NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.4 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.4.

- Relations and Functions Class 12 Ex 1.1
- Relations and Functions Class 12 Ex 1.2
- Relations and Functions Class 12 Ex 1.3

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 12 |

Subject |
Maths |

Chapter |
Chapter 1 |

Chapter Name |
Relations and Functions |

Exercise |
Ex 1.4 |

Number of Questions Solved |
13 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.4

**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 = ab²**

**(iv) On Z**

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

^{+}, define * by a * b = a**Solution:**

(i) If a > b, a * b = a – b > 0, which belongs to Z

^{+}.

But if a < b, a * b = a – b < 0, which does not belong to Z

^{+}

=> * given operation is not a binary operation.

(ii) For all a and b belonging to Z

^{-1}, ab also belongs to Z

^{+}.

∴ The operation *, defined by a * b = ab is a binary operation.

(iii) For all a and b belonging to R, ab² also belongs to R.

∴ The operation * defined by a * b = ab² is binary operation.

(iv) For all a and b belonging to Z

^{+}, |a – b| also belongs to Z

^{+1}

∴ The operation a * b = |a – b| is a binary operation.

(v) On Z

^{+}defined by a * b = a

a, b ∈ Z

^{+}= a ∈ Z

^{+}

∴ The operation * is a binary operation.

**Question 2.**

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

**(i) OnZ, define a * b = a – b**

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

**(iii) On Q, define a * b = **

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

**(v) On Z**

^{+}, define a * b = ab**(vi) On R- {-1}, define a * b =**

**Solution:**

(i) On Z, operation * is defined as

(a) a * b = a – b => b * a = b – a

But a – b ≠ b – a ==> a * b ≠ b * a

Defined operation is not commutative

(b) a – (b – c) ≠ (a – b) – c

Binary operation * as defined is not associative.

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

(a) ab + 1 = ba + 1, a * b = b * a

Defined binary operation is commutative.

(b) (a*b)*c = (ab + 1)*c = (ab + 1)c + 1 = abc + c + 1

a * (b * c) = a * (bc + 1) = a(bc + 1)+ 1 = abc + a+ 1

=> a * (b * c) ≠ (a * b) * c

∴ Binary operation defined is not associative.

(iii) (a) On Q, operation * is defined as a * b =

∴ a * b = b * a

∴ Operation binary defined is commutative.

be abc

(b) a * (b * c) = a * = and

(a * b) * c = * c

=> (a * b) * c = * c

Defined binary operation is associative.

(iv) On Z^{+} operation * is defined as a * b = 2^{ab}

(a) a * b = 2^{ab}, b * a = 2^{ba} = 2^{ab}

=> a * b = b * a .

∴ Binary operation defined as commutative.

(b) a * (b * c) = a * 2^{ba} = 2^{a.2bc}

(a * b) * c = 2^{ab} * c = 2^{2ab}

Thus (a * b) * c ≠ a * (b * c)

∴ Binary operation * as defined as is not associative.

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

(a) b * a = b^{a} .

∴ a^{b} ≠ b^{a} = a * b ≠ b * a.

* is not commutative.

(b) (a*b)*c = a^{b}*c = (a^{b})^{c} = a^{bc} a*(b* c)

= a*b^{c} = a^{bc}

Thus (a * b) * c ≠ (a * b * c)

∴ Operation * is not associative.

(vi) Neither commutative nor associative.

**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:**

Operation ^ table on the set {1, 2, 3, 4, 5} is as follows.

**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)**

**Solution:**

(i) From the given table, we find

2*3 = 1, 1*4 = 1

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

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

(ii) Let a, b ∈ {1,2,3,4,5} From the given table, we find

a*a = a

a*b = b*a = 1 when a or b or are odd and a b.

2 * 4 = 4 * 2 = 2, when a and b are even and a ≠ b

Thus a * b = b * c

∴Binary operation * given is commutative.

(iii) Binary operation * given is commutative (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 the exp no. 4 above? Justify your answer.**

**Solution:**

The set is {1,2,3,4, 5} and a * b = HCF of a and b.

Let us prepare the table of 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**

**Solution:**

Binary operation * defined as a * b = 1 cm. of a and b.

(i) 5 * 7 = 1 cm of 5 and 7 = 35

20 * 16= 1 cm of 20 and 16 = 80

(ii) a * b= 1 cm of a and b b * a = 1 cm of b and a

=> a * b = b * a, 1 cm of a, b and b, a are equal

∴ Binary operation * is commutative.

(iii) a * (b * c) = 1 cm of a, b, c and (a * b) * c = 1 cm of a, b, c

=> a * (b * c) = (a * b) * c

=> Binary operation * given is associative.

(iv) Identity of * in N is 1

1 * a = a * 1 = a = 1 cm of 1 and a.

(v) Let * : N x N—> N defined as a * b = 1.com of (a, b)

For a = 1, b = 1, a * b = l b * a

Otherwise a * b ≠ 1

∴ Binary operation * is not invertible

=> 1 is invertible for operation *

**Question 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.**

**Solution:**

The given set = {1,2,3,4,5} Binary operation is defined as a * b = 1 cm of a and b. 4 * 5 = 20 which does not belong to the given set {1, 2, 3, 4, 5}.

