All Exams >
Civil Engineering (CE) >
Engineering Mathematics >
All Questions
All questions of Set Theory & Algebra for Civil Engineering (CE) Exam
How many properties can be held by a group?- a)2
- b)3
- c)5
- d)4
Correct answer is option 'C'. Can you explain this answer?
How many properties can be held by a group?
a)
2
b)
3
c)
5
d)
4
|
Yash Yoyo answered |
Group does not contain commutative prorperty
therefore answer D is correct
therefore answer D is correct
If a relation R on the set {1, 2, 3} be defined by R = {(1, 2)}, then R is- a)reflexive
- b)transitive
- c)symmetric
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
If a relation R on the set {1, 2, 3} be defined by R = {(1, 2)}, then R is
a)
reflexive
b)
transitive
c)
symmetric
d)
none of these
|
Swati Patel answered |
Explanation:
To determine if the relation R on the set {1, 2, 3} defined by R = {(1, 2)} is transitive, we need to check if for every (a, b) and (b, c) in R, the pair (a, c) is also in R.
In this case, R = {(1, 2)}. So, we need to check if (1, 2) and (2, 3) are both in R.
Since (1, 2) is in R, we check if (2, 3) is in R. However, there is no pair in R with 2 as the first element. Therefore, (1, 2) and (2, 3) are not both in R, and the relation R is not transitive.
Hence, the correct answer is option 'B' - R is not transitive.
To determine if the relation R on the set {1, 2, 3} defined by R = {(1, 2)} is transitive, we need to check if for every (a, b) and (b, c) in R, the pair (a, c) is also in R.
In this case, R = {(1, 2)}. So, we need to check if (1, 2) and (2, 3) are both in R.
Since (1, 2) is in R, we check if (2, 3) is in R. However, there is no pair in R with 2 as the first element. Therefore, (1, 2) and (2, 3) are not both in R, and the relation R is not transitive.
Hence, the correct answer is option 'B' - R is not transitive.
A relation R is said to be circular if aRb and bRc together imply cRa. Which of the following options is/are correct?- a)If a relation S is transitive and circular, then S is an equivalence relation.
- b)If a relation S is reflexive and symmetric, then S is an equivalence relation.
- c)if a relation S is reflexive and circular, then S is an equivalence relation.
- d)if a relation S is circular and symmetric, then S is an equivalence relation.
Correct answer is option 'C'. Can you explain this answer?
A relation R is said to be circular if aRb and bRc together imply cRa. Which of the following options is/are correct?
a)
If a relation S is transitive and circular, then S is an equivalence relation.
b)
If a relation S is reflexive and symmetric, then S is an equivalence relation.
c)
if a relation S is reflexive and circular, then S is an equivalence relation.
d)
if a relation S is circular and symmetric, then S is an equivalence relation.
|
Sravya Rane answered |
Understanding Circular Relations
A circular relation R implies that if aRb and bRc hold true, then it must also follow that cRa. This property fundamentally shapes how we assess the nature of the relation.
Examining Option C: Reflexive and Circular
- Reflexive: A relation S is reflexive if for every element a in the set, aSa holds true.
- Circular: Given the definition, if we have aSa and aSb, then we can derive that bSa through the circular property.
Proof of Equivalence
1. Reflexivity: Since S is reflexive, for any element a, aSa holds true.
2. Symmetry: If aSb is true, then by the circular property, bSa must also hold true.
3. Transitivity: Assume aSb and bSc are true. By the circular property, we can derive cSa.
Since S satisfies all three properties—reflexivity, symmetry, and transitivity— it qualifies as an equivalence relation.
Why Other Options are Incorrect
- Option A (Transitive and Circular): A circular relation does not guarantee reflexivity or symmetry, which are essential for equivalence.
- Option B (Reflexive and Symmetric): Although these properties hold, transitivity is not ensured by circularity alone.
- Option D (Circular and Symmetric): Without reflexivity, we cannot conclude that S is an equivalence relation.
Conclusion
Thus, among the options provided, only option C correctly proves that a relation S, when both reflexive and circular, is indeed an equivalence relation.
A circular relation R implies that if aRb and bRc hold true, then it must also follow that cRa. This property fundamentally shapes how we assess the nature of the relation.
Examining Option C: Reflexive and Circular
- Reflexive: A relation S is reflexive if for every element a in the set, aSa holds true.
- Circular: Given the definition, if we have aSa and aSb, then we can derive that bSa through the circular property.
Proof of Equivalence
1. Reflexivity: Since S is reflexive, for any element a, aSa holds true.
2. Symmetry: If aSb is true, then by the circular property, bSa must also hold true.
3. Transitivity: Assume aSb and bSc are true. By the circular property, we can derive cSa.
Since S satisfies all three properties—reflexivity, symmetry, and transitivity— it qualifies as an equivalence relation.
Why Other Options are Incorrect
- Option A (Transitive and Circular): A circular relation does not guarantee reflexivity or symmetry, which are essential for equivalence.
- Option B (Reflexive and Symmetric): Although these properties hold, transitivity is not ensured by circularity alone.
- Option D (Circular and Symmetric): Without reflexivity, we cannot conclude that S is an equivalence relation.
Conclusion
Thus, among the options provided, only option C correctly proves that a relation S, when both reflexive and circular, is indeed an equivalence relation.
A cyclic group is always _________- a)abelian group
- b)monoid
- c)semigroup
- d)subgroup
Correct answer is option 'A'. Can you explain this answer?
A cyclic group is always _________
a)
abelian group
b)
monoid
c)
semigroup
d)
subgroup
|
Sarthak Menon answered |
Cyclic Group:
A cyclic group is a type of mathematical group that is generated by a single element. In other words, every element in the group can be expressed as a power of a single element, called the generator. The generator can be any element within the group.
Abelian Group:
An abelian group, also known as a commutative group, is a group in which the group operation is commutative. This means that for any two elements a and b in the group, the product of a and b is the same as the product of b and a.
Explanation:
To determine whether a cyclic group is always an abelian group, we need to understand the properties of a cyclic group.
1. Cyclic Group:
- A cyclic group is generated by a single element, called the generator.
- Every element in the group can be expressed as a power of the generator.
- The generator can be any element within the group.
- The group operation is defined as multiplication or exponentiation of the generator.
2. Abelian Group:
- An abelian group is a group in which the group operation is commutative.
- For any two elements a and b in the group, the product of a and b is the same as the product of b and a.
- In other words, the order of the elements does not affect the result of the group operation.
3. Relation between Cyclic and Abelian Groups:
- Since a cyclic group is generated by a single element, the group operation is commutative.
- The generator is the only element in the group, and any combination of the generator will result in the same element.
- Therefore, every element in the cyclic group commutes with every other element.
- This property makes all cyclic groups abelian groups.
Conclusion:
Based on the properties of cyclic groups and abelian groups, we can conclude that a cyclic group is always an abelian group. The commutative property of the group operation in a cyclic group ensures that the order of the elements does not affect the result, making it an abelian group. Hence, option 'A' is the correct answer.
A cyclic group is a type of mathematical group that is generated by a single element. In other words, every element in the group can be expressed as a power of a single element, called the generator. The generator can be any element within the group.
Abelian Group:
An abelian group, also known as a commutative group, is a group in which the group operation is commutative. This means that for any two elements a and b in the group, the product of a and b is the same as the product of b and a.
Explanation:
To determine whether a cyclic group is always an abelian group, we need to understand the properties of a cyclic group.
1. Cyclic Group:
- A cyclic group is generated by a single element, called the generator.
- Every element in the group can be expressed as a power of the generator.
- The generator can be any element within the group.
- The group operation is defined as multiplication or exponentiation of the generator.
2. Abelian Group:
- An abelian group is a group in which the group operation is commutative.
- For any two elements a and b in the group, the product of a and b is the same as the product of b and a.
- In other words, the order of the elements does not affect the result of the group operation.
3. Relation between Cyclic and Abelian Groups:
- Since a cyclic group is generated by a single element, the group operation is commutative.
- The generator is the only element in the group, and any combination of the generator will result in the same element.
- Therefore, every element in the cyclic group commutes with every other element.
- This property makes all cyclic groups abelian groups.
Conclusion:
Based on the properties of cyclic groups and abelian groups, we can conclude that a cyclic group is always an abelian group. The commutative property of the group operation in a cyclic group ensures that the order of the elements does not affect the result, making it an abelian group. Hence, option 'A' is the correct answer.
Let, R = {(a, b): a,b ∈ Z and (a + b) is even}, then R is - a)Reflexive relation on Z
- b)Equivalence relation on Z
- c)transitive relation on Z
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
Let, R = {(a, b): a,b ∈ Z and (a + b) is even}, then R is
a)
Reflexive relation on Z
b)
Equivalence relation on Z
c)
transitive relation on Z
d)
None of the above
|
Anmol Choudhary answered |
Understanding the Relation R
R = {(a, b): a, b ∈ Z and (a + b) is even} defines a relation on the set of integers Z based on the evenness of the sum of two integers.
