OP Malhotra Class 12 Maths Solutions Chapter 3 Binary Operations Ex 3(a)

Utilizing OP Malhotra Class 12 Solutions Chapter 3 Binary Operations Ex 3(a) as a study aid can enhance exam preparation.

S Chand Class 12 ICSE Maths Solutions Chapter 3 Binary Operations Ex 3(a)

Question 1.
Let * be a binary operation on N given by a * b = HCF (a, b), a, b ∈N. Write the value of 22 * 4.
Solution:
(i) Given * be a binary operation on N
given by a * b = HCF(a, b) ∀a,b ∈ N
∴ 22 * 4 = HCF (22, 4) = 2

Question 2.
Let * be a binary operation defined by a * b = 2a + b – 3, find 3 * 4.
Solution:
Given * be a binary operation defined by
a * b = 2a + b – 3
∴ 3 * 4 = 2 x 3 + 4 – 3
= 6 + 4 – 3 = 7.

Question 3.
If a * b = 3a + 4b – 2, find the value of 8 * 2.
Solution:
Given a * b = 3a + 4b – 2
∴ 8 * 2 = 3 * 8 + 4 * 2 – 2
= 24 + 8 – 2 = 30.

Question 4.
If the binary operation * on the set of integers Z, is defined by a * b = a + 3b², then find the value 2 * 4.
Solution:
Given binary operation * on the set of integers Z is defined by
a * b = a + 3b²
∴ 2 * 4 = 2 + 3 * 4² = 2 + 48 = 50

Question 5.
Let * be a binary operation on N given by a* b = LCM of (a, b) for a, b ∈ N. Find 5 * 7.
Solution:
Let * be a binary operation on N given by
a* b = LCM of (a, b)
Thus, 5 * 7 = LCM of (5, 7) = 35

Question 6.
The binary operation * : R * R → R is defined as a * b = 2a + b. Find (2 * 3) * 4.
Solution:
The binary operation * : R * R → R is defined as
a * b = 2a + b
∴ (2 * 3) * 4 = (2 * 2 + 3) * 4
= 7 * 4 = 2 * 7 + 4 = 18

Question 7.
Let * be a binary operation, on the set of all non-zero numbers, given by a * b = \(\frac { ab }{ 5 }\) for all a, b ∈ R – {0}. Find the value of *, given that 2* (** 5) 10.
Solution:
Let * be a binary operation on set of all non-zero numbers given by
a * b = \(\frac { ab }{ 5 }\) ∀a, b ∈ R – {0}
and 2 * (x * 5) = 10
⇒ 2 * \(\frac { 5x }{ 5 }\) = 10 ⇒ 2 * x = 10
⇒ \(\frac { 2x }{ 5 }\) = 10 ⇒ x = 25

Question 8.
Check whether the following are binary operation or not.
(i) S = {0, 1} under
(a) multiplication, (b) addition
(ii) S = {1, 3, 5, 7,….} undergo
(a) multiplication (b) addition
(iii) S = {2, 4, 6, 8,…} under
(a) multiplication (b) addition
Solution:
(i) Given S = {0, 1}
Since 0 x 0 = 0; 0 x 1 = 1 x o = 0; 1 x 1 = 1
Thus the operation x is defined by
a x b = ab ∈ S ∀a, b ∈ S
∴ operation * is binary under multiplication.
Since 1 * 1 = 1 + 1 = 2 ∉ S
∴ operation * is defined by
a * b = a + b ∉ S ∀ a b ∈ S
Thus * is not a binary operation under addition.

(ii) Given S = {1, 3, 5, 7,….}
we know that product of two odd natural numbers is an odd natural number
∴ a * b = a * b ∈S
Thus * is a binary operation under multiplication.
We know that sum of two odd numbers is even.
Now 1 * 3 = 1 + 3 = 4 ∉ S,
where 1, 3 ∈ S
Thus * is not a binary operation under addition.

(iii) Given S = {2, 4, 6, 8,…}
clearly product of two even numbers be an even number
∴ a * b = a * b ∈ S
Thus * is a binary operation under multiplication in given set S.
clearly sum of two even nuumbers be an even number
∴ a * b = a + b ∈ S
Thus * is a binary operation under addition in S.

