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

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

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

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

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

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

(i) On Z, define a * b = a – b
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 $$a * b=\frac{a b}{2}$$

(iv) On Z+, define a*b = 2“*
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
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.

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

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 ∧

 ∧ 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

(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.

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.
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* ?
(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

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.
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?
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
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

(ii) a * b = a2 + b2
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
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
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

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

(vi) a*b = ab2
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
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
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
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
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.

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

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 .

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?