Evaluating Properties of R
1. Reflexive Relation:
- A relation is reflexive if for every element a in Z, (a, a) is in R.
- Here, (a + a) = 2a, which is always even.
- Thus, R is reflexive.
2. Symmetric Relation:
- A relation is symmetric if for every (a, b) in R, (b, a) is also in R.
- Since (a + b) is even, (b + a) = (a + b) is also even.
- Therefore, R is symmetric.
3. Transitive Relation:
- A relation is transitive if whenever (a, b) and (b, c) are in R, then (a, c) is also in R.
- If (a + b) is even and (b + c) is even, then:
- a + b = 2m (even)
- b + c = 2n (even)
- Adding these gives (a + c) + 2b = 2m + 2n, which is even, indicating (a + c) is even.
- Thus, R is transitive.
Conclusion: Equivalence Relation
Since R is reflexive, symmetric, and transitive, it satisfies all properties of an equivalence relation. Thus, the correct answer is option 'B', indicating that R is an equivalence relation on Z.
This classification helps in understanding the structure of integers when grouped by evenness of sums, providing insight into their mathematical behavior.
R = {(a, b): a, b ∈ Z and (a + b) is even} defines a relation on the set of integers Z based on the evenness of the sum of two integers.
Evaluating Properties of R
1. Reflexive Relation:
- A relation is reflexive if for every element a in Z, (a, a) is in R.
- Here, (a + a) = 2a, which is always even.
- Thus, R is reflexive.
2. Symmetric Relation:
- A relation is symmetric if for every (a, b) in R, (b, a) is also in R.
- Since (a + b) is even, (b + a) = (a + b) is also even.
- Therefore, R is symmetric.
3. Transitive Relation:
- A relation is transitive if whenever (a, b) and (b, c) are in R, then (a, c) is also in R.
- If (a + b) is even and (b + c) is even, then:
- a + b = 2m (even)
- b + c = 2n (even)
- Adding these gives (a + c) + 2b = 2m + 2n, which is even, indicating (a + c) is even.
- Thus, R is transitive.
Conclusion: Equivalence Relation
Since R is reflexive, symmetric, and transitive, it satisfies all properties of an equivalence relation. Thus, the correct answer is option 'B', indicating that R is an equivalence relation on Z.
This classification helps in understanding the structure of integers when grouped by evenness of sums, providing insight into their mathematical behavior.
Consider the following statements :
1. The set of all irrational numbers between 2 and 5 is an infinite set.
2. The set of all odd integers less than 100 is a finite set.
Which of the statements given above is/are correct?- a)1 only
- b)2 only
- c)Both 1 and 2
- d)Neither 1 nor 2
Correct answer is option 'A'. Can you explain this answer?
Consider the following statements :
1. The set of all irrational numbers between 2 and 5 is an infinite set.
2. The set of all odd integers less than 100 is a finite set.
Which of the statements given above is/are correct?
1. The set of all irrational numbers between 2 and 5 is an infinite set.
2. The set of all odd integers less than 100 is a finite set.
Which of the statements given above is/are correct?
a)
1 only
b)
2 only
c)
Both 1 and 2
d)
Neither 1 nor 2
|
|
Samarth Ghoshal answered |
Statement 1: The set of all irrational numbers between 2 and 5 is an infinite set.
An irrational number is a number that cannot be expressed as a fraction of two integers. It is a decimal that goes on forever without repeating.
To determine whether the set of all irrational numbers between 2 and 5 is infinite, we need to consider the nature of irrational numbers.
- There are infinitely many irrational numbers between any two rational numbers.
- The set of irrational numbers is not countable, meaning it cannot be listed in a finite or countable sequence.
Since the set of irrational numbers between 2 and 5 is a subset of the set of all irrational numbers, it follows that the set of irrational numbers between 2 and 5 is also infinite.
Therefore, statement 1 is correct.
Statement 2: The set of all odd integers less than 100 is a finite set.
To determine whether the set of all odd integers less than 100 is finite, we need to consider the definition of a finite set.
A finite set is a set that has a specific number of elements and can be counted or listed.
In this case, the set of all odd integers less than 100 includes numbers like 1, 3, 5, 7, 9, ... up to 99.
Since we can list or count all the elements in this set, it is a finite set.
Therefore, statement 2 is incorrect.
Conclusion:
- Statement 1 is correct as the set of all irrational numbers between 2 and 5 is infinite.
- Statement 2 is incorrect as the set of all odd integers less than 100 is a finite set.
Hence, the correct answer is option 'A' - 1 only.
An irrational number is a number that cannot be expressed as a fraction of two integers. It is a decimal that goes on forever without repeating.
To determine whether the set of all irrational numbers between 2 and 5 is infinite, we need to consider the nature of irrational numbers.
- There are infinitely many irrational numbers between any two rational numbers.
- The set of irrational numbers is not countable, meaning it cannot be listed in a finite or countable sequence.
Since the set of irrational numbers between 2 and 5 is a subset of the set of all irrational numbers, it follows that the set of irrational numbers between 2 and 5 is also infinite.
Therefore, statement 1 is correct.
Statement 2: The set of all odd integers less than 100 is a finite set.
To determine whether the set of all odd integers less than 100 is finite, we need to consider the definition of a finite set.
A finite set is a set that has a specific number of elements and can be counted or listed.
In this case, the set of all odd integers less than 100 includes numbers like 1, 3, 5, 7, 9, ... up to 99.
Since we can list or count all the elements in this set, it is a finite set.
Therefore, statement 2 is incorrect.
Conclusion:
- Statement 1 is correct as the set of all irrational numbers between 2 and 5 is infinite.
- Statement 2 is incorrect as the set of all odd integers less than 100 is a finite set.
Hence, the correct answer is option 'A' - 1 only.
Condition for monoid is __________- a)(a+e)=a
- b)(a*e)=(a+e)
- c)a=(a*(a+e)
- d)e)=ab) (a*e)=(a+e)c) a=(a*(a+e)
Correct answer is option 'D'. Can you explain this answer?
Condition for monoid is __________
a)
(a+e)=a
b)
(a*e)=(a+e)
c)
a=(a*(a+e)
d)
e)=ab) (a*e)=(a+e)c) a=(a*(a+e)
|
|
Gitanjali Menon answered |
Monoid is an algebraic structure that consists of a set together with an associative binary operation and an identity element. In order for a set to form a monoid, it must satisfy the following conditions:
1. Closure: For any two elements 'a' and 'b' in the set, their binary operation 'a * b' must also be in the set.
2. Associativity: For any three elements 'a', 'b', and 'c' in the set, the operation must satisfy the associative property, i.e., (a * b) * c = a * (b * c).
3. Identity element: There must exist an element 'e' in the set such that for any element 'a' in the set, the operation 'a * e' and 'e * a' both yield 'a'.
Let's analyze the given options:
a) (a * e) = a
This option represents the existence of an identity element 'e' in the set, which is required for a monoid.
b) (a * e) = (a * c)
This option represents the closure property, as it states that for any element 'a' and 'c' in the set, their binary operation yields the same result.
c) a = (a * (a * e))
This option represents the closure property, as it states that for any element 'a' in the set, the operation 'a * (a * e)' yields the same result as 'a'.
d) (a * e) = a and a = (a * (a * e))
This option represents both the identity element and closure property, as it satisfies conditions (a) and (c) mentioned above.
From the analysis, it is clear that option 'D' satisfies all the conditions required for a monoid, making it the correct answer.
1. Closure: For any two elements 'a' and 'b' in the set, their binary operation 'a * b' must also be in the set.
2. Associativity: For any three elements 'a', 'b', and 'c' in the set, the operation must satisfy the associative property, i.e., (a * b) * c = a * (b * c).
3. Identity element: There must exist an element 'e' in the set such that for any element 'a' in the set, the operation 'a * e' and 'e * a' both yield 'a'.
Let's analyze the given options:
a) (a * e) = a
This option represents the existence of an identity element 'e' in the set, which is required for a monoid.
b) (a * e) = (a * c)
This option represents the closure property, as it states that for any element 'a' and 'c' in the set, their binary operation yields the same result.
c) a = (a * (a * e))
This option represents the closure property, as it states that for any element 'a' in the set, the operation 'a * (a * e)' yields the same result as 'a'.
d) (a * e) = a and a = (a * (a * e))
This option represents both the identity element and closure property, as it satisfies conditions (a) and (c) mentioned above.
From the analysis, it is clear that option 'D' satisfies all the conditions required for a monoid, making it the correct answer.
A non empty set A is termed as an algebraic structure ________- a)with respect to binary operation *
- b)with respect to ternary operation ?
