## Rajasthan Board RBSE Class 12 Maths Chapter 1 Composite Functions Ex 1.3

Question 1.

Determine whether each of the following operation define a binary operation on the given set or not. Also, Justify your answer.

(i) a*b = a, on N

(ii) a*b = a + b – 3, on N

(iii) a*b = a + 3b, on N

(iv) a*b = a/b, on Q

(v) a*b = a – b, on R

Solution:

(i) a*b = a, on N

Here, * is a binary operation because

a, b ∈ N

= a*b = a ∈ N

Here, a*b ∈ N

Hence, * is a binary operation.

(ii) a*b = a + b – 3 ∈ N

Here, a*b is not a binary operation because,

1 ∈ N, 2 ∈ N

then 1 + 2 – 3 = 0 ∈ N

(iii) a*b= a + 3b ∈ N

Here, * is a binary operation because

1 ∈ N, 2 ∈ N

then 1 + 3 × 2 = 1 + 6 = 7 ∈N

(iv) a*b = \(\frac { a }{ b }\) ∈ Q

Here, a*b is not a binary operation because,

Let a = 22 ∈ Q and b= 7 ∈ Q

But

(v) a*b= a – b ∈ R

Here, * is a binary operation because,

a ∈ R, b ∈ R

⇒ a – b ∈ R, ∀ a, b ∈ R

Question 2.

Determine which of the following binary operation is commutative and which is associative :

(i) * on N defined as a*b = 2^{ab}

(ii) * on N defined as a*b = a + b + a^{a}b

(iii) * on Z defined as a*b = a – b

(iv) * on Q defined as a*b = ab + 1

(v) * on R defined as a*b = a + b – 7

Solution:

(i) Given a*b = 2^{ab}

Commutativity: Let a, b ∈ N

a*b = 2^{ab}

= 2^{b.a}

= b*a

So, a*b = b*a

∴ * is a commutative operation.

Associativity: Let a, b, c ∈ N

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

= 2^{c}*2^{(ab)} = 2^{c +ab}

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

2^{ab+c} ≠ 2^{a+bc}

It is clear that (a*b)*c ≠ a*(b*c)

So, (a*b)*c is not an associative operation.

Hence, a*b = 2^{ab} is commutative but not associative.

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

Commutativity: Let a, b ∈ N

a*b = a + b + a^{2}b

b*a = b + a + b^{2}a

a*b ≠ b*a .

So, * is not a commutative operation.

Associativity: Let a, b, c ∈ N

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

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

t is clear that (a*b)*c ≠ a*(b*c)

So * is not an associative operation.

Hence, a*b = a + b + a^{2}b is neither commutative nor associative.

(iii) Given, a*b = a – b

Commutativity:

a*b = a – b, (a, b ∈ Z)

b*a = b – a, (a, b ∈ Z)

a*b* ≠ b*a

So * is not a commutative operation.

Associativity :

(a*b)*c = (a – b)*c

= a – b-c

a*(b*c) = a*(b-c)

= a – b + c

∵ (a*b)*c ≠ a*(b*c)

So, it is not associative operation.

It is clear that

a*b = a – b is neither commutative nor associative.

(iv) Given, a*b = ab + 1

Commutativity: Let a, b ∈ Q

a*b = ab + 1 and : b*a= ba + 1

⇒ a*b= b*a

∴ It is commutative.

∴ Addition and multiplication of rational number is commutative.

Associativity: Let a, b, c ∈ Q

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

= ab + 1 + c

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

= (a*b)*c ≠ (b*c)*a

So, * is not associative.

It is clear from above that a*b = ab + 1 is commutative but not associative.

(v) Given, a*b = a + b – 7

Commutativity: In R,

a*b = a + b – 7

= b + a – 7

= b*a’

Associativity :

(a*b)*c = (a + b – 7)*c

= (a + b – 7) + c – 7

= a + b + c – 14

a*(b*c) = a*(b + c – 7)

= a + (b + c – 7) – 7

= a + b + c – 14

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

Hence, it is clear that a*b = a + b – 7 are commutative and associative.

Question 3.

If * be an operation on Z, defined as a*b = a + b + 1, ∀ a, b ∈ Z then prove that * is commutative and associative, find its identity element. Also, find inverse element of any integer in Z.

Solution:

Given a*b = a + b + 1, ∀ a, b ∈ z

Commutativity :

a*b = a + b + 1

a*b = b + a + 1

= b*a

∴ a*b = b*a

∴ * is commutative operation.

Associativity :

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

= a + b + 1 + c +1

a + b + c + 2

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

= a + b + c + 1 + 1

= a + b + c + 2

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

∴ * is associative operation

Identity : If e is identity element, then

a*e = a

⇒ a + e + 1 = a

⇒ e = -1

So, – 1 ∈ Z is identity element.

Inverse : Let inverse of a is x, then by definition

a*x = -1 [∵ – 1 is identity]

⇒ a + x + 1 = -1

⇒ x = -(a + 2) ∈ Z

Inverse element a^{-1} = -(a + 2).

Question 4.

If * be a binary operation defined on R – {1}. as a*b = a + b – ab, ∀ a, b ∈ R – {1}

prove that * is commutative and associative. Find identity element and also find inverse of a.

Solution:

If a, b ∈ R – {1} by definition

a*b = a + b – ab

= b + a – ba

= b*a

∴ * is a binary operation.

Again, (a*b)*c = (a + b – ab)*c

= (a + b – ab) + c – (a + b – ab)c

= a + b – ab + c – ac – bc + abc

= a + b +c – ab – bc – ac + abc …(i)

and a*(b*c) = a*(b + c – bc)

= a + (b + c – bc) – a (b + c – bc)

= a + b + c – bc – ab – ac + abc

= a + b + c – ab – bc – ac + abc … (ii)

From eqs. (i) and (ii),

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

∴ * is associative operation.

Let e is identity of *, then for a ∈ R,

a*e = a (from definition of identity)

⇒ a + e – ae = a

⇒ e(1 – a)= 0

⇒ e = 0 ∈ R – {1}

∵ 1 – a ≠ 0

∴ 0 is identity of *.

Let b is inverse of a.

a*b = e

a + b – ab = 0.e

b – ab = -a

b(1 – a) = -a

Question 5.

Four functions are defined on set R_{o}, Such that,

f_{1}(x) = x, f_{2}(x) = -x, f_{3}(x) = 1/x, f_{4}(x) = – 1/x Construct the composition table for the composition of functions f_{1}, f_{2}, f_{3}, f_{4}. Also, find identity element and inverse of every element.

Solution:

Given,

f_{1}(x) = x, f_{2}(x) = – x,

∴ Table for above operations

Hence, it is clear that f_{1} is identity function of f_{1}, f_{2}, f_{3}, f_{4}.

Hence, inverse element is itself too.

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