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 *a ** *b *= *a *− *b*

(*ii*) On Z^{+}, define * by *a ** *b *= *ab*

(*iii*) On R, define * by *a ** *b *= *ab*^{2}

(*iv*) On Z^{+}, define * by *a ** *b *= |*a *− *b*|

(*v*) On Z^{+}, define * by *a ** *b *= *a*

**Solution 1:**

(*i*) On Z^{+}, * is defined by *a *** **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* **a *** **b* = *ab*

We can observe that,

For each *a*, *b* ∈ Z^{+}, there is a unique element *ab* in Z^{+}

Hence, * carries each pair (*a*, *b*) to a unique element *a *** **b *= *ab *in Z^{+}

Therefore, * is a binary operation

(*iii*) On R, * is defined by* **a *** **b* = *ab*^{2}

We can observe that,

For each *a*, *b* ∈ R, there is a unique element *ab*^{2} in R

Thus, * carries each pair (*a*, *b*) to a unique element *a *** **b *= *ab*^{2 }in R

Therefore, * is a binary operation.

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

We can observe that,

For each *a*, *b* ∈ Z^{+}, there is a unique element |*a − b*| in Z^{+}.

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

Therefore, * is a binary operation

(*v*) On Z^{+}, * is defined by* **a *** **b* = *a*

We can observe that,

Thus,* carries each pair (*a*, *b*) to a unique element *a *** **b *= *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 *a ** *b *= *a *− *b*

(*ii*) On Q, define *a ** *b *= *ab *+ 1

(*iii*) On Q, define *a ** *b *= *ab*/2

(*iv*) On Z^{+}, define *a ** *b *= 2^{ab}

(*v*) On Z^{+}, define *a ** *b *= *a ^{b}*

(*vi*) On R − {−1}, define *a ** *b *= *a* / (*b* + 1)

**Solution 2:**

(*i*) On Z, * is defined by* **a *** **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

⇒ *a *** **b* = *a *** **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*∨*b *= min {*a*, *b*}. Write the operation table of the operation∨.

**Solution 3:**

*a *∨_{ }*b* = min {*a*, *b*}

*a*, *b* ∈ {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 *a *** **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 *a **′ *b *= H.C.F. of *a *and *b*. Is the operation *′ same as the operation * defined in Exercise 4 above? Justify your answer.

**Solution 5:**

*a **′ *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* * *b *= L.C.M. of *a *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:**

*a *** **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* = *b *** **a*

Hence, the operation * is commutative.

(*iii*)

We have,

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

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

Thus, (*a *** **b*) * *c* = *a* * (*b *** **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*)

*a *** **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 *.

**Download NCERT Solutions for Class 12 Maths: Chapter 1 Relations and Functions in PDF format**