- c)with respect to binary operation +
- d)with respect to unary operation –
Correct answer is option 'A'. Can you explain this answer?
A non empty set A is termed as an algebraic structure ________
a)
with respect to binary operation *
b)
with respect to ternary operation ?
c)
with respect to binary operation +
d)
with respect to unary operation –
|
Sanvi Kapoor answered |
A non empty set A is called an algebraic structure w.r.t binary operation “*” if (a*b) belongs to S for all (a*b) belongs to S. Therefore “*” is closure operation on ‘A’.
Comprehension:
Read the following information and answer the three items that follow:
Consider the function f(x) = g(x) + h(x)
What is the period of the function g(x) = sin(4x/5) ?- a)5π/2
- b)5π/4
- c)π/2
- d)4π
Correct answer is option 'A'. Can you explain this answer?
Comprehension:
Read the following information and answer the three items that follow:
Consider the function f(x) = g(x) + h(x)
What is the period of the function g(x) = sin(4x/5) ?
Read the following information and answer the three items that follow:
Consider the function f(x) = g(x) + h(x)
What is the period of the function g(x) = sin(4x/5) ?
a)
5π/2
b)
5π/4
c)
π/2
d)
4π
|
|
Sanvi Kapoor answered |
Given:
g(x) = sin(4x/5)
As we know, period of sin x is 2π.
So, period of g(x) = sin(4x/5) is
As we know, period of sin x is 2π.
So, period of g(x) = sin(4x/5) is
∴ Period of the function g(x) = 
What is the nature of relation R, if R is defined as R = {(x, y) : 2x + y = 41, x, y ∈ N}? - a)reflexive
- b)symmetric
- c)transitive
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
What is the nature of relation R, if R is defined as R = {(x, y) : 2x + y = 41, x, y ∈ N}?
a)
reflexive
b)
symmetric
c)
transitive
d)
None of these
|
|
Sanya Agarwal answered |
Let us check all the conditions:
Reflexivity:
Let x be an arbitrary element of R. For Reflexive, let y + x
i.e. for any x ∈ R
⇒ 2x + x = 41
Let x be an arbitrary element of R. For Reflexive, let y + x
i.e. for any x ∈ R
⇒ 2x + x = 41
Thus, it is NOT reflexive.
Symmetry:
Let (x, y) ∈ R. Then,
2x + y = 41 Which is not equal to 2y + x = 41
i.e. (y , x) ∉ R
So, R is NOT symmetric.
Let (x, y) ∈ R. Then,
2x + y = 41 Which is not equal to 2y + x = 41
i.e. (y , x) ∉ R
So, R is NOT symmetric.
Transitivity:
Let (x, y) and (y, z) ∈ R
⇒ 2x + y = 41 and 2y + z = 41
which is not equal to 2x + z = 41
i.e. (x , z) ∉ R
Thus, R is NOT transitive.
Let (x, y) and (y, z) ∈ R
⇒ 2x + y = 41 and 2y + z = 41
which is not equal to 2x + z = 41
i.e. (x , z) ∉ R
Thus, R is NOT transitive.
Let A be a set of k (k>0) elements. Which is larger between the number of binary relations (say, Nr) on A and the number of functions (say, Nf) from A to A?- a)number of relations
- b)number of functions
- c)the element set
- d)number of subsets of the relation
Correct answer is option 'A'. Can you explain this answer?
Let A be a set of k (k>0) elements. Which is larger between the number of binary relations (say, Nr) on A and the number of functions (say, Nf) from A to A?
a)
number of relations
b)
number of functions
c)
the element set
d)
number of subsets of the relation
|
|
Sanvi Kapoor answered |
For a set with k elements the number of binary relations should be 2(n*n) and the number of functions should be nn. Now, 2(n*n) => n2log (2) [taking log] and nn => nlog (n) [taking log]. It is known that n2log (2) > nlog (n). Hence, the number of binary relations > the number of functions i.e, Nr > Nf.
If A = {1, 2, 3, 4, 5, 7, 8, 9} and B = {2, 4, 6, 7, 9} then find the number of proper subsets of A ∩ B ?
- a)15
- b)16
- c)32
- d)31
Correct answer is option 'A'. Can you explain this answer?
If A = {1, 2, 3, 4, 5, 7, 8, 9} and B = {2, 4, 6, 7, 9} then find the number of proper subsets of A ∩ B ?
a)
15
b)
16
c)
32
d)
31
|
|
Siddharth Bajaj answered |
A proper subset of A is any subset of A that is not equal to A itself.
To find the number of proper subsets of A, we first need to find the total number of subsets of A. We can do this by noting that each element of A can either be included or excluded from a subset, giving us 2 choices for each element. Therefore, the total number of subsets of A is 2^8 = 256.
However, this includes the subset containing all elements of A, which is not a proper subset. Therefore, we need to subtract 1 from 256 to get the number of proper subsets:
256 - 1 = 255
Therefore, there are 255 proper subsets of A.
To find the number of proper subsets of A, we first need to find the total number of subsets of A. We can do this by noting that each element of A can either be included or excluded from a subset, giving us 2 choices for each element. Therefore, the total number of subsets of A is 2^8 = 256.
However, this includes the subset containing all elements of A, which is not a proper subset. Therefore, we need to subtract 1 from 256 to get the number of proper subsets:
256 - 1 = 255
Therefore, there are 255 proper subsets of A.
A monoid is called a group if _______- a)(a*a)=a=(a+c)
- b)c)
- c)b) (a*c)=(a+c)c) (a+c)=a
- d)(a*c)=(c*a)=e
Correct answer is option 'D'. Can you explain this answer?
A monoid is called a group if _______
a)
(a*a)=a=(a+c)
b)
c)
c)
b) (a*c)=(a+c)c) (a+c)=a
d)
(a*c)=(c*a)=e
|
|
Sanya Agarwal answered |
A monoid(B,*) is called Group if to each element there exists an element c such that (a*c)=(c*a)=e. Here e is called an identity element and c is defined as the inverse of the corresponding element.
If A = {x, y, z}, then the number of subsets in powerset of A is- a)6
- b)8
- c)7
- d)9
Correct answer is option 'B'. Can you explain this answer?
If A = {x, y, z}, then the number of subsets in powerset of A is
a)
6
b)
8
c)
7
d)
9
|
Sagnik Sen answered |
Explanation:
Given:
A = {x, y, z}
Finding the Power set:
To find the power set of A, we need to consider all possible subsets of A, including the empty set and A itself.
Number of Elements in Set A:
Since A has 3 elements, the number of subsets in the power set can be calculated using the formula 2^n, where n is the number of elements in the set.
Calculating the Number of Subsets:
In this case, n = 3. Therefore, the number of subsets in the power set of A is 2^3 = 8.
Answer:
Therefore, the correct answer is option B, which states that there are 8 subsets in the power set of A.
Given:
A = {x, y, z}
Finding the Power set:
To find the power set of A, we need to consider all possible subsets of A, including the empty set and A itself.
Number of Elements in Set A:
Since A has 3 elements, the number of subsets in the power set can be calculated using the formula 2^n, where n is the number of elements in the set.
Calculating the Number of Subsets:
In this case, n = 3. Therefore, the number of subsets in the power set of A is 2^3 = 8.
Answer:
Therefore, the correct answer is option B, which states that there are 8 subsets in the power set of A.
Relation R is defined asR = {(a, b) | (a - b) = km for some fixed integer m and a, b, k ∈ z}, then R is- a)Reflective but not symmetric
- b)Symmetric but not transitive
- c)Transitive but not reflective
- d)An equivalence relation
Correct answer is option 'D'. Can you explain this answer?
Relation R is defined as
R = {(a, b) | (a - b) = km for some fixed integer m and a, b, k ∈ z}, then R is
a)
Reflective but not symmetric
b)
Symmetric but not transitive
c)
Transitive but not reflective
d)
An equivalence relation
|
|
Gowri Sharma answered |
Understanding Relation R
Relation R is defined as R = {(a, b) | (a - b) = km for some fixed integer m and a, b, k ∈ Z}. This definition implies that the difference between a and b is a multiple of a fixed integer m. To determine whether R is an equivalence relation, we need to check three properties: reflexivity, symmetry, and transitivity.
Reflexivity
- A relation R is reflexive if for every element a, (a, a) ∈ R.
- For any integer a, (a - a) = 0, which is indeed a multiple of any integer m.
- Thus, R is reflexive.
Symmetry
- A relation R is symmetric if for any (a, b) ∈ R, (b, a) must also be in R.
- If (a, b) ∈ R, then (a - b) = km for some integer k. Therefore, (b - a) = -km, which is also a multiple of m.
- Thus, R is symmetric.
Transitivity
- A relation R is transitive if whenever (a, b) ∈ R and (b, c) ∈ R, then (a, c) must also be in R.
