NCERT Solutions

Given: \(3x – y = 0 \implies y = 3x\).
The relation R is: \(R = \{(1, 3), (2, 6), (3, 9), (4, 12)\}\).

• Reflexive: \((1, 1) \notin R\) because \(3(1) – 1 \neq 0\). Not reflexive.
• Symmetric: \((1, 3) \in R\) but \((3, 1) \notin R\) because \(3(3) – 1 \neq 0\). Not symmetric.
• Transitive: \((1, 3) \in R\) and \((3, 9) \in R\). However, \((1, 9) \notin R\) because \(3(1) – 9 \neq 0\). Not transitive.

Since \(x \in N\) and \(x < 4\), possible values for \(x\) are 1, 2, 3.
\(R = \{(1, 6), (2, 7), (3, 8)\}\).

• Reflexive: \((1, 1) \notin R\). Not reflexive.
• Symmetric: \((1, 6) \in R\) but \((6, 1) \notin R\). Not symmetric.
• Transitive: There are no pairs \((x, y)\) and \((y, z)\) in R. The condition for transitivity is not contradicted. Transitive.

• Reflexive: Every number is divisible by itself. \((x, x) \in R\). Reflexive.
• Symmetric: \((1, 2) \in R\) (2 is divisible by 1), but \((2, 1) \notin R\) (1 is not divisible by 2). Not symmetric.
• Transitive: If \(y\) is divisible by \(x\) and \(z\) is divisible by \(y\), then \(z\) is divisible by \(x\). Transitive.

• Reflexive: \(x – x = 0\), which is an integer. Reflexive.
• Symmetric: If \(x – y\) is an integer, then \(y – x = -(x – y)\) is also an integer. Symmetric.
• Transitive: If \(x – y\) is an integer and \(y – z\) is an integer, their sum \((x – y) + (y – z) = x – z\) is also an integer. Transitive.

• (a) Work at same place: Reflexive, Symmetric, Transitive. Equivalence.
• (b) Live in same locality: Reflexive, Symmetric, Transitive. Equivalence.
• (c) x is 7cm taller than y: Neither Reflexive, nor Symmetric, nor Transitive (x taller than y, y taller than z implies x is 14cm taller than z).
• (d) x is wife of y: Neither Reflexive, nor Symmetric, nor Transitive.
• (e) x is father of y: Neither Reflexive, nor Symmetric, nor Transitive.

• Reflexive: Take \(a = \frac{1}{2}\). \(\frac{1}{2} \le (\frac{1}{2})^2 \Rightarrow 0.5 \le 0.25\) (False). Not reflexive.
• Symmetric: Take \((1, 4)\). \(1 \le 16\) (True). But \((4, 1)\): \(4 \le 1^2\) (False). Not symmetric.
• Transitive: Take \(a = 3, b = 2, c = 1.5\).
  \((3, 2) \in R\) as \(3 \le 4\).
  \((2, 1.5) \in R\) as \(2 \le 2.25\).
  Check \((3, 1.5)\): \(3 \le (1.5)^2 \Rightarrow 3 \le 2.25\) (False). Not transitive.

\(R = \{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)\}\).

• Reflexive: \((1, 1) \notin R\). No.
• Symmetric: \((1, 2) \in R\) but \((2, 1) \notin R\). No.
• Transitive: \((1, 2) \in R\) and \((2, 3) \in R\), but \((1, 3) \notin R\). No.

• Reflexive: \(a \le a\) is always true. Reflexive.
• Transitive: If \(a \le b\) and \(b \le c\), then \(a \le c\). Transitive.
• Symmetric: \((2, 4) \in R\) (\(2 \le 4\)). But \((4, 2) \notin R\) (\(4 \not\le 2\)). Not symmetric.

• Reflexive: Take \(a = \frac{1}{2}\). \(\frac{1}{2} \not\le (\frac{1}{2})^3\) (\(0.5 \not\le 0.125\)). Not reflexive.
• Symmetric: \((1, 2) \in R\) (\(1 \le 8\)). But \((2, 1) \notin R\) (\(2 \not\le 1\)). Not symmetric.
• Transitive: Take \(a = 10, b = 3, c = 2\).
  \((10, 3) \in R\) (\(10 \le 27\)).
  \((3, 2) \in R\) (\(3 \le 8\)).
  Check \((10, 2)\): \(10 \le 2^3 \Rightarrow 10 \le 8\) (False). Not transitive.

