# NCERT Solutions for Class 12 Maths Relations and Functions

Hi Students, Welcome to **Amans Maths Blogs (AMB)**. In this post, you will get the * NCERT Solutions for Class 12 Maths Relations and Functions Exercise 1.4*. This

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*NCERT Solutions*## NCERT Solutions for Class 12 Maths Relations and Functions Exercise 1.4

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.4: Ques No 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**

**NCERT Solutions:**

Given that ∗ defined by a ∗ b = a – b in Z^{+}.

Since the image of (1, 2) under * is 1 * 2 = 1 − 2 = −1 ∉ Z^{+}, the operation * is not a binary operation.

**(ii) On Z ^{+}, define **

**∗**

**by a**

**∗**

**b = ab**

**NCERT Solutions:**

Given that ∗ defined by a ∗ b = ab in Z^{+}.

Since for each a, b ∈ Z^{+}, * carries each pair (a, b) to a unique element a * b = ab in Z^{+}, then the operation * is a binary operation.

**(iii) On Z ^{+}, define **

**∗**

**by a**

**∗**

**b = ab**

^{2}**NCERT Solutions:**

Given that ∗ defined by a ∗ b = ab^{2} in R.

Since for each a, b ∈ R, * carries each pair (a, b) to a unique element a * b = ab^{2} in R, then the operation * is a binary operation.

**(iv) On Z ^{+}, define **

**∗**

**by a**

**∗**

**b = |a – b|**

**NCERT Solutions:**

Given that ∗ defined by a ∗ b = | a – b | in Z^{+}.

Since for each a, b ∈ Z^{+}, * carries each pair (a, b) to a unique element a * b = | a – b | in Z^{+}, then the operation * is a binary operation.

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

**∗**

**by a**

**∗**

**b = a**

**NCERT Solutions:**

Given that ∗ defined by a ∗ b = a in Z^{+}.

Since for each a, b ∈ Z^{+}, * carries each pair (a, b) to a unique element a * b = a in Z^{+}, then the operation * is a binary operation.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.4: Ques No 1.**

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

**(i) On Z ^{+}, define **

**∗**

**by a**

**∗**

**b = a – b**

**NCERT Solutions:**

Given that ∗ defined by a ∗ b = a – b in Z.

Let a = 2, b = 3 in Z. Then,

2 * 3 = 2 – 3 = – 1 and 3 * 2 = 3 – 2 = 1

We get that 2 * 3 ≠ 3 * 2

Thus, the given operation * is not commutative.

Let a = 2, b = 3, c = 4 in Z. Then,

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

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

We get that (2 * 3) * 4 ≠ 2 * (3 * 4)

Thus, the given operation * is not associative.

**(ii) On Z ^{+}, define **

**∗**

**by a**

**∗**

**b = ab + 1**

**NCERT Solutions:**

Given that ∗ defined by a ∗ b = ab + 1 in Q.

Let a = 2, b = 3 in Q. Then,

2 * 3 = 2×3 + 1 = 6 + 1 = 7 and 3 * 2 = 3×2 + 1 = 6 + 1 = 7

We get that 2 * 3 = 3 * 2

Thus, the given operation * is commutative.

Let a = 2, b = 3, c = 4 in Q. Then,

(2 * 3) * 4 = (2×3 + 1) * 4 = 7 * 4 = 7×4 + 1 = 29

2 * (3 * 4) = 2 * (3×4 + 1) = 2 * 13 = 2×13 + 1 = 27

We get that (2 * 3) * 4 ≠ 2 * (3 * 4)

Thus, the given operation * is not associative.

**(iii) On Z ^{+}, define **

**∗**

**by a**

**∗**

**b = ab/2**

**NCERT Solutions:**

Given that ∗ defined by a ∗ b = ab/2 in Q.

Let a = 2, b = 3 in Q. Then, 2 * 3 = (2 × 3)/2 = 3 and 3 * 2 = (3 × 2) / 2 = 3

We get that 2 * 3 = 3 * 2. Thus, the given operation * is commutative.

