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

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

Ex 1.4 Class 12 Maths 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.

Ex 1.4 Class 12 Maths 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 = \ frac { ab }{ 2 }
∴ a * b = b * a
∴ Operation binary defined is commutative.
be abc
(b) a * (b * c) = a * \ frac { bc }{ 2 } = \ frac { abc }{ 4 } and
(a * b) * c = \ frac { bc }{ 2 } * c \ frac { abc }{ 4 }
=> (a * b) * c = \ frac { bc }{ 2 } * c \ frac { abc }{ 4 }
Defined binary operation is associative.

(iv) On Z+ operation * is defined as a * b = 2ab
(a) a * b = 2ab, b * a = 2ba = 2ab
=> a * b = b * a .
∴ Binary operation defined as commutative.
(b) a * (b * c) = a * 2ba = 2a.2bc
(a * b) * c = 2ab * c = 22ab
Thus (a * b) * c ≠ a * (b * c)
∴ Binary operation * as defined as is not associative.

(v) On Z+, a * b = ab
(a) b * a = ba .
∴ ab ≠ ba = a * b ≠ b * a.
* is not commutative.
(b) (a*b)*c = ab*c = (ab)c = abc a*(b* c)
= a*bc = abc
Thus (a * b) * c ≠ (a * b * c)
∴ Operation * is not associative.

(vi) Neither commutative nor associative.

Ex 1.4 Class 12 Maths 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.
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions 3

Ex 1.4 Class 12 Maths 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)
tiwari academy class 12 maths Chapter 1 Relations and Functions 4
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.

Ex 1.4 Class 12 Maths 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 *.
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions 5

Ex 1.4 Class 12 Maths 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 *

Ex 1.4 Class 12 Maths 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.

Ex 1.4 Class 12 Maths 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.

Ex 1.4 Class 12 Maths 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.

Ex 1.4 Class 12 Maths 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 = \ frac { ab }{ 4 }
∴ a * e = \ frac { ae }{ 4 } ≠ 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.

Ex 1.4 Class 12 Maths 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.

Ex 1.4 Class 12 Maths 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.

Ex 1.4 Class 12 Maths 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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