- If (a, b) = km and (b, c) = k'm, it follows that (a - c) = (a - b) + (b - c) = (k + k')m, which is a multiple of m.
- Thus, R is transitive.
Conclusion
Since R satisfies all three properties (reflexivity, symmetry, transitivity), it is indeed an equivalence relation. Therefore, the correct answer is option 'D', confirming that R is an equivalence relation.
Relation R is defined as R = {(a, b) | (a - b) = km for some fixed integer m and a, b, k ∈ Z}. This definition implies that the difference between a and b is a multiple of a fixed integer m. To determine whether R is an equivalence relation, we need to check three properties: reflexivity, symmetry, and transitivity.
Reflexivity
- A relation R is reflexive if for every element a, (a, a) ∈ R.
- For any integer a, (a - a) = 0, which is indeed a multiple of any integer m.
- Thus, R is reflexive.
Symmetry
- A relation R is symmetric if for any (a, b) ∈ R, (b, a) must also be in R.
- If (a, b) ∈ R, then (a - b) = km for some integer k. Therefore, (b - a) = -km, which is also a multiple of m.
- Thus, R is symmetric.
Transitivity
- A relation R is transitive if whenever (a, b) ∈ R and (b, c) ∈ R, then (a, c) must also be in R.
- If (a, b) = km and (b, c) = k'm, it follows that (a - c) = (a - b) + (b - c) = (k + k')m, which is a multiple of m.
- Thus, R is transitive.
Conclusion
Since R satisfies all three properties (reflexivity, symmetry, transitivity), it is indeed an equivalence relation. Therefore, the correct answer is option 'D', confirming that R is an equivalence relation.
Matrix multiplication is a/an _________ property.- a)Commutative
- b)Associative
- c)Additive
- d)Disjunctive
Correct answer is option 'B'. Can you explain this answer?
Matrix multiplication is a/an _________ property.
a)
Commutative
b)
Associative
c)
Additive
d)
Disjunctive
|
|
Sanya Agarwal answered |
The set of two M*M non-singular matrices form a group under matrix multiplication operation. Since matrix multiplication is itself associative, it holds associative property.
A group (M,*) is said to be abelian if ___________- a)(x+y)=(y+x)
- b)(x*y)=(y*x)
- c)(x+y)=x
- d)(y*x)=(x+y)
Correct answer is option 'B'. Can you explain this answer?
A group (M,*) is said to be abelian if ___________
a)
(x+y)=(y+x)
b)
(x*y)=(y*x)
c)
(x+y)=x
d)
(y*x)=(x+y)
|
|
Sanya Agarwal answered |
A group (M,*) is said to be abelian if (x*y) = (x*y) for all x, y belongs to M. Thus Commutative property should hold in a group.
If A = {1, 2, 3, 4, 5}, then the number of subsets of A which contain element 2 but not 4, is- a)2
- b)4
- c)6
- d)8
Correct answer is option 'D'. Can you explain this answer?
If A = {1, 2, 3, 4, 5}, then the number of subsets of A which contain element 2 but not 4, is
a)
2
b)
4
c)
6
d)
8
|
|
Srestha Datta answered |
Number of subsets of A which contain element 2 but not 4
To find the number of subsets of set A that contain element 2 but not 4, we need to consider that the set A has 5 elements: {1, 2, 3, 4, 5}.
Step 1: Count the total number of subsets of A.
Since A has 5 elements, the total number of subsets of A can be calculated using the formula 2^n, where n is the number of elements in the set.
In this case, n = 5, so the total number of subsets of A is 2^5 = 32.
Step 2: Count the number of subsets that contain both 2 and 4.
To count the number of subsets that contain both 2 and 4, we treat these two elements as a single entity. So, we have 4 remaining elements to consider: {1, 3, 5}.
Using the same formula as before, the total number of subsets that contain both 2 and 4 is 2^4 = 16.
Step 3: Count the number of subsets that contain only 2 and not 4.
To count the number of subsets that contain only 2 and not 4, we need to subtract the subsets that contain both 2 and 4 from the total number of subsets.
So, the number of subsets that contain only 2 and not 4 is 32 - 16 = 16.
Step 4: Finalize the answer.
The question asks for the number of subsets that contain element 2 but not 4. From Step 3, we know that there are 16 subsets that contain only 2 and not 4. However, we need to consider that the empty set is also a subset of A that satisfies this condition. So, we add 1 to the count, making it a total of 16 + 1 = 17 subsets.
Therefore, the correct answer is option D: 8.
To find the number of subsets of set A that contain element 2 but not 4, we need to consider that the set A has 5 elements: {1, 2, 3, 4, 5}.
Step 1: Count the total number of subsets of A.
Since A has 5 elements, the total number of subsets of A can be calculated using the formula 2^n, where n is the number of elements in the set.
In this case, n = 5, so the total number of subsets of A is 2^5 = 32.
Step 2: Count the number of subsets that contain both 2 and 4.
To count the number of subsets that contain both 2 and 4, we treat these two elements as a single entity. So, we have 4 remaining elements to consider: {1, 3, 5}.
Using the same formula as before, the total number of subsets that contain both 2 and 4 is 2^4 = 16.
Step 3: Count the number of subsets that contain only 2 and not 4.
To count the number of subsets that contain only 2 and not 4, we need to subtract the subsets that contain both 2 and 4 from the total number of subsets.
So, the number of subsets that contain only 2 and not 4 is 32 - 16 = 16.
Step 4: Finalize the answer.
The question asks for the number of subsets that contain element 2 but not 4. From Step 3, we know that there are 16 subsets that contain only 2 and not 4. However, we need to consider that the empty set is also a subset of A that satisfies this condition. So, we add 1 to the count, making it a total of 16 + 1 = 17 subsets.
Therefore, the correct answer is option D: 8.
Let L denote the set of all straight lines in a plane. Let a relation R be l R m if l is perpendicular to m ∀ l, m ∈ L. Then R is:- a)reflexive
- b)symmetric
- c)transitive
- d)equivalence
Correct answer is option 'B'. Can you explain this answer?
Let L denote the set of all straight lines in a plane. Let a relation R be l R m if l is perpendicular to m ∀ l, m ∈ L. Then R is:
a)
reflexive
b)
symmetric
c)
transitive
d)
equivalence
|
|
Sanya Agarwal answered |
Let two lines ℓ1, ℓ2 ∈ L
Now (ℓ1, ℓ2) ∈ R
Only when ℓ1 ⊥ ℓ2
1. For reflexivity:
Let ℓ1 ∈ L
But (ℓ1, ℓ1) ∉ R
Since no line is perpendicular to itself
R is not reflexive
Now (ℓ1, ℓ2) ∈ R
Only when ℓ1 ⊥ ℓ2
1. For reflexivity:
Let ℓ1 ∈ L
But (ℓ1, ℓ1) ∉ R
Since no line is perpendicular to itself
R is not reflexive
2. For symmetric:
Let, (ℓ1, ℓ2) ∈ L
Now, if ℓ1 ⊥ ℓ2 then this implies that ℓ2 is also perpendicular to ℓ1
Hence, (ℓ1, ℓ1) ∈ L
R is symmetric.
Let, (ℓ1, ℓ2) ∈ L
Now, if ℓ1 ⊥ ℓ2 then this implies that ℓ2 is also perpendicular to ℓ1
Hence, (ℓ1, ℓ1) ∈ L
R is symmetric.
3. For Transitive:
Let (ℓ1, ℓ2) ∈ L and (ℓ2, ℓ3) ∈ L
ℓ1 ⊥ ℓ2 and ℓ2 ⊥ ℓ3
Let (ℓ1, ℓ2) ∈ L and (ℓ2, ℓ3) ∈ L
ℓ1 ⊥ ℓ2 and ℓ2 ⊥ ℓ3
ℓ1 is not perpendicular to ℓ3
(ℓ1, ℓ3) ∉ R
∴ Not transitive
(ℓ1, ℓ3) ∉ R
∴ Not transitive
Let m ∈ Z and consider the relation Rm defined by a Rm b if and only if a ≡ b mod m. Then Rm is -- a)reflexive relation
- b)symmetric relation
- c)transitive relation
- d)equivalence relation
Correct answer is option 'D'. Can you explain this answer?
Let m ∈ Z and consider the relation Rm defined by a Rm b if and only if a ≡ b mod m. Then Rm is -
a)
reflexive relation
b)
symmetric relation
c)
transitive relation
d)
equivalence relation
|
Jay Menon answered |
Be a positive integer. If m is even, then m can be written in the form m = 2k for some positive integer k. If m is odd, then m can be written in the form m = 2k + 1 for some positive integer k.
Let R be a relation defined as xRy if and only if 2x + 3y = 20, where x, y ∈ N. How many elements of the form (x, y) are there in R?- a)2
- b)3
- c)4
- d)6
Correct answer is option 'B'. Can you explain this answer?
