2nd PUC Maths Question Bank Chapter 1 Relations and Functions Ex 1.4

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Karnataka 2nd PUC Maths Question Bank Chapter 1 Relations and Functions Ex 1.4

2nd PUC Maths Relations and Functions NCERT Text Book Questions and Answers Ex 1.4

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
Answer:
2 ∈ Z+, 5 ∈ Z+, 2 – 5 = -3 ∉ Z+
hence not a binary operation.

(ii) On Z+, define * by a * b =ab
Answer:
∀ a ∈ Z+, b ∈ Z+ ab ∈ Z+, hence a * b ∈ Z+
∴ * is a binary operation.

(iii) On R, define * by a*b = ab2
Answer:
a ∈ R, b ∈ R then ab2 ∈ R
∴ a*b ∈ R
∴ * is a binary operation.

(iv) On Z+, define * by a * 6 = |a – b|
Answer:
a ∈ Z+, b ∈ Z+, |a – b| ∈ Z+
∴ a * b ∈ Z+
* is a binary operation.

(v) On Z+, define* by a * b = a
Answer:
a ∈ Z+, b ∈ Z+ then a ∈ Z+
∴ a * b ∈ Z+
∴ * is a binary operation.

KSEEB Solutions

Question 2.
For each binary * operation defined below,determine * whether is commutative or associative.

(i) On Z, define a * b = a – b
Answer:
a = 3, b = 2, a – b = 3 – 2= 1
b-a = 2-3 = -1
a – b ≠ b – a, hence not commutative
a = 3, b = 2, c = 1
(a * b) * c = (3 – 2) * 1 = 1 * 1 = 0
a*(b * c) = 3* (2 * 1) = 3*(2 – 1) = 3 * 1 = 3 – 1 = 2
(a * b) * c ≠ a * (b * c), hence * is not associative

(ii) On Q, define a*b = ab + 1
Ans:
a ∈ Q, b ∈ Q
a * b = ab + 1
b * a = ab + 1
∴  a * b = b * a, hence commutative ( 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)
hence * is not associative

(iii) On Q,define \(a * b=\frac{a b}{2}\)
Answer:
2nd PUC Maths Question Bank Chapter 1 Relations and Functions Ex 1.4 1

(iv) On Z+, define a*b = 2“*
Answer:
a g Z+, be Z+,
a*b = 2ab
b * a = 2ba = 2ab              a * b = b * a
hence * is commutative
(a * b) * c = 2ab * c = 2ab * c = 22alx
a*(b*c) = a* 2ac = 2a2
hence (a * b) * c * a * (b * c)
.: * is not Associative.

(v) On Z+, define a * b = ab
Answer:
a ∈ Z+, b ∈ Z+, a * b = ab
b * a = ba
a * b ≠ b * a, hence * is not commutative
(a * b) * c = ab * c = (ab)c = abc
a* (b * c) = a * bc = (a)bc
(a * b) * c ≠ a* (b * c)
hence * is not associative.

KSEEB Solutions

(vi) On R – {- 1}, define \(a * b=\frac{a}{b+1}\)
Answer:
2nd PUC Maths Question Bank Chapter 1 Relations and Functions Ex 1.4 2

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 ∧
Answer:

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 (Table 1.2).
(i) Compute (2 * 3) * 4 and 2 * (3 * 4)
(ii) * Is commutative?
(iii) Compute (2 * 3)*(4 * 5).
(Hint: use the following table)

Table 1.2

* 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

Answer:
(i) (2 * 3) * 4 = 1 * 4= 1
2 *  (3 * 4) = 2 * 1 = 1

(ii) Since the table is symmetric with main diagonal it is commutative.

(iii) (2 * 3) * (4 * 5) = 1 * 1 = 1.

KSEEB Solutions

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.
Answer:
a, b ∈ {1,2, 3,4, 5}, a* b = H.C.F. of a and b we composition table for *’

*’ 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

Both the composition tables are exactly same hence the operation* and *’ are same.

Question 6.
Let* be the binary operation on N given by a * b = L.C.M. of a and 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* ?
Answer:
(i) 5 *7 = L.C.M. of 5, 7 = 35
20 *16 = L.C.M. of 20, 16 = 80.

(ii) a * b = L.C.M. of (a, b) = L.C.M. (b,a)
= b * a
hence * is commutative.

(iii) (a * b) * c = L.C.M. of a, b, c
a* (b*c) = L.C.M. of a, b, c
hence * is associative.

(iv) Let ‘e’ be the identity element, than
a * e = e * a = a ∀ a ∈ N
L.C.M. (a, e) = L.C.M. (e, a) = a
⇒ e divides a ∀a ∈N
⇒ e = 1
∴ 1 is the identity element.

(v) Let a ∈ N be invertible.
∴  be N such that a * b = b * a = L.C.M. (a, b) = 1
⇒ a = 1, b = 1
∴ only invertible element in N is 1

KSEEB Solutions

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.
Answer:
L.C.M. of 3, 5 = 15
3 * 5 = L.C.M. of (3, 5) = 15 g {1,2, 3,4,5}
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?
Answer:
a ∈ N, b ∈ N, a ∈ b = H.C.F. of a and b
= H.C.F. of b and a = b * a
hence * is commutative
a ∈ N, b ∈ N, e ∈ N.
a* (b * e) = H.C.F. a, b, c
= (a * b) * c * is associative.
Let e ∈ N be the identity element
a * e = e * a = a
H.C.F. a and e = a
∴  a divides e ∀ a ∈ N
which is not possible as there does not exist such a number e, which is divisible by every number.
∴ identity element does not exist.

