# NCERT Solutions for CBSE Class 12 Mathematics ‒ Chapter 1: Relations and Functions (Part VI)

NCERT Solutions for CBSE Class 12 Mathematics, Chapter 1: Relations & Functions are available in this article. Here, you will get solutions to the questions of exercise 1.4 from question number 1 to question number 7. Most of the questions given in this exercise are related to binary operations. These questions are important CBSE Class 12 Maths board exam.

NCERT Solutions for CBSE Class 12th Maths, Chapter 1: Relations and Functions (Exercise 1.4) are given below

Question 1:

Determine whether or not each of the definition of given below gives a binary operation. In the event that * is not a binary operation, give justification for this.

(i) On Z+, define * by − b

(ii) On Z+, define * by ab

(iii) On R, define * by ab2

(iv) On Z+, define * by = |− b|

(v) On Z+, define * by a

Solution 1:

(i) On Z+, * is defined by * b = a − b

The image of (1, 2) under * is 1 * 2 = 1 − 2 = −1 ∉ Z+

Thus, it is not a binary operation

(ii) On Z+, * is defined by * b = ab

We can observe that,

For each ab ∈ Z+, there is a unique element ab in Z+

Hence, * carries each pair (ab) to a unique element * ab in Z+

Therefore, * is a binary operation

(iii) On R, * is defined by * b = ab2

We can observe that,

For each ab ∈ R, there is a unique element ab2 in R

Thus, * carries each pair (ab) to a unique element * abin R

Therefore, * is a binary operation.

(iv) On Z+, * is defined by * b = |a − b|

We can observe that,

For each ab ∈ Z+, there is a unique element |a − b| in Z+.

Thus, * carries each pair (ab) to a unique element * = |a − b| in Z+

Therefore, * is a binary operation

(v) On Z+, * is defined by * b = a

We can observe that,

Thus,* carries each pair (ab) to a unique element * a in Z+

Therefore, * is a binary operation

NCERT Exemplar Class 12 Mathematics – Chapter 1 Relations and Functions

Question 2:

For each binary operation * defined below, determine whether * is commutative or associative.

(i) On Z, define − b

(ii) On Q, define ab + 1

(iii) On Q, define ab/2

(iv) On Z+, define = 2ab

(v) On Z+, define ab

(vi) On R − {−1}, define a / (b + 1)

Solution 2:

(i) On Z, * is defined by * b = a − b

We can observe that,

1 * 2 = 1 − 2 = 1

2 * 1 = 2 − 1 = 1

∴1 * 2 ≠ 2 * 1; where 1, 2 ∈ Z

Thus, the operation * is not commutative.

Also,

(1 * 2) * 3 = (1 − 2) * 3 = −1 * 3 = −1 − 3 = −4

1 * (2 * 3) = 1 * (2 − 3) = 1 * −1 = 1 − (−1) = 2

∴ (1 * 2) * 3 ≠ 1 * (2 * 3); where 1, 2, 3 ∈ Z

Thus, the operation * is not associative.

(ii) On Q, * is defined by a * b = ab + 1

We know that,

ab = ba

ab + 1 = ba + 1

⇒ * b = * b

Hence, the operation * is commutative.

We can observe that,

(1 * 2) * 3 = (1 × 2 + 1) * 3 = 3 * 3 = 3 × 3 + 1 = 10

1 * (2 * 3) = 1 * (2 × 3 + 1) = 1 * 7 = 1 × 7 + 1 = 8

∴ (1 * 2) * 3 ≠ 1 * (2 * 3); where 1, 2, 3 ∈ Q

Hence, the operation * is not associative.

(iii)

(iv)

(v)

(vi)

Question 3:

Consider the binary operation ∨ on the set {1, 2, 3, 4, 5} defined by a= min {ab}. Write the operation table of the operation∨.

Solution 3:

b = min {ab}

ab ∈ {1, 2, 3, 4, 5}.

Hence, the operation table for the given operation ∨is:

 ∨ 1 2 3 4 5 1 1 1 1 1 1 2 1 2 2 2 2 3 1 2 3 3 3 4 1 2 3 4 4 5 1 2 3 4 5

Question 4:

Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table.

(i) Compute (2 * 3) * 4 and 2 * (3 * 4)

(ii) Is * commutative?

(iii) Compute (2 * 3) * (4 * 5).

(Hint: use the following table)

 * 1 2 3 4 5 1 1 1 1 1 1 2 1 2 1 2 1 3 1 1 3 1 1 4 1 2 1 4 1 5 1 1 1 1 5

Solution 4:

(i)

2 * (3 * 4) = 2 * 1 = 1

(2 * 3) * 4 = 1 * 4 = 1

(ii)

For every a, b ∈ {1, 2, 3, 4, 5}, we have * b = b * a

Thus, the operation * is commutative.

(iii)

(2 * 3) = 1

(4 * 5) = 1

(2 * 3) * (4 * 5) = 1 * 1 = 1

Question 5:

Let *′ be the binary operation on the set {1, 2, 3, 4, 5} defined by *′ = H.C.F. of and b. Is the operation *′ same as the operation * defined in Exercise 4 above? Justify your answer.

Solution 5:

*′ b = H.C.F of a and b

The operation table for the operation *′ can be represented as:

 *′ 1 2 3 4 5 1 1 1 1 1 1 2 1 2 1 2 1 3 1 1 3 1 1 4 1 2 1 4 1 5 1 1 1 1 5

We can see that the operation tables for the operations * and *′ are the same.

Hence, the operation *′ is same as the operation*.

Question 6:

Let * be the binary operation on N given by a * = L.C.M. of and b. Find

(i) 5 * 7, 20 * 16

(ii) Is * commutative?

(iii) Is * associative?

(iv) Find the identity of * in N

(v) Which elements of N are invertible for the operation *?

Solution 6:

* b = L.C.M. of a and b

(i)

5 * 7 = L.C.M. of 5 and 7

= 35

20 * 16 = L.C.M of 20 and 16

= 80

(ii)

We known that,

L.C.M of a and b = L.C.M of b and a

Thus, a * b = * a

Hence, the operation * is commutative.

(iii)

We have,

a * (b * c) = a * (LCM of b and c) = L.C.M of ab, and c

(* b) * = (L.C.M of a and b) * c = LCM of ab, and c

Thus, (* b) * c = a * (* c)

Hence, the operation * is associative.

(iv)

L.C.M. 1 and a = a = L.C.M. of a and 1

1 * a = a = a * 1

Thus, 1 is the identity of * in N

(v)

* b = e = b * a

Here, e = 1

L.C.M of a and b = 1 = L.C.M of b and a

This is possible only when a and b are equal to 1.

Thus, 1 is the only invertible element of N with respect to the operation *.

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