Let R be a relation defined as xRy if and only if 2x + 3y = 20, where x, y ∈ N. How many elements of the form (x, y) are there in R?
a)
2
b)
3
c)
4
d)
6
|
|
Sanvi Kapoor answered |
Concept:
If x ∈ R then we express it by writing xRy and say that " x is related to y with relation R"
Thus, (x, y) ∈ R ⇔ xRy
If x ∈ R then we express it by writing xRy and say that " x is related to y with relation R"
Thus, (x, y) ∈ R ⇔ xRy
Calculation:
Given
2x + 3y = 20
The relation R can be written as
R = {(1,6), (4, 4), (7, 2)}
There are 3 elements in the form (x, y) are there in R.
Given
2x + 3y = 20
The relation R can be written as
R = {(1,6), (4, 4), (7, 2)}
There are 3 elements in the form (x, y) are there in R.
Let A and B be two non-empty relations on a set S. Which of the following statements is false?- a)A and B are transitive ⇒ A∩B is transitive
- b)A and B are symmetric ⇒ A∪B is symmetric
- c)A and B are transitive ⇒ A∪B is not transitive
- d)A and B are reflexive ⇒ A∩B is reflexive
Correct answer is option 'C'. Can you explain this answer?
Let A and B be two non-empty relations on a set S. Which of the following statements is false?
a)
A and B are transitive ⇒ A∩B is transitive
b)
A and B are symmetric ⇒ A∪B is symmetric
c)
A and B are transitive ⇒ A∪B is not transitive
d)
A and B are reflexive ⇒ A∩B is reflexive
|
|
Anshul Chakraborty answered |
The statement "A and B are transitive" is not necessarily false. It is possible for both relations A and B to be transitive.
An algebraic structure _________ is called a semigroup.- a)(P, *)
- b)(Q, +, *)
- c)(P, +)
- d)(+, *)
Correct answer is option 'A'. Can you explain this answer?
An algebraic structure _________ is called a semigroup.
a)
(P, *)
b)
(Q, +, *)
c)
(P, +)
d)
(+, *)
|
|
Gitanjali Menon answered |
Algebraic Structure:
An algebraic structure is a set with one or more operations defined on it, satisfying certain properties. These structures are used to study various mathematical concepts, such as numbers, vectors, and functions.
Semigroup:
A semigroup is an algebraic structure consisting of a set and a binary operation defined on that set. The binary operation combines two elements of the set to produce another element of the set. This operation is generally denoted by '*'.
Option A: (P, *)
In this option, the set is denoted by 'P' and the binary operation is denoted by '*'. This structure satisfies the requirements of a semigroup because:
1. Closure Property: For any two elements 'a' and 'b' in set P, the result of the operation 'a * b' is also an element of set P. Therefore, the operation '* 'is closed on set P.
2. Associativity Property: For any three elements 'a', 'b', and 'c' in set P, the operation '(a * b) * c' is equal to 'a * (b * c)'. In other words, the order in which the operations are performed does not matter.
Since option A satisfies both the closure and associativity properties, it is a valid semigroup.
Option B: (Q, , *)
In this option, the set is denoted by 'Q', and there are two binary operations, denoted by ' ' and '*'. However, a semigroup requires only one binary operation. Therefore, option B cannot be a semigroup.
Option C: (P, )
In this option, the set is denoted by 'P', but there is no binary operation defined. Without a binary operation, it is not possible to define any algebraic structure. Therefore, option C cannot be a semigroup.
Option D: ( , *)
In this option, there is no set defined. A semigroup requires a set on which the binary operation is defined. Therefore, option D cannot be a semigroup.
Conclusion:
The correct answer is option A, (P, *), as it satisfies the requirements of a semigroup by having a set 'P' and a binary operation '*'.
An algebraic structure is a set with one or more operations defined on it, satisfying certain properties. These structures are used to study various mathematical concepts, such as numbers, vectors, and functions.
Semigroup:
A semigroup is an algebraic structure consisting of a set and a binary operation defined on that set. The binary operation combines two elements of the set to produce another element of the set. This operation is generally denoted by '*'.
Option A: (P, *)
In this option, the set is denoted by 'P' and the binary operation is denoted by '*'. This structure satisfies the requirements of a semigroup because:
1. Closure Property: For any two elements 'a' and 'b' in set P, the result of the operation 'a * b' is also an element of set P. Therefore, the operation '* 'is closed on set P.
2. Associativity Property: For any three elements 'a', 'b', and 'c' in set P, the operation '(a * b) * c' is equal to 'a * (b * c)'. In other words, the order in which the operations are performed does not matter.
Since option A satisfies both the closure and associativity properties, it is a valid semigroup.
Option B: (Q, , *)
In this option, the set is denoted by 'Q', and there are two binary operations, denoted by ' ' and '*'. However, a semigroup requires only one binary operation. Therefore, option B cannot be a semigroup.
Option C: (P, )
In this option, the set is denoted by 'P', but there is no binary operation defined. Without a binary operation, it is not possible to define any algebraic structure. Therefore, option C cannot be a semigroup.
Option D: ( , *)
In this option, there is no set defined. A semigroup requires a set on which the binary operation is defined. Therefore, option D cannot be a semigroup.
Conclusion:
The correct answer is option A, (P, *), as it satisfies the requirements of a semigroup by having a set 'P' and a binary operation '*'.
Let A = {1, 2, 3} and consider the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1,3)}. Then R is- a)reflexive but not symmetric
- b)reflexive but not transitive
- c)symmetric and transitive
- d)neither symmetric, nor transitive
Correct answer is option 'A'. Can you explain this answer?
Let A = {1, 2, 3} and consider the relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1,3)}. Then R is
a)
reflexive but not symmetric
b)
reflexive but not transitive
c)
symmetric and transitive
d)
neither symmetric, nor transitive
|
Aditi Chakraborty answered |
Understanding the Relation R
Let’s analyze the relation \( R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\} \) defined on the set \( A = \{1, 2, 3\} \).
Reflexivity
- A relation is reflexive if every element in the set is related to itself.
- In \( R \), we have:
- \( (1, 1) \)
- \( (2, 2) \)
- \( (3, 3) \)
- All elements of \( A \) are present in \( R \) as pairs related to themselves.
- Thus, R is reflexive.
Symmetry
- A relation is symmetric if for every \( (a, b) \) in \( R \), \( (b, a) \) is also in \( R \).
- Checking pairs:
- For \( (1, 2) \), \( (2, 1) \) is not in \( R \).
- For \( (2, 3) \), \( (3, 2) \) is not in \( R \).
- For \( (1, 3) \), \( (3, 1) \) is not in \( R \).
- Since not all pairs satisfy the condition, R is not symmetric.
Transitivity
- A relation is transitive if whenever \( (a, b) \) and \( (b, c) \) are in \( R \), then \( (a, c) \) must also be in \( R \).
- Check:
- From \( (1, 2) \) and \( (2, 3) \), \( (1, 3) \) is in \( R \) (valid).
- However, there are no other combinations that violate transitivity.
- Thus, R is transitive.
Conclusion
- Hence, the correct classification for the relation \( R \) is that it is reflexive but not symmetric. Therefore, the correct answer is option 'A'.
Let’s analyze the relation \( R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\} \) defined on the set \( A = \{1, 2, 3\} \).
Reflexivity
- A relation is reflexive if every element in the set is related to itself.
- In \( R \), we have:
- \( (1, 1) \)
- \( (2, 2) \)
- \( (3, 3) \)
- All elements of \( A \) are present in \( R \) as pairs related to themselves.
- Thus, R is reflexive.
Symmetry
- A relation is symmetric if for every \( (a, b) \) in \( R \), \( (b, a) \) is also in \( R \).
- Checking pairs:
- For \( (1, 2) \), \( (2, 1) \) is not in \( R \).
- For \( (2, 3) \), \( (3, 2) \) is not in \( R \).
- For \( (1, 3) \), \( (3, 1) \) is not in \( R \).
- Since not all pairs satisfy the condition, R is not symmetric.
Transitivity
- A relation is transitive if whenever \( (a, b) \) and \( (b, c) \) are in \( R \), then \( (a, c) \) must also be in \( R \).
- Check:
- From \( (1, 2) \) and \( (2, 3) \), \( (1, 3) \) is in \( R \) (valid).
- However, there are no other combinations that violate transitivity.
- Thus, R is transitive.
Conclusion
- Hence, the correct classification for the relation \( R \) is that it is reflexive but not symmetric. Therefore, the correct answer is option 'A'.
Consider R and S be two equivalence relations, which of the following is true regarding the R and S- a)The Largest equivalence relation which is inside R and S is R∩S
- b)The smallest equivalence relation which is inside R and S is R∩S
- c)R∩S can never be an equivalence relation
- d)R ∪ S is always an equivalence relation
Correct answer is option 'A'. Can you explain this answer?
