Exercise 1.4 | Set 2 builds upon the concepts introduced in the first set, focusing on more advanced aspects of function composition and inverse functions. This set challenges students to apply their understanding to more complex scenarios, including composite functions with trigonometric, exponential, and logarithmic elements. It also delves deeper into the properties of inverse functions and their applications in solving equations. The problems in this set are designed to enhance students' analytical skills, preparing them for higher-level mathematical reasoning in calculus and beyond. By mastering these concepts, students develop a robust foundation for tackling intricate mathematical problems and gain insights into the interconnected nature of various functions.
Content of this article has been merged with Chapter 1 Relations And Functions - Exercise 1.4 as per the revised syllabus of NCERT.
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 = a2 + b2
(iii) a ∗ b = a + ab
(iv) a ∗ b = (a – b)2
(v) a ∗ b = ab / 4
(vi) a ∗ b = ab2
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 = a2 + b2
RHS = b * a = b2 + a2
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 * (b2 + c2) = a2 + (b2 + c2)2
RHS = (a * b) * c = (a2 + b2) * c = (a2 + b2)2 + c2
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 = ab2
RHS = b * a = ba2
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(bc2)2
RHS = (a * b) * c = (ab2) * c = ab2c2
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 = 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:
On N, * is defined as a * b = a3 + b3
Commutative:
If a, b belongs to Z, a * b = b * a
LHS = a * b = a3 + b3
RHS = b * a = b3 + a3
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 * (b3 + c3) = a3 + (b3 + c3)3
RHS = (a * b) * c = (a3 + b3) * c = (a3 + b3)3 + c3
Since, LHS is not equal to RHS
Therefore, * is not Associative
Thus, Option (B) is correct.
Summary
Exercise 1.4 | Set 2 elevates students' understanding of function composition and inverse functions to a more sophisticated level. By working through problems involving a diverse range of functions, including trigonometric, exponential, and rational functions, students develop a nuanced grasp of how different types of functions interact when composed. The set emphasizes the importance of determining domains and ranges in composite functions, a crucial skill in advanced mathematics. Additionally, it reinforces the concept of inverse functions, pushing students to not only find inverses but also to verify and apply them in various contexts. This comprehensive approach ensures that students are well-prepared for the rigors of higher mathematics, where these concepts form the bedrock of more advanced topics in calculus, complex analysis, and abstract algebra.