Let a = 2, b = 3, c = 4 in Q. Then,

We get that (2 * 3) * 4 = 2 * (3 * 4). Thus, the given operation * is associative.

**(iv) On Z ^{+}, define **

**∗**

**by a**

**∗**

**b = 2**

^{ab}**NCERT Solutions:**

Given that ∗ defined by a ∗ b = 2^{ab} in Z^{+}.

Let a = 2, b = 3 in Z. Then, 2 * 3 = 2^{(2}^{×}^{3)} = 2^{6} = 64 and 3 * 2 = 2^{(3}^{×}^{2)} = 2^{6} = 64

We get that 2 * 3 = 3 * 2. Thus, the given operation * is commutative.

Let a = 2, b = 3, c = 4 in Z^{+}. Then,

(2 * 3) * 4 = 2^{(2}^{×}^{3)} * 4 = 64 * 4 = 2^{(64}^{×}^{4) }= 2^{256}

2 * (3 * 4) = 2 * 2^{(3}^{×}^{4)} = 2 * 4096 = 2^{(2}^{×}^{4096)} = 2^{8192}

We get that (2 * 3) * 4 ≠ 2 * (3 * 4)

Thus, the given operation * is not associative.

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

**∗**

**by a**

**∗**

**b = a**

^{b}**NCERT Solutions:**

Given that ∗ defined by a ∗ b = a^{b} in Z^{+}.

Let a = 2, b = 3 in Z. Then, 2 * 3 = 2^{3} = 8 and 3 * 2 = 3^{2} = 9

We get that 2 * 3 ≠ 3 * 2. Thus, the given operation * is not commutative.

Let a = 2, b = 3, c = 4 in Z^{+}. Then,

(2 * 3) * 4 = 2^{3} * 4 = 8 * 4 = 8^{4 }= 4096 = 2^{12}

2 * (3 * 4) = 2 * 3^{4} = 2 * 81 = 2^{81}

We get that (2 * 3) * 4 ≠ 2 * (3 * 4). Thus, the given operation * is not associative.

**(vi) On Z ^{+}, define **

**∗**

**by a**

**∗**

**b = a/(b + 1)**

**NCERT Solutions:**

Given that ∗ defined by a ∗ b = a/(b + 1) in R – {–1}.

Let a = 2, b = 3 in R – {–1}. Then, 2 * 3 = 2/(3 + 1) = ½ and 3 * 2 = 3/(2 + 1) = 1

We get that 2 * 3 ≠ 3 * 2. Thus, the given operation * is not commutative.

Let a = 2, b = 3, c = 4 in R – {–1}. Then,

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.4: Ques No 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 ∧.

**NCERT Solutions:**

Given that the binary operation ∧ on the set {1, 2, 3, 4, 5} defined by a ∧ b = min {a, b}.

Thus, the operation table for the given operation is as below.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.4: Ques No 4.**

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

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

**NCERT Solutions:**

From the given table, we have 2 * 3 = 1, 1 * 4 = 1, 3 * 4 = 1 and 2 * 1 = 1. Thus, (2 * 3) = 1 * 4 = 1 and 2 * (3 * 4) = 2 * 1 = 1.

**(ii) Is * commutative?**

**NCERT Solutions:**

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

**(iii) Compute (2 ∗ 3) ∗ (4 ∗ 5). **

**NCERT Solutions:**

From the given table, we have 2 * 3 = 1, 4 * 5 = 1, and 1 * 1 = 1. Thus, (2 * 3) * (4 * 5) = 1 * 1 = 1.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.4: Ques No 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.

**NCERT Solutions:**

Given that the binary operation ∗′ on the set {1, 2, 3, 4, 5} defined by a ∗′ b = HCF of a and b.

Thus, the operation table for the given operation ∗′ is as below.

In the question no 4, the operation table for the given operation ∗ is as below.

