) ∈ R, a ≤ a². For a ≥ 0, a² ≥ a (e.g., 2² = 4 ≥ 2); for a < 0, a² ≥ 0 > a (e.g., (–2)² = 4 ≥ –2); for a = 0, 0 ≤ 0. Reflexive.
• Symmetric: If (a, b) ∈ R, a ≤ b². For (b, a) ∈ R, b ≤ a². This is not always true (e.g., a = –2, b = 1: –2 ≤ 1² = 1 is true, but 1 ≰ (–2)² = 4). Not symmetric.
• Transitive: If (a, b) ∈ R and (b, c) ∈ R, a ≤ b² and b ≤ c². For (a, c) ∈ R, a ≤ c². This is not always true (e.g., a = –2, b = 0, c = 1: –2 ≤ 0² = 0, 0 ≤ 1² = 1, but –2 ≰ 1² = 1). Not transitive.
• Conclusion: Reflexive, not symmetric or transitive.

Question 6

Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.

Solution (Ref: PAGE6, Question 6)

• Reflexive: For (1, 1) ∈ R, not present in R. Similarly, (2, 2), (3, 3) ∉ R. Not reflexive.
• Symmetric: If (1, 2) ∈ R, then (2, 1) ∈ R. Symmetric.
• Transitive: If (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R. For (1, 2) ∈ R, (2, 1) ∈ R, but (1, 1) ∉ R. Not transitive.
• Conclusion: Symmetric, not reflexive or transitive.

Question 7

Show that the relation R in the set A of all the books in a library of a college, given by R = {(x, y) : x and y have the same number of pages}, is an equivalence relation.

Solution (Ref: PAGE6, Question 7)

• Reflexive: For (x, x) ∈ R, x has the same number of pages as x. Reflexive.
• Symmetric: If (x, y) ∈ R, x and y have the same number of pages, so y and x do. Symmetric.
• Transitive: If (x, y) ∈ R and (y, z) ∈ R, x and y have the same number of pages, and y and z do, so x and z do. Transitive.
• Conclusion: Equivalence relation.

Question 8

Show that the relation R in the set A = {1, 2, 3, 4, 5} given by R = {(a, b) : |a – b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.

Solution (Ref: PAGE6, Question 8)

• Reflexive: |a – a| = 0, even. Reflexive.
• Symmetric: If |a – b| is even, then |b – a| = |a – b| is even. Symmetric.
• Transitive: If |a – b| is even and |b – c| is even, then a – b = 2k, b – c = 2m. So, a – c = (a – b) + (b – c) = 2k + 2m = 2(k + m), even. Transitive.
• Equivalence: Reflexive, symmetric, transitive.
• Subsets:
  • For {1, 3, 5}: |1 – 3| = 2, |3 – 5| = 2, |1 – 5| = 4, all even. Related.
  • For {2, 4}: |2 – 4| = 2, even. Related.
  • Between {1, 3, 5} and {2, 4}: e.g., |1 – 2| = 1, |3 – 4| = 1, odd. Not related.
• Conclusion: Equivalence relation; {1, 3, 5} and {2, 4} form equivalence classes.

Question 9

Show that each of the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by
(i) R = {(a, b) : |a – b| is a multiple of 4}
(ii) R = {(a, b) : a = b}
is an equivalence relation. Find the set of all elements related to 1 in each case.

Solution (Ref: PAGE6, Question 9)
(i) R = {(a, b) : |a – b| is a multiple of 4}

• Reflexive: |a – a| = 0 = 4·0, reflexive.
• Symmetric: If |a – b| = 4k, then |b – a| = |a – b| = 4k. Symmetric.
• Transitive: If |a – b| = 4k and |b – c| = 4m, then a – b = ±4k, b – c = ±4m. Thus, a – c = (a – b) + (b – c) = ±4k ± 4m = 4(±k ± m), so |a – c| is a multiple of 4. Transitive.
• Equivalence: Yes.
• Elements related to 1: |x – 1| = 4k, x = 1 + 4k, 0 ≤ x ≤ 12. For k = 0, x = 1; k = 1, x = 5; k = 2, x = 9; k = 3, x = 13 (not in A). So, {1, 5, 9}.

(ii) R = {(a, b) : a = b}

• Reflexive: a = a, reflexive.
• Symmetric: If a = b, then b = a. Symmetric.
• Transitive: If a = b and b = c, then a = c. Transitive.
• Equivalence: Yes.
• Elements related to 1: a = 1, so {1}.

Question 10

Give an example of a relation which is
(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive.

Solution
On A = {1, 2, 3}:
(i) Symmetric, not reflexive or transitive: R = {(1, 2), (2, 1)}.

• Symmetric: (1, 2) implies (2, 1).
• Not reflexive: (1, 1) ∉ R.
• Not transitive: (1, 2), (2, 1) ∈ R, but (1, 1) ∉ R.

(ii) Transitive, not reflexive or symmetric: R = {(1, 2), (2, 3), (1, 3)}.

• Transitive: (1, 2), (2, 3) imply (1, 3).
• Not reflexive: (1, 1) ∉ R.
• Not symmetric: (1, 2) ∈ R, but (2, 1) ∉ R.

(iii) Reflexive, symmetric, not transitive: R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}.

• Reflexive: (1, 1), (2, 2), (3, 3) ∈ R.
• Symmetric: (1, 2) implies (2, 1), (2, 3) implies (3, 2).
• Not transitive: (1, 2), (2, 3) ∈ R, but (1, 3) ∉ R.

(iv) Reflexive, transitive, not symmetric: R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}.