Consider R and S be two equivalence relations, which of the following is true regarding the R and S
a)
The Largest equivalence relation which is inside R and S is R∩S
b)
The smallest equivalence relation which is inside R and S is R∩S
c)
R∩S can never be an equivalence relation
d)
R ∪ S is always an equivalence relation
|
|
Sanya Agarwal answered |
Option D is wrong as transitivity may break on taking the union of two equivalence relations
For two equivalence relation R and S
For two equivalence relation R and S
- The largest equivalence relation in R and S is R∩S
- The smallest equivalence relation which contains R and S is (R∪S)∞ {transitive closure}
So option A is correct.
Option C is trivially false as option A is correct.
Option C is trivially false as option A is correct.
Let R be a relation between A and B. R is asymmetric if and only if ________- a)Intersection of D(A) and R is empty, where D(A) represents diagonal of set
- b)R-1 is a subset of R, where R-1 represents inverse of R
- c)Intersection of R and R-1 is D(A)
- d)D(A) is a subset of R, where D(A) represents diagonal of set
Correct answer is option 'A'. Can you explain this answer?
Let R be a relation between A and B. R is asymmetric if and only if ________
a)
Intersection of D(A) and R is empty, where D(A) represents diagonal of set
b)
R-1 is a subset of R, where R-1 represents inverse of R
c)
Intersection of R and R-1 is D(A)
d)
D(A) is a subset of R, where D(A) represents diagonal of set
|
Rutuja Pillai answered |
Asymmetric Relation and its Characteristics:
An asymmetric relation between sets A and B is a relation that does not have any pairs where both elements are related to each other. In other words, if (a, b) is in relation R, then (b, a) cannot be in R.
To understand why option 'A' is the correct answer, let's examine the characteristics of an asymmetric relation:
1. No symmetric pairs: In an asymmetric relation, there are no pairs where both elements are related to each other. This means that if (a, b) is in R, then (b, a) cannot be in R.
Explanation of Option 'A':
Option 'A' states that the intersection of the diagonal of set A (D(A)) and relation R is empty. Let's break down this statement to understand why it is true for an asymmetric relation:
- Diagonal of set A (D(A)): The diagonal of set A, denoted as D(A), is a set of ordered pairs where both elements are the same. For example, if A = {1, 2, 3}, then D(A) = {(1, 1), (2, 2), (3, 3)}. It represents the pairs where an element is related to itself.
- Intersection of D(A) and R: The intersection of D(A) and R represents the pairs where both elements are related to each other in the relation R.
- Empty intersection for asymmetric relation: In an asymmetric relation, there are no symmetric pairs, i.e., no pairs where both elements are related to each other. Therefore, the intersection of D(A) and R will be empty for an asymmetric relation.
Hence, option 'A' is the correct answer because an asymmetric relation will have an empty intersection between the diagonal of set A and relation R.
Other Options Explanation:
- Option 'B': R-1 represents the inverse of relation R. If R is asymmetric, it means that (a, b) is in R implies (b, a) is not in R. However, this does not imply that the inverse of R, R-1, is a subset of R.
- Option 'C': The intersection of R and R-1 represents the pairs that are both in R and its inverse R-1. This intersection does not necessarily correspond to the diagonal of set A.
- Option 'D': D(A) represents the diagonal of set A, and it consists of pairs where an element is related to itself. However, an asymmetric relation does not require all elements of set A to be related to themselves, so D(A) being a subset of R is not a characteristic of an asymmetric relation.
An asymmetric relation between sets A and B is a relation that does not have any pairs where both elements are related to each other. In other words, if (a, b) is in relation R, then (b, a) cannot be in R.
To understand why option 'A' is the correct answer, let's examine the characteristics of an asymmetric relation:
1. No symmetric pairs: In an asymmetric relation, there are no pairs where both elements are related to each other. This means that if (a, b) is in R, then (b, a) cannot be in R.
Explanation of Option 'A':
Option 'A' states that the intersection of the diagonal of set A (D(A)) and relation R is empty. Let's break down this statement to understand why it is true for an asymmetric relation:
- Diagonal of set A (D(A)): The diagonal of set A, denoted as D(A), is a set of ordered pairs where both elements are the same. For example, if A = {1, 2, 3}, then D(A) = {(1, 1), (2, 2), (3, 3)}. It represents the pairs where an element is related to itself.
- Intersection of D(A) and R: The intersection of D(A) and R represents the pairs where both elements are related to each other in the relation R.
- Empty intersection for asymmetric relation: In an asymmetric relation, there are no symmetric pairs, i.e., no pairs where both elements are related to each other. Therefore, the intersection of D(A) and R will be empty for an asymmetric relation.
Hence, option 'A' is the correct answer because an asymmetric relation will have an empty intersection between the diagonal of set A and relation R.
Other Options Explanation:
- Option 'B': R-1 represents the inverse of relation R. If R is asymmetric, it means that (a, b) is in R implies (b, a) is not in R. However, this does not imply that the inverse of R, R-1, is a subset of R.
- Option 'C': The intersection of R and R-1 represents the pairs that are both in R and its inverse R-1. This intersection does not necessarily correspond to the diagonal of set A.
- Option 'D': D(A) represents the diagonal of set A, and it consists of pairs where an element is related to itself. However, an asymmetric relation does not require all elements of set A to be related to themselves, so D(A) being a subset of R is not a characteristic of an asymmetric relation.
If f(x) satisfies the relation 2f(x) + f(1 - x) = x2 for all real x, then f(x) is- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
If f(x) satisfies the relation 2f(x) + f(1 - x) = x2 for all real x, then f(x) is
a)
b)
c)
d)
|
Pioneer Academy answered |
Given,
2f(x) + f(1 - x) = x2 ___(i)
Replacing x by (1 - x),
⇒ 2f(1 - x) + f(x) = (1 - x)2 ___(ii)
Applying 2 × (i) - (ii),
⇒ 4f(x) + 2f(1 - x) - 2f(1 - x) - f(x) = 2x2 - (1 - x)2
⇒ 3f(x) = 2x2 - (1 + x2 - 2x)
⇒ 3f(x) = x2 - 1 + 2x
Replacing x by (1 - x),
⇒ 2f(1 - x) + f(x) = (1 - x)2 ___(ii)
Applying 2 × (i) - (ii),
⇒ 4f(x) + 2f(1 - x) - 2f(1 - x) - f(x) = 2x2 - (1 - x)2
⇒ 3f(x) = 2x2 - (1 + x2 - 2x)
⇒ 3f(x) = x2 - 1 + 2x
A cyclic group can be generated by a/an ________ element.- a)singular
- b)non-singular
- c)inverse
- d)multiplicative
Correct answer is option 'A'. Can you explain this answer?
A cyclic group can be generated by a/an ________ element.
a)
singular
b)
non-singular
c)
inverse
d)
multiplicative
|
|
Aditya Jain answered |
Generating a Cyclic Group with a Singular Element
Cyclic groups are algebraic structures that are generated by a single element, called a generator. This generator can produce all other elements in the group through repeated application of the group operation. In the context of a cyclic group, the generator is a singular element that can generate the entire group.
Definition of a Singular Element
In mathematics, a singular element refers to an element that is unique or one-of-a-kind in a particular context. In the case of a cyclic group, the singular element serves as the generator that can produce all other elements in the group.
Role of a Singular Element in Generating a Cyclic Group
When a singular element is chosen as the generator of a cyclic group, it can be used to generate all other elements in the group through repeated application of the group operation. This process creates a cycle of elements that eventually leads back to the generator, forming a cyclic group.
Importance of Using a Singular Element
Using a singular element as the generator of a cyclic group is crucial for ensuring that the generated group is truly cyclic. By starting with a singular element, the generator can produce a complete cycle of elements that form a closed group under the group operation.
In conclusion, a cyclic group can be generated by a singular element, which serves as the unique generator that can produce all other elements in the group. This property highlights the importance of choosing a singular element to create a truly cyclic group in mathematics.
Determine the characteristics of the relation aRb if a2 = b2.- a)Transitive and symmetric
- b)Reflexive and asymmetry
- c)Trichotomy, antisymmetry, and irreflexive
- d)Symmetric, Reflexive, and transitive
Correct answer is option 'D'. Can you explain this answer?
Determine the characteristics of the relation aRb if a2 = b2.
a)
Transitive and symmetric
b)
Reflexive and asymmetry
c)
Trichotomy, antisymmetry, and irreflexive
d)
Symmetric, Reflexive, and transitive
|
Rounak Choudhary answered |
Characteristics of the relation aRb if a^2 = b^2:
To determine the characteristics of the relation aRb where a^2 = b^2, we need to analyze the properties of the relation based on the given condition.
Symmetric:
- A relation is symmetric if for every (a, b) that belongs to the relation, (b, a) also belongs to the relation.
