# RS Aggarwal Solutions for Class 12 Maths Chapter 1: Relations Exercise 1A

In this exercise, students will mainly get conceptual knowledge about relations. The problems help students understand the method of determining domain and range of a given set of relations. The solutions for the exercise wise problems are prepared by experts at BYJUâ€™S in the best possible way understandable by students. To boost exam preparation of students RS Aggarwal Solutions Class 12 Maths Chapter 1 Relations Exercise 1A PDF is available here for free download.

## RS Aggarwal Solutions for Class 12 Maths Chapter 1: Relations Exercise 1A Download PDF

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Exercise 1B Solutions

### Access RS Aggarwal Solutions for Class 12 Maths Chapter 1: Relations Exercise 1A

1. Find the domain and range of the relation

R = {(-1, 1), (1, 1), (-2, 4), (2, 4)}.

Solution:

The domain of the relation

dom (R) = {-1, 1, -2, 2}

The range of the relation

range (R) = {1, 4}

2. Let R = {(a, a3): a is a prime number less than 5}.

Find the range of R.

Solution:

By substituting a as 2 and 3 we get

R = {(2, 8), (3, 27)

We know that range of R

range (R) = {8, 27}

3. Let R = {(a, a3): a is a prime number less than 10}.

Find (i) R (ii) dom (R) (iii) range (R).

Solution:

(i) R = {(2, 8), (3, 27), (5, 125), (7, 343)}

(ii) dom (R) = {2, 3, 5, 7}

(iii) range (R) = {8, 27, 125, 343}

4. Let R = {x, y): x + 2y = 8} be a relation on N.

Write the range of R.

Solution:

By substituting the values of x and y we get

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

The range of R for the relation on N

range (R) = {3, 2, 1}

5. Let R = {(a, b): a, b âˆˆ N and a + 3b = 12}.

Find the domain and range of R.

Solution:

By substituting values of a and b we get

R = {(3, 3), (6, 2), (9, 1)}

So we get

dom (R) = {3, 6, 9}

range (R) = {3, 2, 1}

6. Let R = {(a, b): b = |a â€“ 1|, a âˆˆ Z and |a| < 3}.

Find the domain and range of R.

Solution:

We know that a is an integer where â€“ 3 < a < 3

So we get

R = {(-2, 3), (-1, 2), (0, 1), (1, 0), (2, 1)}

dom (R) = {-2, -1, 0, 1, 2}

range (R) = {3, 2, 1, 0}

7. Let R = {(a, 1/a): a âˆˆ N and 1 < a < 5}.

Find the domain and range of R.

Solution:

By substituting values we get

R = {(2, 1/2), (3, 1/3), (4, 1/4)}

So we get

dom (R) = {2, 3, 4}

range (R) = {1/2, 1/3, 1/4}

8. Let R = {(a, b): a, b âˆˆ N and b = a + 5, a < 4}.

Find the domain and range of R.

Solution:

By substituting values we get

R = {(1, 6), (2, 7), (3, 8)}

So we get

dom (R) = {1, 2, 3}

range (R) = {6, 7, 8}

9. Let S be the set of all sets and let R = {(A, B): A âŠ‚ B)}, i.e. A is a proper subset of B. Show that R is (i) transitive (ii) not reflexive (iii) not symmetric.

Solution:

(i) Transitive

Consider A, B and C âˆˆ S, where (A, B) and (B, C) âˆˆ R

We get

(A, B) âˆˆ R => A âŠ‚ B â€¦â€¦. (1)

(B, C) âˆˆ R => B âŠ‚ C â€¦â€¦.. (2)

Using both the equations we get

A âŠ‚ C => (A, C) âˆˆ R

Hence, R is a transitive relation S.

(ii) Non reflexive

We know that

A âŠ„ A where (A, A) âˆˆ R

Hence, R is non reflexive.

(iii) Non symmetric

Consider A âŠ‚ B where (A, B) âˆˆ R

We know that B âŠ„ A

So (B, A) âˆ‰ R

We get (A, B) âˆˆ R and (B, A) âˆ‰ R

Hence, R is non symmetric.

10. Let A be the set of all points in a plane and let O be the origin. Show that the relation R = {(P, Q): P, Q âˆˆ A and OP = OQ} is an equivalence relation.

Solution:

Consider O as the origin of the plane

So R = {(P, Q): OP = OQ)

By considering properties of relation R

Symmetric:

Consider P and Q as the two points in set A where (P, Q) âˆˆ R

We can write it as

OP = OQ where (Q, P) âˆˆ R

So we get (P, Q) âˆˆ R and (Q, P) âˆˆ R for P, Q âˆˆ A

Hence, R is symmetric.

Reflexivity:

Consider P as any point in set A where OP = OP

We know that (P, P) âˆˆ R for all P âˆˆ A

Hence, R is reflexive.

Transitivity:

Consider P, Q and S as three points in a set A where (P, Q) âˆˆ R and (Q, S) âˆˆ R

We know that OP = OQ and OQ = OS

So we get OP = OS where (P, S) âˆˆ R

Hence, R is transitive

Therefore, R is an equivalence relation.

Consider P as a fixed point in set A and let Q be a point in set A where (P, Q) âˆˆ R

We know that OP = OQ where Q moves in the plane that its distance from O.

So we get O (0, 0) = OP

So the locus of Q is a circle having centre at O and OP as the radius

Therefore, the set of all points which is related to P passes through the point P having O as centre.

11. On the set S of all real numbers, define a relation R = {(a, b): a â‰¤ b}.

Show that R is (i) reflexive, (ii) transitive (iii) not symmetric.

Solution:

(i) Reflexivity

Consider a as an arbitrary element on the set S

So we get a â‰¤ a where (a, a) âˆˆ R

Hence, R is reflective.

(ii) Transitivity

Consider a, b and c âˆˆ S where (a, b) and (b, c) âˆˆ S

We get

(a, b) âˆˆ R => a â‰¤ b and (b, c) âˆˆ R => b â‰¤ c

Based on the above equation we get

(a, c) âˆˆ R => a â‰¤ c

Hence, R is transitive.

(iii) Non symmetry

We know that

(5, 6) âˆˆ R => 5 â‰¤ 6

In the same way

(6, 5) âˆˆ R => 6 â‰° 5

Hence, R is non symmetric.

12. Let A = {1, 2, 3, 4, 5, 6} and let R = {(a, b): a, b âˆˆ A and b = a + 1}.

Show that R is (i) not reflexive, (ii) not symmetric and (iii) not transitive.

Solution:

By using roster form

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

(i) Non reflexive

We know that

(1, 1) âˆˆ R where 1 âˆˆ A

Hence, R is non reflexive.

(ii) Non symmetric

We know that

(1, 2) âˆˆ R but (2, 1) âˆ‰ R

Hence, R is non symmetric.

(iii) Non transitive

We know that

(1, 2) âˆˆ R and (2, 3) âˆˆ R

In the same way (1, 3) âˆ‰ R

Hence, R is non transitive.