Class 12-NCERT Solutions-Chapter-1-Relations and Functions-Ex 1.1 -

NCERT Solutions Class-12-Chapter-1-Relations and Functions

Excercise-1.1

Note: Understanding the properties of relations (Reflexive, Symmetric, and Transitive) is key to solving these problems.

Determine whether each of the following relations are reflexive, symmetric and transitive:

Relation \( R \) in the set \( A = \{1, 2, 3, \dots, 13, 14\} \) defined as \( R = \{(x, y) : 3x - y = 0\} \)
Relation \( R \) in the set \( \mathbf{N} \) of natural numbers defined as \( R = \{(x, y) : y = x + 5 \text{ and } x < 4\} \)
Relation \( R \) in the set \( A = \{1, 2, 3, 4, 5, 6\} \) as \( R = \{(x, y) : y \text{ is divisible by } x\} \)
Relation \( R \) in the set \( \mathbf{Z} \) of all integers defined as \( R = \{(x, y) : x - y \text{ is an integer}\} \)
Relation \( R \) in the set \( A \) of human beings in a town at a particular time given by:
(a) \( R = \{(x, y) : x \text{ and } y \text{ work at the same place}\} \)
(b) \( R = \{(x, y) : x \text{ and } y \text{ live in the same locality}\} \)
(c) \( R = \{(x, y) : x \text{ is exactly 7 cm taller than } y\} \)
(d) \( R = \{(x, y) : x \text{ is wife of } y\} \)
(e) \( R = \{(x, y) : x \text{ is father of } y\} \)

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Solution:

(i) \( A = \{1, 2, \dots, 14\} \) and \( 3x - y = 0 \Rightarrow y = 3x \).

The pairs are \( R = \{(1, 3), (2, 6), (3, 9), (4, 12)\} \).

Reflexive: Since \( (1, 1) \notin R \), it is not reflexive.
Symmetric: \( (1, 3) \in R \) but \( (3, 1) \notin R \). Not symmetric.
Transitive: \( (1, 3) \in R \) and \( (3, 9) \in R \), but \( (1, 9) \notin R \). Not transitive.

Therefore, R is neither reflexive, nor symmetric, nor transitive.

(ii) \( R = \{(x, y) : y = x + 5, x \in \mathbf{N}, x < 4\} \).

Possible values for \( x \) are 1, 2, 3. The pairs are \( R = \{(1, 6), (2, 7), (3, 8)\} \).

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. Since the premise for transitivity cannot be met, it is vacuously transitive.

(iii) \( A = \{1, 2, 3, 4, 5, 6\} \). \( y \) is divisible by \( x \).

Reflexive: Every number is divisible by itself. \( (x, x) \in R \). It is 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 \( x \). It is transitive.

(iv) \( x - y \) is an integer.

Reflexive: \( x - x = 0 \), which is an integer. It is reflexive.
Symmetric: If \( x - y \) is an integer, then \( -(x - y) = y - x \) is also an integer. It is symmetric.
Transitive: If \( x - y \) and \( y - z \) are integers, their sum \( (x - y) + (y - z) = x - z \) is also an integer. It is transitive.

Therefore, R is an equivalence relation.

(v)

(a) Work at same place: Reflexive, Symmetric, and Transitive (Equivalence).

(b) Live in same locality: Reflexive, Symmetric, and Transitive (Equivalence).

(c) Exactly 7 cm taller:
- Not Reflexive (x is not 7cm taller than x).
- Not Symmetric (If x is taller than y, y cannot be taller than x).
- Not Transitive (If x is 7cm taller than y, and y is 7cm taller than z, x is 14cm taller than z).

(d) Wife of: Not Reflexive, Not Symmetric (If x is wife of y, y is husband of x), Not Transitive.

(e) Father of: Not Reflexive, Not Symmetric, Not Transitive.

Show that the relation \( R \) in the set \( \mathbf{R} \) of real numbers, defined as \( R = \{(a, b) : a \leq b^2\} \) is neither reflexive nor symmetric nor transitive.

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Solution:

Reflexive: Let \( a = \frac{1}{2} \). \( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \). Since \( \frac{1}{2} > \frac{1}{4} \), \( (\frac{1}{2}, \frac{1}{2}) \notin R \). Not reflexive.
Symmetric: Let \( a = 1, b = 2 \). \( 1 \leq 2^2 \) (True). But \( 2 \leq 1^2 \) (False). Not symmetric.
Transitive: Let \( a = 2, b = -2, c = -1 \).
\( 2 \leq (-2)^2 \Rightarrow 2 \leq 4 \) (True).
\( -2 \leq (-1)^2 \Rightarrow -2 \leq 1 \) (True).
Check \( (a, c) \): \( 2 \leq (-1)^2 \Rightarrow 2 \leq 1 \) (False).
Not transitive.