• Reflexive: (1, 1), (2, 2), (3, 3) ∈ R.
• Transitive: (1, 2), (2, 3) imply (1, 3).
• Not symmetric: (1, 2) ∈ R, but (2, 1) ∉ R.

(v) Symmetric, transitive, not reflexive: R = {(1, 2), (2, 1), (1, 1), (2, 2)}.

• Symmetric: (1, 2) implies (2, 1).
• Transitive: (1, 2), (2, 1) imply (1, 1); (2, 1), (1, 2) imply (2, 2).
• Not reflexive: (3, 3) ∉ R.

Question 11

Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.

Solution

• Reflexive: Distance from P to origin = distance from P to origin. Reflexive.
• Symmetric: If distance from P to origin = distance from Q to origin, then distance from Q to origin = distance from P to origin. Symmetric.
• Transitive: If distance from P to origin = distance from Q to origin, and distance from Q to origin = distance from R to origin, then distance from P to origin = distance from R to origin. Transitive.
• Equivalence: Yes.
• Set related to P ≠ (0, 0): If distance from P to origin is r, then all points Q with distance r from origin lie on the circle with center (0, 0) and radius r, passing through P.

Question 12

Show that the relation R defined in the set A of all triangles as R = {(T₁, T₂) : T₁ is similar to T₂}, is an equivalence relation. Consider three right angle triangles T₁ with sides 3, 4, 5, T₂ with sides 5, 12, 13, and T₃ with sides 6, 8, 10. Which triangles among T₁, T₂, and T₃ are related?

Solution (Ref: PAGE3, Example 2, similar problem)

• Reflexive: T₁ is similar to T₁ (same angles and proportional sides). Reflexive.
• Symmetric: If T₁ is similar to T₂, then T₂ is similar to T₁. Symmetric.
• Transitive: If T₁ is similar to T₂ and T₂ is similar to T₃, then T₁ is similar to T₃. Transitive.
• Equivalence: Yes.
• Triangles: Check similarity (proportional sides):
  • T₁: 3, 4, 5; T₃: 6, 8, 10. Ratios: 6/3 = 2, 8/4 = 2, 10/5 = 2. T₁ and T₃ are similar.
  • T₂: 5, 12, 13. Ratios with T₁: 5/3, 12/4 = 3, 13/5 ≠ 3. Not similar.
  • T₂ and T₃: 5/6, 12/8 = 1.5, 13/10 ≠ 1.5. Not similar.
• Conclusion: T₁ and T₃ are related.

Question 13

Show that the relation R defined in the set A of all polygons as R = {(P₁, P₂) : P₁ and P₂ have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4, and 5?

Solution

• Reflexive: P₁ has the same number of sides as P₁. Reflexive.
• Symmetric: If P₁ has the same number of sides as P₂, then P₂ has the same number as P₁. Symmetric.
• Transitive: If P₁ and P₂ have the same number of sides, and P₂ and P₃ do, then P₁ and P₃ do. Transitive.
• Equivalence: Yes.
• Set related to T: T has 3 sides, so the set is all triangles (polygons with 3 sides).

Question 14

Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L₁, L₂) : L₁ is parallel to L₂}. Show that R is an equivalence relation. Find the set of all elements related to the line y = 2x + 4.

Solution

• Reflexive: L₁ is parallel to L₁. Reflexive.
• Symmetric: If L₁ is parallel to L₂, then L₂ is parallel to L₁. Symmetric.
• Transitive: If L₁ is parallel to L₂ and L₂ is parallel to L₃, then L₁ is parallel to L₃. Transitive.
• Equivalence: Yes.
• Set related to y = 2x + 4: All lines parallel to y = 2x + 4 have slope 2, so {y = 2x + c, c ∈ R}.

Question 15

Let R be the relation in the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}. Choose the correct answer.
(A) R is reflexive and symmetric but not transitive.
(B) R is reflexive and transitive but not symmetric.
(C) R is symmetric and transitive but not reflexive.
(D) R is an equivalence relation.

Solution (Ref: PAGE7, Question 15)

• Reflexive: Check (1, 1), (2, 2), (3, 3), (4, 4). All present. Reflexive.
• Symmetric: If (a, b) ∈ R, check (b, a). For (1, 2), (2, 1) ∉ R; for (1, 3), (3, 1) ∉ R; for (3, 2), (2, 3) ∉ R. Not symmetric.
• Transitive: Check if (a, b), (b, c) ∈ R implies (a, c) ∈ R:
  • (1, 2), (2, 2) → (1, 2) ∈ R.
  • (1, 3), (3, 3) → (1, 3) ∈ R.
  • (1, 3), (3, 2) → (1, 2) ∈ R.
  • (3, 2), (2, 2) → (3, 2) ∈ R. Transitive.
• Conclusion: Reflexive and transitive, not symmetric. Answer: (B).

Question 16

Let R be the relation in the set N given by R = {(a, b) : a = b – 2, b > 6}. Choose the correct answer.
(A) (2, 4) ∈ R
(B) (3, 8) ∈ R
(C) (6, 8) ∈ R
(D) (8, 7) ∈ R

Solution

• For (a, b) ∈ R, a = b – 2 and b > 6.
• (A) (2, 4): 4 ≯ 6. False.
• (B) (3, 8): 3 = 8 – 2, 8 > 6. True.
• (C) (6, 8): 6 = 8 – 2, 8 > 6. True.
• (D) (8, 7): 8 ≠ 7 – 2, 7 > 6. False.
• Conclusion: Both (B) and (C) are correct, but typically one option is chosen. Since (6, 8) is a clear example, (C) is likely intended.

Notes:

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