NCERT Solutions Class 12 maths Chapter-1 (Relation And Functions)Exercise 1.4
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.
Exercise 1.4
Set-1
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
Solution:
If a, b belongs to Z+
a * b = a – b which may not belong to Z+
For eg: 1 – 3 = -2 which doesn’t belongs to Z+
Therefore, * is not a Binary Operation on Z+
(ii) On Z+, define * by a * b = ab
Solution:
If a, b belongs to Z+
a * b = ab which belongs to Z+
Therefore, * is Binary Operation on Z+
(iii) On R, define * by a * b = ab²
Solution:
If a, b belongs to R
a * b = ab^{2 } which belongs to R
Therefore, * is Binary Operation on R
(iv) On Z+, define * by a * b = |a – b|
Solution:
If a, b belongs to Z+
a * b = |a – b| which belongs to Z+
Therefore, * is Binary Operation on Z+
(v) On Z+, define * by a * b = a
Solution:
If a, b belongs to Z+
a * b = a which belongs to Z+
Therefore, * is Binary Operation on Z+
Question 2:
For each binary operation * defined below, determine whether * is binary, commutative or associative.
(i) On Z, define a * b = a – b
Solution:
a) Binary:
If a, b belongs to Z
a * b = a – b which belongs to Z
Therefore, * is Binary Operation on Z
b) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = a – b
RHS = b * a = b – a
Since, LHS is not equal to RHS
Therefore, * is not Commutative
c) Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a – b + c
RHS = (a – b) * c = a – b- c
Since, LHS is not equal to RHS
Therefore, * is not Associative
(ii) On Q, define a * b = ab + 1
Solution:
a) Binary:
If a, b belongs to Q, a * b = ab + 1 which belongs to Q
Therefore, * is Binary Operation on Q
b) Commutative:
If a, b belongs to Q, a * b = b * a
LHS = a * b = ab + 1
RHS = b * a = ba + 1 = ab + 1
Since, LHS is equal to RHS
Therefore, * is Commutative
c) Associative:
If a, b, c belongs to Q, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (bc + 1) = abc + a + 1
RHS = (a * b) * c = abc + c + 1
Since, LHS is not equal to RHS
Therefore, * is not Associative
(iii) On Q, define a ∗ b = ab/2
Solution:
a) Binary:
If a, b belongs to Q, a * b = ab/2 which belongs to Q
Therefore, * is Binary Operation on Q
b) Commutative:
If a, b belongs to Q, a * b = b * a
LHS = a * b = ab/2
RHS = b * a = ba/2
Since, LHS is equal to RHS
Therefore, * is Commutative
c) Associative:
If a, b, c belongs to Q, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (bc/2) = (abc)/2
RHS = (a * b) * c = (ab/2) * c = (abc)/2
Since, LHS is equal to RHS
Therefore, * is Associative
(iv) On Z+, define a * b = 2^{ab}
Solution:
a) Binary:
If a, b belongs to Z+, a * b = 2^{ab} which belongs to Z+
Therefore, * is Binary Operation on Z+
b) Commutative:
If a, b belongs to Z+, a * b = b * a
LHS = a * b = 2^{ab}
RHS = b * a = 2^{ba} = 2^{ab}
Since, LHS is equal to RHS
Therefore, * is Commutative
c) Associative:
If a, b, c belongs to Z+, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * 2^{bc = }2^{a * 2^(bc)}
RHS = (a * b) * c = 2^{ab} * c = 2^{2abc}
Since, LHS is not equal to RHS
Therefore, * is not Associative
(v) On Z+, define a * b = a^{b}
Solution:
a) Binary:
If a, b belongs to Z+, a * b = a^{b} which belongs to Z+
Therefore, * is Binary Operation on Z+
b) Commutative:
If a, b belongs to Z+, a * b = b * a
LHS = a * b = a^{b}
RHS = b * a = b^{a}
Since, LHS is not equal to RHS
Therefore, * is not Commutative
c) Associative:
If a, b, c belongs to Z+, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * b^{c} = a^{b^c}
RHS = (a * b) * c = a^{b} * c = a^{bc}
Since, LHS is not equal to RHS
Therefore, * is not Associative
(vi) On R – {– 1}, define a ∗ b = a / (b + 1)
Solution:
a) Binary:
If a, b belongs to R, a * b = a / (b+1) which belongs to R
Therefore, * is Binary Operation on R
b) Commutative:
If a, b belongs to R, a * b = b * a
LHS = a * b = a / (b + 1)
RHS = b * a = b / (a + 1)
Since, LHS is not equal to RHS
Therefore, * is not Commutative
c) Associative:
If a, b, c belongs to A, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * b / (c+1) = a(c+1) / b+c+1
RHS = (a * b) * c = (a / (b+1)) * c = a / (b+1)(c+1)
Since, LHS is not equal to RHS
Therefore, * 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 | 2 | 3 | 4 | 5 |
1 | 1 | 1 | 1 | 1 | 1 |
2 | 1 | 2 | 2 | 2 | 2 |
3 | 1 | 2 | 3 | 3 | 3 |
4 | 1 | 2 | 3 | 4 | 4 |
5 | 1 | 2 | 3 | 4 | 5 |
Question 4:
Consider a binary operation ∗ on the set {1, 2, 3, 4, 5} given by the following multiplication table.
