Looking for NCERT Solutions for Class 12 Maths Exercise 1.4? If yes then you are in the right place. Here we have provided you with the Question and Answers of Exercise 1.4. These solutions are written by expert teachers and are so accurate to rely on.

Chapter | 1. Relations and Functions |

Exercise | 1.4 |

Class | Twelve |

Subject | Maths |

Category | NCERT Solutions for Class 12 |

NCERT Solutions for Class 12 Maths are available online for students to use and refer to while studying for their exams. The solutions provide step-by-step explanations for each problem in the textbook, making it easy for students to understand the concepts. With Class 12 Maths NCERT Solutions, students can be sure that they are getting the most accurate and up-to-date information available. The solutions are regularly updated by experts in the field, so students can be confident that they are using the most current information.

## NCERT Solutions for Class 12 Maths Exercise 1.4

### Relations and Functions Solutions

**Q** **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**Answer)**

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

It is not a binary operation as the image of (1, 2) under * is 1 * 2 = 1 − 2

= −1 ∉ Z^{+}.

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

It is seen that for each a, b ∈ Z^{+}, there is a unique element ab in Z^{+}.

This means that * 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}.

It is seen that for each a, b ∈ R, there is a unique element ab^{2} in R.

This means that * 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|.

It is seen that for each a, b ∈ Z^{+}, there is a unique element |a − b| in Z^{+}.

This means that * 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.

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

Therefore, * is a binary operation.

**Q 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**

**Answer)**

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

It can be observed that 1 * 2 = 1 − 2 = 1 and 2 * 1 = 2 − 1 = 1.

∴1 * 2 ≠ 2 * 1; where 1, 2 ∈ **Z**

Hence, the operation * is not commutative.

Also we have:

(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**

Hence, the operation * is not associative.

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

It is known that:*ab* = *ba* &mnForE; *a, b* ∈ **Q**

⇒ *ab* + 1 = *ba *+ 1 &mnForE; *a, b* ∈ **Q**

⇒ *a *** b* = *a *** b* &mnForE; *a, b* ∈ **Q**

Therefore, the operation * is commutative.

It can be observed 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**

Therefore, the operation * is not associative.

**(iii) **On **Q**, * is defined by *a *** b* = ab/2

It is known that:*ab* = *ba* &mnForE; *a, b* ∈ **Q**

⇒ ab/2 = ba/2 &mn ForE; *a, b* ∈ **Q**

⇒ *a *** b* = *b *** a* &mnForE; *a, b* ∈ **Q**

Therefore, the operation * is commutative.

For all *a, b*,* c* ∈ **Q**, we have:

Therefore, the operation * is associative.

**(iv)** On **Z**^{+}, * is defined by *a *** b* = 2^{ab}.

It is known that:

*ab* = *ba* &mnForE; *a, b* ∈ **Z**^{+}

⇒ 2^{ab} = 2^{ba} &mnForE; *a, b* ∈ **Z**^{+}

⇒ *a *** b* = *b *** a* &mnForE; *a, b* ∈ **Z**^{+}

Therefore, the operation * is commutative.

It can be observed that:

∴(1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ **Z**^{+}

Therefore, the operation * is not associative.

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

It can be observed that:

1 x2 = 1^{2}=1 and 2x 1 =2^{1} = 2

∴ 1 * 2 ≠ 2 * 1 ; where 1, 2 ∈ **Z**^{+}

Therefore, the operation * is not commutative.

It can also be observed that:

∴(2 * 3) * 4 ≠ 2 * (3 * 4) ; where 2, 3, 4 ∈ **Z**^{+}

Therefore, the operation * is not associative.

**(vi)** On **R**, * − {−1} is defined by a.b = a/b+1

It can be observed that 1 x 2 = 1/2+1 = 1/2 and 2 x 1 = 2/1+1 = 2/2

∴1 * 2 ≠ 2 * 1 ; where 1, 2 ∈ **R **− {−1}

Therefore, the operation * is not commutative.

It can also be observed that:

∴ (1 * 2) * 3 ≠ 1 * (2 * 3) ; where 1, 2, 3 ∈ **R **− {−1}

Therefore, the operation * is not associative.

**Q 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∨.**

**Answer)** The binary operation ∨ on the set {1, 2, 3, 4, 5} is defined as a ∨ b = min {a, b} & mn For E; a, b ∈ {1, 2, 3, 4, 5}.

Thus, the operation table for the given operation ∨ can be given as:

V | 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 |

**Q** **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 |

**Answer)**

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

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

**(ii)** For every a, b ∈{1, 2, 3, 4, 5}, we have a * b = b * a. Therefore, the operation * is commutative.

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

∴(2 * 3) * (4 * 5) = 1 * 1 = 1

**Q 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.**

**Answer)** The binary operation *′ on the set {1, 2, 3 4, 5} is defined as a *′ b = H.C.F of a and b.

The operation table for the operation *′ can be given 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 observe that the operation tables for the operations * and *′ are the same.

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

**Q 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 *?**

**Answer) **The binary operation * on N is defined as 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)** It is known that:

L.C.M of a and b = L.C.M of b and a &mnForE; a, b ∈ N.

∴a * b = b * a

Thus, the operation * is commutative.

**(iii)** For a, b, c ∈ N, we have:

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

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

∴(a * b) * c = a * (b * c)

Thus, the operation * is associative.

**(iv)** It is known that:

L.C.M. of a and 1 = a = L.C.M. 1 and a &mnForE; a ∈ N

⇒ a * 1 = a = 1 * a &mnForE; a ∈ N

Thus, 1 is the identity of * in N.