• Reflexive: \((1, 1), (2, 2), (3, 3) \notin R\). No.
• Symmetric: \((1, 2) \in R\) and \((2, 1) \in R\). Yes.
• Transitive: \((1, 2) \in R\) and \((2, 1) \in R\), but \((1, 1) \notin R\). No.

• Reflexive: A book has same pages as itself. \((x, x) \in R\).
• Symmetric: If \(x\) has same pages as \(y\), then \(y\) has same pages as \(x\).
• Transitive: If \(x\) has same pages as \(y\), and \(y\) as \(z\), then \(x\) has same pages as \(z\).

• Reflexive: \(|a – a| = 0\) (even). Yes.
• Symmetric: \(|a – b| = |b – a|\). If one is even, so is the other. Yes.
• Transitive: If \(|a-b|\) is even and \(|b-c|\) is even, then \(a-b\) and \(b-c\) are both even or both odd. Their sum \((a-c)\) is even. Yes.

{1, 3, 5}: All are odd. Difference of any two is even. Related.
{2, 4}: All are even. Difference of any two is even. Related.
Cross Relation: Odd – Even = Odd. So {1, 3, 5} cannot relate to {2, 4}.

• (i): We need \(x \in A\) such that \(|x – 1| = 4k\).
  \(|x – 1| = 0 \implies x = 1\)
  \(|x – 1| = 4 \implies x = 5\)
  \(|x – 1| = 8 \implies x = 9\)
  Set: {1, 5, 9}
• (ii): We need \(x \in A\) such that \(x = 1\).
  Set: {1}

• (i) Symmetric only: Perpendicular lines.
• (ii) Transitive only: “Greater than” (\(a > b\)).
• (iii) Reflexive & Symmetric: “Friend of”.
• (iv) Reflexive & Transitive: “Is subset of”.
• (v) Symmetric & Transitive: Relation on \(\{1, 2, 3\}\): \(R = \{(1,1), (1,2), (2,1), (2,2)\}\).

Let \(O\) be the origin. Condition: \(OP = OQ\).

• Reflexive: \(OP = OP\). True.
• Symmetric: \(OP = OQ \implies OQ = OP\). True.
• Transitive: \(OP = OQ\) and \(OQ = OR \implies OP = OR\). True.

Set related to \(P \neq (0,0)\): All points \(Q\) such that \(OQ = OP\) (constant). This forms a circle passing through P with origin as centre.

Similarity implies corresponding sides are proportional.

• Check T1 and T3: \( \frac{6}{3} = 2, \frac{8}{4} = 2, \frac{10}{5} = 2 \). Ratios are equal.
• Therefore, T1 is related to T3.

The relation is defined for polygons having the same number of sides. Since T is a triangle (3 sides), the set of related elements is the set of all triangles in A.

Parallel lines have the same slope. The given line has slope \(m = 2\).
The set of all related lines is \(y = 2x + c\), where \(c\) is any real constant.

• Reflexive: \((1,1), (2,2), (3,3), (4,4)\) are present. Yes.
• Symmetric: \((1, 2) \in R\) but \((2, 1) \notin R\). No.
• Transitive: We have \((1,3)\) and \((3,2)\). Is \((1,2)\) present? Yes. Transitivity holds. Yes.

We need \(b > 6\) and \(a = b – 2\).

• (A) \((2, 4)\): \(b=4\) (Not > 6). Incorrect.
• (B) \((3, 8)\): \(b=8\). \(a = 8-2 = 6\). Given \(a=3\). Incorrect.
• (C) \((6, 8)\): \(b=8\). \(a = 8-2 = 6\). Matches. Correct.
• (D) \((8, 7)\): \(b=7\). \(a = 7-2 = 5\). Given \(a=8\). Incorrect.