- In this case, if a^2 = b^2, then b^2 = a^2. Therefore, the relation is symmetric as it satisfies the condition.
Reflexive:
- A relation is reflexive if for every element 'a' in the set, (a, a) belongs to the relation.
- In this case, a^2 = a^2 is true for every element 'a' in the set. Therefore, the relation is reflexive as it satisfies the condition.
Transitive:
- A relation is transitive if for every (a, b) and (b, c) that belongs to the relation, (a, c) also belongs to the relation.
- In this case, if a^2 = b^2 and b^2 = c^2, then a^2 = c^2. Therefore, the relation is transitive as it satisfies the condition.
Therefore, the relation aRb if a^2 = b^2 is symmetric, reflexive, and transitive, which corresponds to option 'D' in the given choices.
To determine the characteristics of the relation aRb where a^2 = b^2, we need to analyze the properties of the relation based on the given condition.
Symmetric:
- A relation is symmetric if for every (a, b) that belongs to the relation, (b, a) also belongs to the relation.
- In this case, if a^2 = b^2, then b^2 = a^2. Therefore, the relation is symmetric as it satisfies the condition.
Reflexive:
- A relation is reflexive if for every element 'a' in the set, (a, a) belongs to the relation.
- In this case, a^2 = a^2 is true for every element 'a' in the set. Therefore, the relation is reflexive as it satisfies the condition.
Transitive:
- A relation is transitive if for every (a, b) and (b, c) that belongs to the relation, (a, c) also belongs to the relation.
- In this case, if a^2 = b^2 and b^2 = c^2, then a^2 = c^2. Therefore, the relation is transitive as it satisfies the condition.
Therefore, the relation aRb if a^2 = b^2 is symmetric, reflexive, and transitive, which corresponds to option 'D' in the given choices.
What is the Cartesian product of A = {1, 2} and B = {a, b}?- a){(1, a), (1,
- b), (2, a), (b, b)}b) {(1, 1), (2, 2), (a, a), (b, b)}
- c){(1, a), (2, a), (1, b), (2, b)}
- d){(1, 1), (a, a), (2, a), (1, b)}
Correct answer is option 'C'. Can you explain this answer?
What is the Cartesian product of A = {1, 2} and B = {a, b}?
a)
{(1, a), (1,
b)
, (2, a), (b, b)}b) {(1, 1), (2, 2), (a, a), (b, b)}
c)
{(1, a), (2, a), (1, b), (2, b)}
d)
{(1, 1), (a, a), (2, a), (1, b)}
|
|
Sanvi Kapoor answered |
A subset R of the Cartesian product A x B is a relation from the set A to the set B.
Comprehension:
Read the following information and answer the three items that follow:
Consider the function f(x) = g(x) + h(x)
What is the period of the function h(x) = cos(2x/3)?- a)π/3
- b)2π/3
- c)3π
- d)4π
Correct answer is option 'C'. Can you explain this answer?
Comprehension:
Read the following information and answer the three items that follow:
Consider the function f(x) = g(x) + h(x)
What is the period of the function h(x) = cos(2x/3)?
Read the following information and answer the three items that follow:
Consider the function f(x) = g(x) + h(x)
What is the period of the function h(x) = cos(2x/3)?
a)
π/3
b)
2π/3
c)
3π
d)
4π
|
|
Sanvi Kapoor answered |
Given:
h(x) = cos(2x/3)
As we know, period of cos x is 2π.
So, period of
∴ Period of the function h(x) = cos(2x/3) is 3π
h(x) = cos(2x/3)
As we know, period of cos x is 2π.
So, period of
∴ Period of the function h(x) = cos(2x/3) is 3π
If 
then find the value of f(tan θ).- a)sin 2θ
- b)cos 2θ
- c)tan 2θ
- d)sin θ
Correct answer is option 'A'. Can you explain this answer?
If then find the value of f(tan θ).
then find the value of f(tan θ).a)
sin 2θ
b)
cos 2θ
c)
tan 2θ
d)
sin θ
|
|
Sanya Agarwal answered |
Substituting x = tan θ, we get:
Directions: Read the following information and answer the three items that follow:
Let f(x) = x2 + 2x – 5 and g(x) = 5x + 30
Consider the following statements:
1. f[g(x)] is a polynomial of degree 3.
2. g[g(x)] is a polynomial of degree 2.Which of the above statements is/are correct?- a)1 only
- b)2 only
- c)Both 1 and 2
- d)Neither 1 nor 2
Correct answer is option 'D'. Can you explain this answer?
Directions: Read the following information and answer the three items that follow:
Let f(x) = x2 + 2x – 5 and g(x) = 5x + 30
Consider the following statements:
1. f[g(x)] is a polynomial of degree 3.
2. g[g(x)] is a polynomial of degree 2.
Let f(x) = x2 + 2x – 5 and g(x) = 5x + 30
Consider the following statements:
1. f[g(x)] is a polynomial of degree 3.
2. g[g(x)] is a polynomial of degree 2.
Which of the above statements is/are correct?
a)
1 only
b)
2 only
c)
Both 1 and 2
d)
Neither 1 nor 2
|
|
Sanvi Kapoor answered |
Given: f(x) = x2 + 2x – 5 and g(x) = 5x + 30
⇒ f[g(x)] = f(5x + 30) = (5x + 30)2 + 2(5x + 30) – 5 = 25x2 + 310x + 955
⇒ f[g(x)] is a polynomial of degree 2.
Hence statement 1 is false.
⇒ g[g(x)] = g(5x + 30) = 5(5x + 30) + 30 = 25x + 180
⇒ g[g(x)] is a polynomial of degree 1.
Hence statement 2 is wrong.
⇒ f[g(x)] = f(5x + 30) = (5x + 30)2 + 2(5x + 30) – 5 = 25x2 + 310x + 955
⇒ f[g(x)] is a polynomial of degree 2.
Hence statement 1 is false.
⇒ g[g(x)] = g(5x + 30) = 5(5x + 30) + 30 = 25x + 180
⇒ g[g(x)] is a polynomial of degree 1.
Hence statement 2 is wrong.
{1, i, -i, -1} is __________- a)semigroup
- b)subgroup
- c)cyclic group
- d)abelian group
Correct answer is option 'C'. Can you explain this answer?
{1, i, -i, -1} is __________
a)
semigroup
b)
subgroup
c)
cyclic group
d)
abelian group
|
|
Sanvi Kapoor answered |
The set of complex numbers {1, i, -i, -1} under multiplication operation is a cyclic group. Two generators i and -i will covers all the elements of this group. Hence, it is a cyclic group.
The binary relation {(1,1), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2)} on the set {1, 2, 3} is __________- a)reflective, symmetric and transitive
- b)irreflexive, symmetric and transitive
- c)neither reflective, nor irreflexive and not transitive
- d)irreflexive and antisymmetric
Correct answer is option 'C'. Can you explain this answer?
The binary relation {(1,1), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2)} on the set {1, 2, 3} is __________
a)
reflective, symmetric and transitive
b)
irreflexive, symmetric and transitive
c)
neither reflective, nor irreflexive and not transitive
d)
irreflexive and antisymmetric
|
Constructing Careers answered |
Not reflexive -> (3,3) not present; not irreflexive -> (1, 1) is present; not symmetric -> (2, 1) is present but not (1, 2); not antisymmetric – (2, 3) and (3, 2) are present; not asymmetric -> asymmetry requires both antisymmetry and irreflexivity. It is also not transitive closure of relation because (3,2) and (2,3) are there in the relation but (3,3) is missing.
The time complexity of computing the transitive closure of a binary relation on a set of n elements should be ________- a)O(n)
- b)O(logn)
- c)O(n(n+(3/2)))
- d)O(n3)
Correct answer is option 'D'. Can you explain this answer?
The time complexity of computing the transitive closure of a binary relation on a set of n elements should be ________
a)
O(n)
b)
O(logn)
c)
O(n(n+(3/2)))
d)
O(n3)
|
|
Sanya Agarwal answered |
Calculation of transitive closure results into matrix multiplication. We can do matrix multiplication in O(n3) time. There are better algorithms that do less than cubic time.
The set O of odd positive integers less than 10 can be expressed by __________.- a){1, 2, 3}
- b){1, 3, 5, 7, 9}
- c){1, 2, 5, 9}
- d){1, 5, 7, 9, 11}
Correct answer is option 'B'. Can you explain this answer?
The set O of odd positive integers less than 10 can be expressed by __________.
a)
{1, 2, 3}
b)
{1, 3, 5, 7, 9}
c)
{1, 2, 5, 9}
d)
{1, 5, 7, 9, 11}
|
|
Sanya Agarwal answered |
Odd numbers less than 10 is {1, 3, 5, 7, 9}.