Question 9.
Examine whether the following are binary operations.
(i) Subtraction on N.
(ii) Addition on irrationals.
(iii) * on R defined by a * b = max {a, b)
(iv) * on R defined by a* b = min {a, b)
Solution:
(i) Clearly operation * on N defined by
a * b = a – b ∀a b ∈N
Now 1, 2 ∈ N but 1 * 2 = 1 – 2 = – 1 ∉ N
Thus * is not a binary operation under subtraction in N.

(ii) Let I_R be the set of all irrational numbers
An operation * on I_R defined by
a * b = a + b ∈ a b ∈ I_R
Here \(\sqrt{2}\), – \(\sqrt{2}\) ∈ I_R.
But \(\sqrt{2}\) + (-\(\sqrt{2}\)) = 0 ∉ I_R
Thus, * is not a binary operation under addition an irrational numbers.

(iii) Given * be an operation on R defined by
a* b = max {a, b}
Since a* b = max {a, b} be either a or b ∈ R ∀ a, b ∈ R
Thus * be an binary operation.

(iv) ∀ a, b ∈ R s.t a * b = min {a, b} be
either a or b ∀ a, b ∈ R
Thus * is binary operation on R.

Question 10.
Show that:
(i) Multiplication is a binary operation on S = {1, -1} but not on 7= {-1, 2};
(ii) Addition is a binary relation on S = {x : x ∈ Z, x < 0} but multiplication is not.
Solution:
(i) Given S = {1, -1}
Since 1 x (- 1) = – 1 ; – 1 x 1 = – 1 ;
1 x 1 = 1 & (-1) x (- 1) = 1
Thus the operation * on S defined by
a * b = a * b ∀ a b∈S
∴ given operation is binary under multiplication on S.
– 1 x 2 = – 2 ∉ T
a * b = a * b ∉ T.
Thus the operation * on T is not a binary operation under multiplication.

(ii) Since sum of two negative integers is a negative integer
∴ a * b = a + b ∀ a b ∈ S
∀a b∈S then a + b∈S ⇒ a * b∈S
∴ addition is a binary operation S
∀ a b∈S, a * b = ab∉S
Since product of two negative integers is a positive integer, – 1, – 2∈S
but (- 1) * (- 2) = (- 1)(- 2) = 2∉S
Thus multiplication is not a binary operation on S.

Question 11.
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⁺, defined * by a * b = a – b
(ii) On Z⁺, defined * by a * b = ab
(iii) On R, defined by a * b = ab²
(iv) On Z⁺, defined * by a * b = |a – b|
(v) On R, defined * by a * b = |a – b|.
(vi) On Z⁺, defined * by a * b = a
(vii) On Z⁺, defined * by a* b = 2^ab.
Solution:
(i) Given * on Z⁺, defined by
a * b = a – b ∉ Z⁺.
As, 1, Z∈Z⁺² but 1 * 2 = 1 – 2 = – 1 ∉ Z⁺
∴ * is not a binary operation on Z⁺

(ii) Given operation * on Z⁺ defined by a * b = ab ∉ Z⁺
Since product of two positive integers is a positive integer.
Thus * be a binary operation on Z⁺

(iii) Given operation * on R defined by a * b = ab²
∀a b∈R ⇒ ab²∈R ⇒ a * b ∈R
Thus * is a binary operation on R.

(iv) Given operation * on Z⁺ defined by a * b = | a – b |
Now a, a∈Z⁺
a * a = | a – a | = 0 ∉ Z⁺
Thus a * b = | a – b | ∉ Z⁺
∴ * is not a binary operation on Z⁺.

(v) Given operation * on R defined by
a * b = | a – b |
∀a b ∈ R, a * b = |(a – b)|∈R
∴ * be a binary operation on R.

(vi) Operation * on Z⁺ defined by
a * b = a.
∀ a b∈Z⁺ s.t a * b = a∈Z⁺
Thus * be a binary operation on Z⁺

(vii) ∀ a, b ∈ Z⁺, a * b = 2ab ∉ Z⁺
[∵ a, b ∈ Z⁺ ⇒ ab∈ Z⁺ ∴ 2ab ∈ Z⁺]
Thus * be a binary operation on Z⁺.