In this article, we share MP Board Class 12th Maths Book Solutions Chapter 1 Relations and Functions Ex 1.4 Pdf, These solutions are solved by subject experts from the latest MP Board books.

## MP Board Class 12th Maths Solutions 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 c is not a binary operation, given 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. Let 3, 4 ∈ Z^{+}

3 * 4 = 3 – 4 = – 1 ∈ Z^{+}

Hence * is not a binary operation.

ii. For every element a, b ∈ Z^{+}, ab ∈ Z^{+}

Hence * is a binary operation.

Example : 3, 4 ∈ Z^{+}, 3 *4 = 3 x 4= 12 ∈ Z^{+}

iii. For a, b ∈ R, ab² ∈ R and is unique.

Hence a * b = ab² is a binary operation,

Example : 5, b ∈ R, 5 * 6 = 5 x 6² = 180 ∈ R

iv. For a, b ∈ Z^{+}, |a – b| ∈ Z^{+}

Hence * is a binary operation.

Example : 3, 4 ∈ Z^{+}, 3 * 4 = |3 – 4| = |- 1| = 1 ∈ Z^{+}

v. For a, b ∈ Z^{+}+, a * b = 2 ∈ Z^{+}

Hence * is a binary operation.

Example : 2, 3 ∈ Z^{+}, 2 * 3 = 2 ∈ Z^{+}

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 = \(\frac { ab }{ 2 }\)

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

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

vi. On R – {- 1}, define a * b = \(\frac { a }{ b+1 }\)

Solution:

i. a * b = a – b; b * a = b – a

∴ a * b * b * a

∴ * is not commutative

Example:

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

a * (b * c) = a – (b – c) = a – b + c

(a * b) * c = (a – b) – c = a – b – c

∴ a * (b * c) ≠ (a * b) * c

∴ * is not associative

ii. a * b = ab + 1

b * a = ba + 1 = ab + 1 since ab – ba in Q.

∴ a * b = b * a.

Hence * is commutative.

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

= a(bc + 1) + 1 = abc + a + 1

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

= (ab + 1)c + 1 = abc + c + 1

∴ a * (b * c) ≠ (a * b) * c

* is not associative.

iv. a * b = 2^{ab}

b * a = 2^{ba} = 2^{ab} since ab = ba in Z^{+}

∴ a * b = b * a

∴ * is commutative

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

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

∴ a * (b * c) ≠ (a * b) * c.

∴ * is not associative

v. a * b = a^{b}

b * a = b^{b} Since a^{b} ≠ b^{a}, a * b ≠ b * a

∴ * is not commutative.

Example :

2 * 3 = 2^{3}3 = 8

3 * 2 = 3² = 9

∴ 2 * 3 ≠ 3 * 2

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

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

∴ a * (b * c) ≠ (a * b) * c

∴ * is not associative

Hence * is not 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:

1 ^ 1 = min{1, 1} = 1, 1 ^ 2 = min{1, 2} = 1, etc.

2 ^ 1 = min{2, 1} = 1, 2 ^ 2 = min{2, 2} = 2, etc.

3 ^ 1 = min{3,1} = 1, 3 ^ 2 = min{3, 2} = 2, etc. and so on.

The operation table of the operation.

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).

Solution:

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

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

ii. The entries in the table are symmetric along the main diagonal. Hence * is commutative.

iii. (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 Question 4 above? Justify your answer.

Solution:

*’ gives the same table as given in. Hence * and *’ are the same operations.

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:

i. 5 * 7 = LCM (5, 7) = 5 x 7 = 35

20 * 16 = LCM (20, 16) = 80

ii. a * b = LCM (a, b)

b * a = LCM (b, a) = LCM (a, b)

∴ a * b = b * a

∴ * is commutative

iii. a * (b * c) = a * LCM (b, c)

= LCM (a, b, c)

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

= LCM (a, b, c)

∴ a * (b * c) = (a * b) * c

∴ * is associative

iv. Let e be the identity element in N.

Then a * e = e * a = a, for all a ∈ N

⇒ LCM (a, e) = a, for all a ∈ N

⇒ e = 1 ∈ N

∴ 1 is the identity element of * in N.

v. Let a be an invertible element in N.

Then there exists an element b in N such that

a * b = 1 = b * a

⇒ LCM (a, b)= 1

a = b= 1

Thus 1 is an invertible element in N.