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.

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Solution:

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

Reflexive: \( (1, 1) \notin R \). Not reflexive.
Symmetric: \( (1, 2) \in R \) but \( (2, 1) \notin R \). Not symmetric.
Transitive: \( (1, 2) \in R \) and \( (2, 3) \in R \), but \( (1, 3) \notin R \). Not transitive.

Therefore, R is neither reflexive, nor symmetric, nor transitive.

Show that the relation \( R \) in \( \mathbf{R} \) defined as \( R = \{(a, b) : a \leq b\} \), is reflexive and transitive but not symmetric.

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Solution:

Reflexive: \( a \leq a \) is always true. Reflexive.
Symmetric: \( 2 \leq 4 \) is true, but \( 4 \leq 2 \) is false. Not symmetric.
Transitive: If \( a \leq b \) and \( b \leq c \), then \( a \leq c \). Transitive.

Check whether the relation \( R \) in \( \mathbf{R} \) defined by \( R = \{(a, b) : a \leq b^3\} \) is reflexive, symmetric or transitive.

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Solution:

Reflexive: Take \( a = \frac{1}{2} \). \( \frac{1}{2} \leq (\frac{1}{2})^3 \Rightarrow 0.5 \leq 0.125 \) (False). Not reflexive.
Symmetric: \( 1 \leq 2^3 \) (True). \( 2 \leq 1^3 \) (False). Not symmetric.
Transitive: Take \( a = 10, b = 3, c = 2 \).
\( 10 \leq 3^3 \Rightarrow 10 \leq 27 \) (True).
\( 3 \leq 2^3 \Rightarrow 3 \leq 8 \) (True).
Check \( (a, c) \): \( 10 \leq 2^3 \Rightarrow 10 \leq 8 \) (False).
Not transitive.

Therefore, R is neither reflexive, nor symmetric, nor transitive.

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.

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Solution:

Reflexive: \( (1, 1) \notin R \). Not reflexive.
Symmetric: \( (1, 2) \in R \) and \( (2, 1) \in R \). Symmetric.
Transitive: \( (1, 2) \in R \) and \( (2, 1) \in R \), but \( (1, 1) \notin R \). Not transitive.

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

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Solution:

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

Therefore, R is an equivalence relation.

Show that the relation \( R \) in the set \( A = \{1, 2, 3, 4, 5\} \) given by \( R = \{(a, b) : |a - b| \text{ is even}\} \), 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 \( \{2, 4\} \).

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Solution:

Reflexive: \( |a - a| = 0 \), which is even. Reflexive.
Symmetric: \( |a - b| = |b - a| \). If one is even, the other is even. Symmetric.
Transitive: If \( |a - b| \) is even (both a,b even or both odd) and \( |b - c| \) is even (both b,c even or both odd), then \( a \) and \( c \) have the same parity (both even or both odd), so \( |a - c| \) is even. Transitive.

Subsets:

Elements of \( \{1, 3, 5\} \) are all odd. The difference between any two odd numbers is even. Thus, they are related.

Elements of \( \{2, 4\} \) are all even. The difference between any two even numbers is even. Thus, they are related.

Difference between an odd number (from first set) and an even number (from second set) is always odd. Therefore, no element of the first set is related to the second.

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

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Solution:

(i) Multiple of 4:

It is reflexive (\( |a-a|=0 \)), symmetric, and transitive. Equivalence relation.

Elements related to 1: We need \( |1 - b| \) to be a multiple of 4.

\( |1 - b| = 0, 4, 8, 12 \dots \)

\( b = 1, 5, 9 \). (Since \( b \leq 12 \)).

Set related to 1 is \( \{1, 5, 9\} \).

(ii) a = b:

It is clearly an equivalence relation.

Elements related to 1: \( a = 1 \Rightarrow b = 1 \).

Set related to 1 is \( \{1\} \).

Q10

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.

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Solution:

(i) \( R = \{(1, 2), (2, 1)\} \) (See Q6).

(ii) \( R = \{(a, b) : a < b\} \) in real numbers.