**(v)** An element a in N is invertible with respect to the operation * if there exists an element b in N, such that a * b = e = b * a.

Here, e = 1

This means that:

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

This case 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 *.

**Q 7) Is * defined on the set {1, 2, 3, 4, 5} by a * b = L.C.M. of a and b a binary operation? Justify your answer.**

**Answer)**

The operation * on the set A = {1, 2, 3, 4, 5} is defined as

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

Then, the operation table for the given operation * can be given as:

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

1 | 1 | 2 | 3 | 4 | 5 |

2 | 2 | 2 | 6 | 4 | 10 |

3 | 3 | 6 | 3 | 12 | 15 |

4 | 4 | 4 | 12 | 4 | 20 |

5 | 5 | 10 | 15 | 20 | 5 |

It can be observed from the obtained table that:

3 * 2 = 2 * 3 = 6 ∉ A, 5 * 2 = 2 * 5 = 10 ∉ A, 3 * 4 = 4 * 3 = 12 ∉ A

3 * 5 = 5 * 3 = 15 ∉ A, 4 * 5 = 5 * 4 = 20 ∉ A

Hence, the given operation * is not a binary operation.

**Q 8) Let * be the binary operation on N defined by a * b = H.C.F. of a and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N?**

**Answer)**

The binary operation * on N is defined as:

a * b = H.C.F. of a and b

It is known that:

H.C.F. of a and b = H.C.F. of b and a &mnForE; a, b ∈ N.

∴a * b = b * a

Thus, the operation * is commutative.

For a, b, c ∈ N, we have:

(a * b)* c = (H.C.F. of a and b) * c = H.C.F. of a, b, and c

a *(b * c)= a *(H.C.F. of b and c) = H.C.F. of a, b, and c

∴(a * b) * c = a * (b * c)

Thus, the operation * is associative.

Now, an element e ∈ N will be the identity for the operation * if a * e = a = e* a ∈ a ∈ N.

But this relation is not true for any a ∈ N.

Thus, the operation * does not have any identity in N.

**Q 9) Let * be a binary operation on the set Q of rational numbers as follows:**

**(i) a * b = a − b****(ii) a * b = a ^{2} + b^{2}**

**(iii) a * b = a + ab**

**(iv) a * b = (a − b)**

^{2}**(v) a * b = ab/4 (vi) a * b = ab**

^{2}**Find which of the binary operations are commutative and which are associative.**

**Answer)**

**Q 10) Find which of the operations given above has identity.**

**Answer)**

Let the identity be I.

**Q** **11) Let A = N × N and * be the binary operation on A defined by**

**(a, b) * (c, d) = (a + c, b + d)****Show that * is commutative and associative. Find the identity element for * on A, if any.**

**Answer)**

A = N × N

* is a binary operation on A and is defined by:

(a, b) * (c, d) = (a + c, b + d)

Let (a, b), (c, d) ∈ A

Then, a, b, c, d ∈ N

We have:

(a, b) * (c, d) = (a + c, b + d)

(c, d) * (a, b) = (c + a, d + b) = (a + c, b + d)

[Addition is commutative in the set of natural numbers]

∴(a, b) * (c, d) = (c, d) * (a, b)

Therefore, the operation * is commutative.

Now, let (a, b), (c, d), (e, f) ∈A

Then, a, b, c, d, e, f ∈ N

We have:

**Q 12) State whether the following statements are true or false. Justify.**

**(i) For an arbitrary binary operation * on a set N, a * a = a ∀ a * N.**

**(ii) If * is a commutative binary operation on N, then a * (b * c) = (c * b) * a**

**Answer)**

**(i) **Define an operation * on N as:

a * b = a + b a, b ∈ N

Then, in particular, for b = a = 3, we have:

3 * 3 = 3 + 3 = 6 ≠ 3

Therefore, statement (i) is false.

**(ii)** R.H.S. = (c * b) * a

= (b * c) * a [* is commutative]

= a * (b * c) [Again, as * is commutative]

= L.H.S.

∴ a * (b * c) = (c * b) * a

Therefore, statement (ii) is true.

**Q 13) Consider a binary operation * on N defined as a * b = a**^{3} + b^{3}. Choose the correct answer.

^{3}+ b

^{3}. Choose the correct answer.

**(A) Is * both associative and commutative?****(B) Is * commutative but not associative?****(C) Is * associative but not commutative?****(D) Is * neither commutative nor associative?**

**Answer)**

On N, the operation * is defined as a * b = a^{3} + b^{3}.

For, a, b, ∈ N, we have:

a * b = a^{3} + b^{3} = b^{3} + a^{3 }= b * a [Addition is commutative in N]

Therefore, the operation * is commutative.

It can be observed that:

(1*2)*3 = (1^{3}+2^{3})*3 = 9 * 3 = 93 + 33 = 729 + 27 = 756

Also, 1*(2*3) = 1*(2^{3 }+3^{3}) = 1*(8 +27) = 1 × 35

= 1^{3} +35^{3} = 1 +(35)^{3} = 1 + 42875 = 42876.

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

Therefore, the operation * is not associative.

Hence, the operation * is commutative, but not associative. Thus, the correct answer is B.

That’s it. These were the solutions of NCERT Class 12 Maths Exercise 1.4. Our team hopes that you have found these solutions helpful for you. If you have any doubt related to this chapter then feel free to comment your doubts below. Our team will try their best to help you with your doubts.