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. (March 2016)

Solution:

a * b = LCM (a, b)

Now 2 * 3 = LCM (2, 3) = 6

But 6 is not an element of the given set.

Hence * 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? (March 2013)

Solution:

a * b = HCF (a, b) for all a, b ∈ N

b * a = HCF (b, a) for all a, b ∈ N

∴ a * b = b * a

∴ * is commutative

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

= HCF (a, b, c)

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

Hence a * (b * c) = (a * b) * c

∴ * is associative

Let e be the identity element in N.

⇒ a * e = e * a = a for all a ∈ N

⇒ HCF (a, e) = HCF (e, a) = a for all a ∈ N

⇒ There is no e e N which makes this true.

⇒ Identity element does not exist in N.

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 = \(\frac { ab }{ 4 }\)

vi. a * b = ab²

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

Solution:

i. a * b = a – b b * a = b – a

Hence a * b ≠ b * a

∴ * is not commutative

a * (b * c) = a* (b – c)

= a – (b – c) = a – b + c

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

Hence (a * b) * c ≠ a * (b * c)

∴ * is not associative.

ii. a * b = a² + b²

b * a = b² + a² = a² + b² = a * b

∴ * is commutative.

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

(a * b) * c = (a² + b²) * c

= (a² + b²)² + c² ≠ a² + (b² + c²)²

∴ * is not associative

iii. a * b = a + ab;

b * a = b + ba

a * b ≠ b * a

∴ * is not commutative.

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

∴ * is not associative.

iv. a * b = (a – b)² = a² – 2ab + b²

b * a = (b – a)² = b² – 2ab + a²

= a² – 2ab + b² = a * b

∴ * is commutative.

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

∴ * is not associative.

v. a * b = \(\frac { ab }{ 4 }\)

b * a = \(\frac { ba }{ 4 }\) = \(\frac { ab }{ 4 }\) = a * b

∴ * is commutative.

vi. a * b = ab² ; b * a = ba²

a * b ± b * a

∴ * is not commutative.

a * (b * c) = a * (bc²) = a(bc²)² = ab²c^{4}

(a * b) * c = (ab²) * c = (ab²)(c²) = ab²c²

a * (b * c) ≠ (a * b) * c

∴ * is not associative.

Question 10.

Show that none of the operations given in Qn.No. 9 (except (v)) has identity.

Solution:

i. Let e be the identity element.

a * e = e * a = a

⇒ a² + e² = e² + a² = a

which is not possible for any e ∈ Q.

∴ There is no identity element.

ii. Let e be the identity element.

a * e = e * a = a

⇒ (a – e)² = (e – a)² = a

which is not possible for any e ∈ Q.

∴ There is no identity element. .

iii. Let e be the identity element.

a * e = e * a = a

⇒ ae² = ea² = a

which is not possible for any e ∈ Q. (SAY 2014)

∴ There is no identity element.

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

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

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

∴ * is commutative.

Let (a, b), (c, d), (e, f) ∈ A.

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

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

= (a + c + e, b + d +f)

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

= (a + c + e, b + d + f)

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

∴ * is associative.

Here identity element does not exist.

Let (e_{1}, e_{2}) ∈ be the identity element for * in A.

∴ (a, b) * (e_{1} + e_{2}) = (e_{1}, e_{2}) * (a, b) = (a, b)

⇒ (a + e_{1}, b + e_{2}) = (e_{1} + a, e_{2} + b) = (a, b)

⇒ a + e_{1} = a and b + e_{2} = b

⇒ e_{1} = 0 and e_{2} = 0

(e_{1}, e_{2}) = (0, 0) ∉ A, since A = N x N

Hence there is no identity element for * in A.

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 for all a ∈ N

ii. If * is a commutative binary operation on N, then

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

Solution:

i. False

Let a * b = a + b, a, b ∈ N

∴ a * a = a + a = 2a ≠ a

ii. True

Since * is commutative b * c = c * b

∴ a * (b * c) = a * (c * b) = (c * b) * a

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. Is * commutative but not associative?

a * b = a³ + b³ = b³ + a³ = b * a

∴ * is commutative.

a * (b * c ) = a * (b³ + c³) = a³ + (b³ + c³)³

(a * b) * c = (a³ + b³) * c = (a³ + b³)³ + c³

* is not associative.