(iii) \( R = \{(1, 2), (2, 1), (1, 1), (2, 2), (3, 3), (2, 3), (3, 2)\} \). Not transitive because \( (1, 2) \) and \( (2, 3) \) are there but \( (1, 3) \) is not.

(iv) \( R = \{(a, b) : a \leq b\} \) (See Q4).

(v) \( R = \{(1, 2), (2, 1), (1, 1)\} \). Symmetric and transitive (on this subset), but not reflexive for entire set if set has more elements like 2.

Q11

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

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Solution:

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

Reflexive: \( OP = OP \). True.
Symmetric: If \( OP = OQ \), then \( OQ = OP \). True.
Transitive: If \( OP = OQ \) and \( OQ = OS \), then \( OP = OS \). True.

Therefore, R is an equivalence relation.

The set of points related to P satisfy \( OP = k \) (constant). This is the definition of a circle with center O and radius \( k \).

Q12

Show that the relation \( R \) defined in the set \( A \) of all triangles as \( R = \{(T_1, T_2) : T_1 \text{ is similar to } T_2\} \), is equivalence relation. Consider three right angle triangles \( T_1 \) with sides 3, 4, 5, \( T_2 \) with sides 5, 12, 13 and \( T_3 \) with sides 6, 8, 10. Which triangles among \( T_1, T_2 \) and \( T_3 \) are related?

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Solution:

Similarity of triangles is Reflexive (every triangle is similar to itself), Symmetric, and Transitive. Hence, it is an equivalence relation.

Sides of \( T_1 \): 3, 4, 5.

Sides of \( T_2 \): 5, 12, 13.

Sides of \( T_3 \): 6, 8, 10.

Comparing \( T_1 \) and \( T_3 \): \( \frac{3}{6} = \frac{4}{8} = \frac{5}{10} = \frac{1}{2} \).

Since the corresponding sides are proportional, \( T_1 \) is related to \( T_3 \).

Q13

Show that the relation \( R \) defined in the set \( A \) of all polygons as \( R = \{(P_1, P_2) : P_1 \text{ and } P_2 \text{ have same number of sides}\} \), 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?

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Solution:

Same number of sides implies Reflexive, Symmetric, and Transitive. Thus, R is an equivalence relation.

The triangle \( T \) has 3 sides. The set of all elements related to \( T \) is the set of all polygons having 3 sides, i.e., the set of all triangles.

Q14

Let \( L \) be the set of all lines in XY plane and \( R \) be the relation in \( L \) defined as \( R = \{(L_1, L_2) : L_1 \text{ is parallel to } L_2\} \). Show that \( R \) is an equivalence relation. Find the set of all lines related to the line \( y = 2x + 4 \).

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Solution:

Reflexive: Every line is parallel to itself.
Symmetric: If \( L_1 \parallel L_2 \), then \( L_2 \parallel L_1 \).
Transitive: If \( L_1 \parallel L_2 \) and \( L_2 \parallel L_3 \), then \( L_1 \parallel L_3 \).

Therefore, R is an equivalence relation.

Lines related to \( y = 2x + 4 \) must be parallel to it, meaning they must have the same slope \( m = 2 \).

The set of such lines is \( y = 2x + k \), where \( k \) is any real number.

Q15

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.

R is reflexive and symmetric but not transitive.
R is reflexive and transitive but not symmetric.
R is symmetric and transitive but not reflexive.
R is an equivalence relation.

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Answer: (B)

Solution:

Reflexive: We have (1,1), (2,2), (3,3), (4,4). It is Reflexive.
Symmetric: We have (1,2) but (2,1) is not in R. It is Not Symmetric.
Transitive: We have (1,3) and (3,2). We also have (1,2). It satisfies transitivity for all pairs present. It is Transitive.

Therefore, R is reflexive and transitive but not symmetric.

Q16

Let \( R \) be the relation in the set \( \mathbf{N} \) given by \( R = \{(a, b) : a = b - 2, b > 6\} \). Choose the correct answer.

\( (2, 4) \in R \)
\( (3, 8) \in R \)
\( (6, 8) \in R \)
\( (8, 7) \in R \)

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Answer: (C)

Solution:

The condition is \( b > 6 \) and \( a = b - 2 \).

(A) \( b=4 \). Not greater than 6. Incorrect.
(B) \( b=8 \). \( a = 8-2 = 6 \). But given \( a=3 \). Incorrect.
(C) \( b=8 \). \( a = 8-2 = 6 \). Correct.
(D) \( b=7 \). \( a = 7-2 = 5 \). But given \( a=8 \). Incorrect.