(Hint: use the following table)
* | 1 | 2 | 3 | 4 | 5 |
1 | 1 | 1 | 1 | 1 | 1 |
2 | 1 | 2 | 1 | 2 | 1 |
3 | 1 | 1 | 3 | 1 | 1 |
4 | 1 | 2 | 1 | 4 | 1 |
5 | 1 | 1 | 1 | 1 | 5 |
(i) Compute (2 ∗ 3) ∗ 4 and 2 ∗ (3 ∗ 4)
Solution:
Here, (2 * 3) * 4 = 1 * 4 = 1
2 * (3 * 4) = 2 * 1 = 1
(ii) Is ∗ commutative?
Solution:
The given composition table is symmetrical about the main diagonal of table. Thus, binary operation ‘*’ is commutative.
(iii) Compute (2 ∗ 3) ∗ (4 ∗ 5).
Solution:
(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 Exercise 4 above? Justify your answer.
Solution:
Let A = {1, 2, 3, 4, 5} and a ∗′ b = HCF of a and b.
*’ | 1 | 2 | 3 | 4 | 5 |
1 | 1 | 1 | 1 | 1 | 1 |
2 | 1 | 2 | 1 | 2 | 1 |
3 | 1 | 1 | 3 | 1 | 1 |
4 | 1 | 2 | 1 | 4 | 1 |
5 | 1 | 1 | 1 | 1 | 5 |
We see that the operation *’ is the same as the operation * in Exercise 4 above.
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
Solution:
If a, b belongs to N
a * b = LCM of a and b
5 * 7 = 35
20 * 16 = 80
(ii) Is ∗ commutative?
Solution:
If a, b belongs to N
LCM of a * b = ab
LCM of b * a = ba = ab
a*b = b*a
Thus, * binary operation is commutative.
(iii) Is ∗ associative?
Solution:
a * (b * c) = LCM of a, b, c
(a * b) * c = LCM of a, b, c
Since, a * (b * c) = (a * b) * c
Thus, * binary operation is associative.
(iv) Find the identity of ∗ in N
Solution:
Let ‘e’ is an identity
a * e = e * a, for a belonging to N
LCM of a * e = a, for a belonging to N
LCM of e * a = a, for a belonging to N
e divides a
e divides 1
Thus, e = 1
Hence, 1 is an identity element
(v) Which elements of N are invertible for the operation ∗?
Solution:
a * b = b * a = identity element
LCM of a and b = 1
a = b = 1
only ‘1’ is invertible element in N.
Exercise 1.4
Set-2
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 operation * on the set {1, 2, 3, 4, 5} is defined as
a * b = L.C.M. of a and b
Let a=3, b=5
3 * 5 = 5 * 3 = L.C.M. of 3 and 5 = 15 which does not belong to the given set
Thus, * 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?
Solution:
If a, b belongs to N
LHS = a * b = HCF of a and b
RHS = b * a = HCF of b and a
Since LHS = RHS
Therefore, * is Commutative
Now, If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = HCF of a, b and c
RHS = (a – b) * c = HCF of a, b and c
Since, LHS = RHS
Therefore, * is Associative
Now, 1 * a = a * 1 ≠ a
Thus, there doesn’t exist any identity element.
Question 9:
Let ∗ be a binary operation on the set Q of rational numbers as follows:
(i) a ∗ b = a – b
(ii) a ∗ b = a^{2} + b^{2}
(iii) a ∗ b = a + ab
(iv) a ∗ b = (a – b)^{2}
(v) a ∗ b = ab / 4
(vi) a ∗ b = ab^{2}
Find which of the binary operations are commutative and which are associative.