It is not a binary operation.

**Question 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?**

**Solution:**

Binary operation on set N is defined as a * b = HCF of a and b

(a) We know HCF of a, b = HCF of b, a

∴ a * b = b * a

∴ Binary operation * is commutative.

(b) a*(b*c) = a* (HCF of b, c) = HCF of a and (HCF of b, c) = HCF of a, b and c

Similarly (a * b) * c = HCF of a, b, and c

=> (a * b) * c = a * (b * c)

Binary operation * as defined above is associative.

(c) 1 * a = a * 1 = 1 ≠ a

∴ There does not exists any identity element.

**Question 9.**

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

**(i) a * b = a – b**

**(ii) a * b = a² + b²**

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

**(iv) a * b = (a – b)²**

**(v) a * b = **

**(vi) a * b = ab²**

Find which of the binary operations are commutative and which are associative.

**Solution:**

Operation is on the set Q.,,

(i) defined as a * b = a – b

(a) Now b * a = b – a

But a – b ≠ b – a

∴ a * b ≠ b * a

∴ Operation * is not commutative.

(b) a * (b * c) = a * (b – c) = a – (b – c) = a – b + c (a * b) * c = (a – b) * c = a – b – c

Thus a * (b * c)¹ (a * b) * c = (a² + b²)² + c²

=> a * (b * c) ≠ (a * b) * c

∴ The operation * as defined is not associative.

(ii) (a) a * b = a² + b * a = b² + a² = a² + b².

a * b = b * a

This binary operation is commutative,

(b) a*(b*c) = a*(b² + c²) = a² + (b²)² + c²)²

=> (a*b)*c = (a² + b²)*c = (a² + b²) + c²

Thus a * (b*c) (a*b) * c

The operation * given is not associative.

(iii) Operation * is defined as a * b = a + ab

(a) b * a = b + ba

a * b ≠ b * a

The operation is not commutative.

(b) a*(b*c) = a*(b + bc)

= a + a(b + bc)

= a + ab + abc (a * b) * c

= (a + ab) * c

= (a + ab) + (a + ab) • c

= a + ab + ac + abc

=> a * (b * c) ≠ (a * b) * c

=> The binary operation is not associative.

(iv) The binary operation is defined as a * b = (a – b)²

(a) b*a = (b – a)² = (a – b)² => a*b = b*a

.’. This binary operation * is commutative.

(b) a*(b*c) = a*(b – c)² = [a – (b – c)²]² (a * b) * c

= (a – b)² * c

= [(a – b)² – c]²

=> a * (b * c) ≠ (a * b) * c

the operation * is not associative.

(v) Commutative and associative.

(vi) Neither commutative nor associative.

**Question 10.**

**Show that none of the operations given above has identity.**

**Solution:**

The binary operation * on set Q is

(i) defined as a*b = a – b

For identity element e, a*e = e*a = a

But a*e = a – e≠a and e*a = e – a≠a

There is no identity element for this operation

(ii) Binary operation * is defined as a * b = a² + b² ≠ a

This operation * has no identity.

(iii) The binary operation is defined as a*b = a+ab

Putting b = e, a + e = a + eb ≠ a

There is no identity element.

(iv) The binary operation is defined as a * b = (a – b)²

Put b = e, a * e = (a – e)² ≠ a for any value of

e∈Q

=> there is no Identity Element.

(v) The operation is a * b =

∴ a * e = ≠ a for any value of e ∈ Q

∴ Operation * has no identity

(vi) The operation * is a * b = ab² Put b = e, a

*e = ae² and e * a = ea² ≠ a for any value of e∈Q

=> There is no Identity Element. Thus, these operations have no Identity.

**Question 11.**

**Let A=N x 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.**

**Solution:**

A = N x N Binary operation * is defined as (a, b) * (c, d) = (a + c, b + d)

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

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

∴ This operation * is commutative

(b) Next(a,b)* [(c,d)*(e,f)]=(a,b)*(c+e,d+f) = ((a + c + e), (b + d + f))

and [(a, b) * (c, d)] * (e, f)=(a+c, b+d) * (e, f) = ((a + c + e, b+d + f))

=> (a, b) * [(c, d) * (e, f)] = [(a, b) * (c, d)] * (e,f)

∴ The binary operation given is associative

(c) Identity element does not exists.

**Question 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**

**Solution:**

(i) A binary operation on N is defined as

a*a=a∀a∈N.

Here operation * is not defined.

∴ Given statement is false.

(ii) * is a binary commutative operation on N. c

* b = b * c

∵ * is commutative

∵ (c * b) * a = (b * c) * a = a * (b * c)

∴ Thus a * (b * c) = (c * b) * a

This statement is true.

**Question 13.**

**Consider a binary operation * on N defined as a * b = a³ + b³. 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?**

**Solution:**

(b)

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