We observe that the operation tables for the operations * and *′ are the same. Thus, the operation *′ is same as the operation*.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.4: Ques No 6.**

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

**(i) Find 5 ****∗**** 7, 20 ****∗**** 16**

**NCERT Solutions:**

Given that the binary operation on N given by a ∗ b = L.C.M. of a and b.

Thus, 5 * 7 = LCM of 5 and 7 = 35 and 20 * 16 = LCM of 20 and 16 = 80

**(ii) Is * commutative?**

**NCERT Solutions:**

Given that the binary operation on N given by a ∗ b = L.C.M. of a and b

Since LCM of (a, b) and LCM of (b, a) is same for all N, then a ∗ b = b * a

Thus, the given operation * is commutative.

**(iii) Is * associative?**

**NCERT Solutions:**

Given that the binary operation on N given by a ∗ b = L.C.M. of a and b

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

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

Hence, the given operation * is associative.

**(iv) ****Find the identity of * in N.**

**NCERT Solutions:**

Given that the binary operation on N given by a ∗ b = L.C.M. of a and b

Since LCM of (a, 1) and LCM of (1, a) is same for all N and that is 1,

then a ∗ 1 = 1 * a = LCM of a and 1 = a

Thus, 1 is the identity of the given operation *.

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

**NCERT Solutions:**

Given that the binary operation on N given by a ∗ b = L.C.M. of a and b

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

Thus, we have c = 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 *.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.4: Ques No 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.

**NCERT Solutions:**

Given that ∗ defined on the set {1, 2, 3, 4, 5} by a ∗ b = L.C.M. of a and b

Thus, the operation table for the given operation ∗ is as below.

From the table, we observe that 6, 10, 12, 15, 20 ∉ A. Thus, the given operation * is not a binary operation.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.4: Ques No 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?

**NCERT Solutions:**

Given that ∗ is the binary operation on N defined by a ∗ b = H.C.F. of a and b.

We know that H.C.F. of a and b = H.C.F. of b and a for all a, b ∈ N.

Thus, we get a * b = b * a. Hence, the given 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

Thus, we get (a * b) * c = a * (b * c). Hence, the given operation * is associative.

Now, an element e ∈ N will be the identity for the operation * if a * e = a = e* a for all a ∈ N. But this relation is not true for any a ∈ N.

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

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.4: Ques No 9.**

Let ∗ be a binary operation on the set Q of rational numbers as follows: Find which of the binary operations are commutative and which are associative.

**(i) a * b = a – b**

**NCERT Solutions:**

Given that ∗ be a binary operation as a ∗ b = a – b

Since a – b ≠ b – a, then a * b ≠ b * a. Thus, the given operation a ∗ b = a – b is not commutative.

For a, b, c, we have, (a * b) * c = (a – b) * c = (a – b) – c = a – b – c

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

Since a – b – c ≠ a – b + c, then (a * b) * c ≠ a * (b * c).

Thus, the given operation a ∗ b = a – b is not associative.

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

**NCERT Solutions:**

Given that ∗ be a binary operation as a ∗ b = a^{2} + b^{2}

Since a^{2} + b^{2} = b^{2} + a^{2}, then a * b = b * a

Thus, the given operation a ∗ b = a^{2} + b^{2} is commutative.

For a, b, c, we have

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

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

Since (a^{2} + b^{2})^{2} + c^{2} ≠ a^{2} + (b^{2} + c^{2})^{2}, then (a * b) * c ≠ a * (b * c).

Thus, the given operation a ∗ b = a^{2} + b^{2 }is not associative.

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

**NCERT Solutions:**

Given that ∗ be a binary operation as a ∗ b = a + ab

Since a + ab ≠ b + ba, then a * b ≠ b * a

Thus, the given operation a ∗ b = a + ab is not commutative.

For a, b, c, we have

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

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

Since a + ab + ac + abc ≠ a + ab + abc, then (a * b) * c ≠ a * (b * c).

Thus, the given operation a ∗ b = a + ab is not associative.