Question 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
Answer:
a * b = a – b,∀ a> be Q
a * b = a – b * a = b – a a * b ≠ b * a
hence 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 ≠ is not associative

KSEEB Solutions

(ii) a * b = a2 + b2
Answer:
a, b ∈ Q ⇒ a * b = a2 + b2
= b2 + a2 = b * a
hence * is commutative
(a * b) * c = (a2 + b2) * c2 = (a2 + b2)2 + c2
a*(b * c) = a* (b2 + c2) = a2 + (b2 + c2)2
hence * is not associative

(iii) a * b = a + ab
Answer:
a * b = a + ab
b * a = b + ab          ∴ a * b ≠ b * a,
hence * is not commutative
(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 + bca
≠ (a * b) * c
hence * is not associative

(iv) a * b = (a – b)2
Answer:
a * b = (a – b)2 = (b – a)2= b * a
hence * is not commutative
(a * b) * c = (a – b)2 * c = ((a-b)2 – c)2 = (a – b)4+ c2 – 2c (a – b)2
a*(b * c) = a* (b – c)2 = [a -(b – c)]2 = a2 + (b – c)2 – 2a (b – c)2
hence not associative

KSEEB Solutions

(v) \(a * b=\frac{a b}{4}\)
Answer:
2nd PUC Maths Question Bank Chapter 1 Relations and Functions Ex 1.4 3
2nd PUC Maths Question Bank Chapter 1 Relations and Functions Ex 1.4 4

(vi) a*b = ab2
Answer:
a ∈ Q, be Q , a * b = ab2
b * a = ba2 ∴ a * b≠ b * a
hence not commutative.
(a * b) * c = (ab2) * c = ab2c2
a * (b * c) = a * (bc2) = ab2c4
(a * b) * c = a * (b)
∴ * is not associative.

Question 10.
Show that none of the operations given above has identity.
(i) Let ‘e’ be the identity in Q such that
Answer:
a * e = e * a = a
a – e = a = e – a
⇒ e = 0 or e = 2a ∀ a ∈ Q
which is not possible. Hence no identity.

(ii) a * b = a2 + b2
Answer:
Let e ∈ Q be the identity element
a * e = a2 + e2 =a e * a = c2 + a2 = a
which is not possible as a2 ≠ a ∀ a ∈ Q
hence identity element doesn’t exist.

(iii) a*b = a + ab
Answer:
Let e ∈ Q be the identity element
a * e = a + ae = a ⇒ ae = 0, ⇒ e = 0
e * a = e + ae = a ⇒ \(e=\frac{a}{1+a}\)
∴\(\mathrm{e}=0 \text { and } \mathrm{e}=\frac{\mathrm{a}}{1+\mathrm{a}} \forall \mathrm{a} \in \mathrm{Q} \) which is not possible. Hence identity element does not exist.

(iv) a * b = (a – b)2
Answer:
Let e be the identity element
a * e = (a – e)2 = a
e * a = (e – a)2 = a
∴ a – e = ≠ a, e – a = ± a
⇒ e = 0 and e = 2a ∀ a ∈ Q which is not possible.
Hence no identity element.

KSEEB Solutions

(v) \(a * b = \frac{a b}{4}\)
Answer:
Let e be the identity element
\(a * e=\frac{a e}{4}=a \text { and } e * a=\frac{e a}{4}=a\)
∴ e = 4
∴ 4 is the identity element.

(vi) a*b = ab2
Answer:
Let e be the identity element
a * e = ae2 = a
e * a = ea2 = a
∴\( \mathrm{e}^{2}=1 \text { and } \mathrm{e}=\frac{1}{\mathrm{a}} \quad \forall \mathrm{a} \in \mathrm{Q}\)
which is not possible. Hence identity element does not exist.

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. Answer:
(a, b) * (c,d) = (a + c, b + d)
= (c + a, d + b) = (c, d) * (a, b)
hence commutative
[(a,b) * (c,d)] * (e, f) = (a + c, b + d) * (e, f) = (a + c + e, b + d + f) = (a, b) * [(c,d) * (e, f)] hence * is associative.
Let (e, f) be the identity element of A
(e, f) * (a, b) = (a,b) * (c, f) = (a, b)
(a + e, b + 0 = (a, b) ⇒ (e = 0, f = 0) ∉ N
hence no identity element.

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.
Answer:
If a * b = a+ b, then
a * a = a + a = 2a ≠ a
∴  statement is false

(ii) If * is a commutative binary operation N, then a*(b*c) = (c*b) * a
Answer:
Since commutative a* (b*c) = (b*c) *a
= (c *b) * a
hence true.

(iii) For every binary operation defined on a set having exactly one element a is necessarily commutative and associative .with a as the identity element and a being the inverse of a.
2nd PUC Maths Question Bank Chapter 1 Relations and Functions Ex 1.4 5
Answer:
A = {a}
* : A x A → A defined by a* a = a ∀ a ∈ A Also from the table it is commutative and associative. Also a is the identity element and also inverse of a .

KSEEB Solutions

Question 13.
Consider a binary operation * on N defined as a * b = a3 + b3. Choose the
(A) *Is both associative and commutative?
(B) *Is commutative but not associative?
(c) *Is associative but not commutative?
(D) *Is neither commutative nor associative?
Answer:
a*b = a3 + b3 = b3 + a3 = b*a
hence commutative
(a*b) * c = (a3 + b3) *c = (a3 + b3)3 + c3
a * (b * c) = a * (b3 + c3) = a3 + (b3 + c3)3
hence not associative
“B “ is the correct answer.

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