Class 12 Maths Exercise 1.4

NCERT solutions for class 12 maths Chapter 1 Relation and Functions are prepared by the academic team of Nashad. We have prepared  NCERT Solutions for all exercises of chapter 1. Given below is a step-by-step solution to all questions given in the NCERT textbook for Chapter 1 Relations and Functions.

NCERT Solutions for Class 12 Maths Exercise 1.4

Solve The Following Questions.

Question1. Determine whether or not each of the definitions 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) 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 = ab².

It is seen that for each a, b ∈ R, there is a unique element ab² in R. 

This means that * carries each pair (a, b) to a unique element a * b = ab² 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.

Question2. 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 :

(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 = 1²=1 and 2x 1 =2¹ = 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.

Question3. 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 :
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:

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) (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

Question5. 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 :
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 operation *′ can be given as:

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

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

Question6. 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 :
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 *.

Question7. 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 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:

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.

Question8. 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 an identity for this binary operation on N?

Solution :
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.

Question9. 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 = ab/4  (vi) a * b = ab²

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

Ans.

10. Find which of the operations given above has an identity.

Solution :
Let the identity be I.

Question 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.

Solution :
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:

Question12. 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) 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.

Question13. 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 :
On N, the operation * is defined as a * b = a³ + b³.

For, a, b, ∈ N, we have:

a * b = a³ + b³ = b³ + a³ = b * a [Addition is commutative in N]

Therefore, the operation * is commutative.

It can be observed that:

 (1*2)*3 = (1³+2³)*3 = 9 * 3 = 93 + 33 = 729 + 27 = 756

Also, 1*(2*3) = 1*(2³ +3³) = 1*(8 +27) = 1 × 35

= 1³ +35³ = 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.