**(iv) a * b = (a – b) ^{2}**

**NCERT Solutions:**

Given that ∗ be a binary operation as a ∗ b = (a – b)^{2}

Since (a – b)^{2} = (b – a)^{2}, then a * b = b * a

Thus, the given operation a ∗ b = (a – b)^{2 }is commutative.

For a, b, c, we have

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

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

Since ((a – b)^{2 }– c)^{2} ≠ (a – (b – c)^{2})^{2}, then (a * b) * c ≠ a * (b * c).

Thus, the given operation a ∗ b = (a – b)^{2 }is not associative.

**(v) a * b = ab/4**

**NCERT Solutions:**

Given that ∗ be a binary operation as a ∗ b = ab/4

Since ab/4 = ba/4, then a * b = b * a

Thus, the given operation a ∗ b = ab/4 is commutative.

For a, b, c, we have

Since abc/16 = abc/16, then (a * b) * c = a * (b * c). Thus, the given operation a ∗ b = ab/4 is associative.

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

**NCERT Solutions:**

Given that ∗ be a binary operation as a ∗ b = ab^{2}

Since ab^{2} ≠ ba^{2}, then a * b ≠ b * a

Thus, the given operation a ∗ b = ab^{2} is not commutative.

For a, b, c, we have

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

a * (b * c) = a * (bc^{2}) = a(bc^{2})^{2 }= ab^{2}c^{4}

Since ab^{2}c^{2} ≠ ab^{2}c^{4}, then (a * b) * c ≠ a * (b * c).

Thus, the given operation a ∗ b = ab^{2 }is not associative.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.4: Ques No 10.**

Show that none of the operations given above has identity.

**NCERT Solutions:**

We know that if a * e = a = e * a for all a ∈ Q, then the element e ∈ Q will be the identity element for the operation *.

However, there is no such element e ∈ Q with respect to each of the six operations given in question no 9.

Thus, none of the six operations in question 9 has identity.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.4: Ques No 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.

**NCERT Solutions:**

Given that A = N × N and ∗ be the binary operation on A defined by

(a, b) ∗ (c, d) = (a + c, b + d).

For a, b, c, d ∈ N, we have

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

Since (a + c, b + d) = (a + c, b + d), then (a, b) ∗ (c, d) = (c, d) * (a, b)

Thus, the given operation (a, b) ∗ (c, d) = (a + c, b + d) is commutative.

For a, b, c, d, e, f ∈ N, we have

{(a, b) ∗ (c, d)} * (e, f) = {a + c, b + d} * (e, f) = (a + c + e, b + d + f)

And,

(a, b) ∗ {(c, d) * (e, f)} = (a, b) * {c + e, d + f}= (a + c + e, b + d + f)

Thus, {(a, b) ∗ (c, d)} * (e, f) = (a, b) ∗ {(c, d) * (e, f)}.

The given operation (a, b) ∗ (c, d) = (a + c, b + d) is associative.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.4: Ques No 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

**NCERT Solutions:**

(i)

First we 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.

Therefore, statement (ii) is true.

**NCERT Solutions for Class 12 Maths Relations and Functions ****Exercise**** 1.4: Ques No 13.**

Consider a binary operation ∗ on N defined as a ∗ b = a^{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?

**NCERT Solutions:**

Given that ∗ be a binary operation as a ∗ b = a^{3} + b^{3}

Since a^{3} + b^{3} = b^{3} + a^{3}, then a * b = b * a

Thus, the given operation a ∗ b = a^{3} + b^{3} is commutative.

For a, b, c, we have

(a * b) * c = (a^{3} + b^{3}) * c = (a^{3} + b^{3})^{3} + c^{3}

a * (b * c) = a * (b^{3} + c^{3}) = a^{3} + (b^{3} + c^{3})^{3}

Since (a^{3} + b^{3})^{3} + c^{3} ≠ a^{3} + (b^{3} + c^{3})^{3}, then (a * b) * c ≠ a * (b * c).

Thus, the given operation a ∗ b = a^{3} + b^{3 }is not associative.

Therefore, the correct option is B.