Solution:
(i) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = a – b
RHS = b * a = b – a
Since, LHS is not equal to RHS
Therefore, * is not Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a – (b – c) = a – b + c
RHS = (a – b) * c = a – b – c
Since, LHS is not equal to RHS
Therefore, * is not Associative
(ii) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = a^{2} + b^{2}
RHS = b * a = b^{2} + a^{2}
Since, LHS is equal to RHS
Therefore, * is Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (b^{2 }+ c^{2}) = a^{2 }+ (b^{2} + c^{2})^{2}
RHS = (a * b) * c = (a^{2} + b^{2}) * c = (a^{2} + b^{2})^{2} + c^{2}
Since, LHS is not equal to RHS
Therefore, * is not Associative
(iii) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = a + ab
RHS = b * a = b + ba
Since, LHS is not equal to RHS
Therefore, * is not Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (b + bc) = a + a(b + bc)
RHS = (a * b) * c = (a + ab) * c = a + ab + (a + ab)c
Since, LHS is not equal to RHS
Therefore, * is not Associative
(iv) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = (a – b)^{2}
RHS = b * a = (b – a)^{2}
Since, LHS is not equal to RHS
Therefore, * is not Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (b – c)^{2} = [a – (b – c)^{2}]^{2}
RHS = (a * b) * c = (a – b)^{2} * c = [(a – b)^{2} – c]^{2}
Since, LHS is not equal to RHS
Therefore, * is not Associative
(v) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = ab / 4
RHS = b * a = ba / 4
Since, LHS is equal to RHS
Therefore, * is Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * bc/4 = abc/16
RHS = (a * b) * c = ab/4 * c = abc/16
Since, LHS is equal to RHS
Therefore, * is Associative
(vi) Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = ab^{2}
RHS = b * a = ba^{2}
Since, LHS is not equal to RHS
Therefore, * is not Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (bc)^{2} = a(bc^{2})^{2}
RHS = (a * b) * c = (ab^{2}) * c = ab^{2}c^{2}
Since, LHS is not equal to RHS
Therefore, * is not Associative
Question 10:
Find which of the operations given above has identity
Solution:
An element e ∈ Q will be the identity element for the operation * if
a * e = a = e * a, for a ∈ Q
for (v) a * b = ab/4
Let e be an identity element
a * e = a = e * a
LHS : ae/4 = a
=> e = 4
RHS : ea/4 = a
=> e = 4
LHS = RHS
Thus, Identity element exists
Other operations doesn’t satisfy the required conditions.
Hence, other operations doesn’t have identity.
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:
Given (a, b) * (c, d) = (a+c, b+d) on A
Let (a, b), (c, d), (e,f) be 3 pairs ∈ A
Commutative :
LHS = (a, b) * (c, d) = (a+c, b+d)
RHS = (c, d) * (a, b) = (c+a, d+b) = (a+c, b+d)
Since, LHS is equal to RHS
Therefore, * is Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = (a, b) * [(c, d) * (e, f)] = (a, b) * (c+e, d+f) = (a+c+e, b+d+f)
RHS = [(a, b) * (c, d)] * (e, f) = (a+c, b+d) * (e, f) = (a+c+e, b+d+f)
Since, LHS is equal to RHS
Therefore, * is Associative
Existence of Identity element:
For a, e ∈ A, a * e = a
(a, b) * (e, e) = (a, b)
(a+e, b+e) = (a, b)
a + e = a
=> e = 0
b + e = b
=> e = 0
As 0 is not a part of set of natural numbers. So, identity function does not exist.
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) Let * be an operation on N, defined as:
a * b = a + b ∀ a, b ∈ N
Let us consider b = a = 6, we have:
6 * 6 = 6 + 6 = 12 ≠ 6
Therefore, this statement is false.
(ii) Since, * is commutative
LHS = a ∗ (b ∗ c) = a * (c * b) = (c * b) * a = RHS
Therefore, this statement is true.
Question 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?
Solution:
On N, * is defined as a * b = a^{3} + b^{3}
Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = a^{3} + b^{3}
RHS = b * a = b^{3} + a^{3}
Since, LHS is equal to RHS
Therefore, * is Commutative
Associative:
If a, b, c belongs to Z, a * (b * c) = (a * b) * c
LHS = a * (b * c) = a * (b^{3} + c^{3}) = a^{3} + (b^{3} + c^{3})^{3}
RHS = (a * b) * c = (a^{3} + b^{3}) * c = (a^{3} + b^{3})^{3} + c^{3}
Since, LHS is not equal to RHS
Therefore, * is not Associative
Thus, Option (B) is correct.
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