Comprehension:
Directions: Read the following information and answer the three items that follow:
Let f(x) = x2 + 2x – 5 and g(x) = 5x + 30
What are the roots of the equation g[f(x)] = 0?- a)1, -1
- b)-1, -1
- c)1, 1
- d)0, 1
Correct answer is option 'A'. Can you explain this answer?
Comprehension:
Directions: Read the following information and answer the three items that follow:
Let f(x) = x2 + 2x – 5 and g(x) = 5x + 30
What are the roots of the equation g[f(x)] = 0?
Directions: Read the following information and answer the three items that follow:
Let f(x) = x2 + 2x – 5 and g(x) = 5x + 30
What are the roots of the equation g[f(x)] = 0?
a)
1, -1
b)
-1, -1
c)
1, 1
d)
0, 1
|
|
Sanya Agarwal answered |
Given: f(x) = x2 + 2x – 5 and g(x) = 5x + 30
⇒ g[f(x)] = g(x2 + 2x – 5) = 5 (x2 + 2x – 5) + 30 = 5x2 + 10x + 5 = 0.
⇒ 5x2 + 10x + 5 = 0
⇒ x2 + 2x + 1 = (x + 1)2 = 0
⇒ x = - 1
⇒ g[f(x)] = g(x2 + 2x – 5) = 5 (x2 + 2x – 5) + 30 = 5x2 + 10x + 5 = 0.
⇒ 5x2 + 10x + 5 = 0
⇒ x2 + 2x + 1 = (x + 1)2 = 0
⇒ x = - 1
If f(x) = 2x3 + 7x2 - 3, find f(x - 1).- a)2x3 + 3x2 - x - 3
- b)x3 - 3x2 - 4x + 3
- c)x3 - x2 + 3x - 1
- d)2x3 + x2 - 8x + 2
Correct answer is option 'D'. Can you explain this answer?
If f(x) = 2x3 + 7x2 - 3, find f(x - 1).
a)
2x3 + 3x2 - x - 3
b)
x3 - 3x2 - 4x + 3
c)
x3 - x2 + 3x - 1
d)
2x3 + x2 - 8x + 2
|
|
Sanya Agarwal answered |
f(x) = 2x3 + 7x2 - 3
∴ f(x - 1) = 2(x - 1)3 + 7(x - 1)2 - 3
⇒ 2(x3 - 3x2 + 3x - 1) + 7(x2 + 1 - 2x) - 3
⇒ 2x3 - 6x2 + 6x - 2 + 7x2 + 7 - 14x - 3
⇒ 2x3 + x2 - 8x + 2
∴ f(x - 1) = 2(x - 1)3 + 7(x - 1)2 - 3
⇒ 2(x3 - 3x2 + 3x - 1) + 7(x2 + 1 - 2x) - 3
⇒ 2x3 - 6x2 + 6x - 2 + 7x2 + 7 - 14x - 3
⇒ 2x3 + x2 - 8x + 2
Let, R = {(a, b): a, b ϵ N and a2 = b}, then what is the relation R- a)Reflexive
- b)Symmetric
- c)Transitive
- d)None of the above
Correct answer is option 'D'. Can you explain this answer?
Let, R = {(a, b): a, b ϵ N and a2 = b}, then what is the relation R
a)
Reflexive
b)
Symmetric
c)
Transitive
d)
None of the above
|
|
Sanvi Kapoor answered |
Here, R = {(a, b): a, b ϵ N and a2 = b}
1. Relation R is not reflexive since, 2 ≠ 22
2. Since, 22 = 4 so, (2, 4) belong to R
But, 4 ≠ 2 and so, R is not symmetric
3. Since, 42 = 16, so (4, 16) belong to R
Also, 162 = 256, so (16, 256) belong to R
But, 42 ≠ 256 so R is not transitive.
So, R satisfies none of the reflexivity, symmetry and transitivity.
1. Relation R is not reflexive since, 2 ≠ 22
2. Since, 22 = 4 so, (2, 4) belong to R
But, 4 ≠ 2 and so, R is not symmetric
3. Since, 42 = 16, so (4, 16) belong to R
Also, 162 = 256, so (16, 256) belong to R
But, 42 ≠ 256 so R is not transitive.
So, R satisfies none of the reflexivity, symmetry and transitivity.
Hence, option (4) is correct.
If a set has n elements then the power set of that set has ______ elements.- a)n2
- b)2n
- c)2n
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
If a set has n elements then the power set of that set has ______ elements.
a)
n2
b)
2n
c)
2n
d)
None of these
|
|
Sanvi Kapoor answered |
Important point:
If a set has n elements then the power set of that set has 2n elements.
Comprehension:
Read the following information and answer the three items that follow:
Consider the function f(x) = g(x) + h(x)
What is the period of the function 
- a)15π
- b)20π
- c)15x/2
- d)10π
Correct answer is option 'A'. Can you explain this answer?
Comprehension:
Read the following information and answer the three items that follow:
Consider the function f(x) = g(x) + h(x)
What is the period of the function
Read the following information and answer the three items that follow:
Consider the function f(x) = g(x) + h(x)
What is the period of the function
a)
15π
b)
20π
c)
15x/2
d)
10π
|
|
Sanvi Kapoor answered |
Given:
We know that period of g(x) is 5x/2 and the period of h(x) is 3π.
As we know, If f and g are two functions with periods p and q, respectively, then the period of the function f + g is LCM (p, q).
So, period of f(x) is
If f (x) = x + 5 and 
then [(f × g)(5)] =- a)1
- b)2.5
- c)2/3
- d)1/25
Correct answer is option 'C'. Can you explain this answer?
If f (x) = x + 5 and then [(f × g)(5)] =
then [(f × g)(5)] =a)
1
b)
2.5
c)
2/3
d)
1/25
|
|
Sanya Agarwal answered |
f(x) = x + 5 ⇒ f(5) = 5 + 5 = 10.
Which of the following is an equivalence relation on the set of all functions from Z to Z ?- a){ (f, g) | f (x) − g (x) = 1 x ϵ Z }
- b){ (f, g) | f (0) = g (0) or f (1) = g (1) }
- c){ (f, g) | f (0) = g (1) and f (1) = g (0) }
- d){ (f, g) | f (x) − g (x) = k for some k ϵ Z }
Correct answer is option 'D'. Can you explain this answer?
Which of the following is an equivalence relation on the set of all functions from Z to Z ?
a)
{ (f, g) | f (x) − g (x) = 1 x ϵ Z }
b)
{ (f, g) | f (0) = g (0) or f (1) = g (1) }
c)
{ (f, g) | f (0) = g (1) and f (1) = g (0) }
d)
{ (f, g) | f (x) − g (x) = k for some k ϵ Z }
|
|
Sanvi Kapoor answered |
(1) { (f, g) | f (x) − g (x) = 1 x ϵ Z }
It is not reflexive. As f(x) – f(x) = 0 it is not 1. It is also not transitive. So, it cannot be an equivalence relation.
It is not reflexive. As f(x) – f(x) = 0 it is not 1. It is also not transitive. So, it cannot be an equivalence relation.
(2) {(f, g) | f (0) = g (0) or f (1) = g (1) }
This relation is not transitive. Suppose f(x) = 0, g(x) = x and h(x) = 1. Here f is not related to h. Only we have a relation given between f and g, g and h. but not between f and h. So, it cannot be an equivalence relation.
This relation is not transitive. Suppose f(x) = 0, g(x) = x and h(x) = 1. Here f is not related to h. Only we have a relation given between f and g, g and h. but not between f and h. So, it cannot be an equivalence relation.
(3) {(f, g) | f (0) = g (1) and f (1) = g (0) }
It is not always true f(0) = f(1) for reflexive case. So, it is not reflexive. Hence, no equivalence relation.
It is not always true f(0) = f(1) for reflexive case. So, it is not reflexive. Hence, no equivalence relation.
(4) { (f, g) | f (x) − g (x) = k for some k ϵ Z }
It is reflexive relation, consider constant as 0. It is also symmetric because the difference will be equal to a constant value. It is also transitive. So, it is an equivalence relation.
It is reflexive relation, consider constant as 0. It is also symmetric because the difference will be equal to a constant value. It is also transitive. So, it is an equivalence relation.
Suppose A is a finite set with n elements. The number of elements and the rank of the largest equivalence relation on A are- a){n,1}
- b){n, n}
- c){n2, 1}
- d){1, n2}
Correct answer is option 'C'. Can you explain this answer?
Suppose A is a finite set with n elements. The number of elements and the rank of the largest equivalence relation on A are
a)
{n,1}
b)
{n, n}
c)
{n2, 1}
d)
{1, n2}
|
|
Sanvi Kapoor answered |
A = {p, q} ∴ n = 4
R = {(p, p), (p, q), (q, p), (q, q)}
∴ Reflexive
The largest equivalence relation is n2 = 22 = 4
Diagraph of this relation will have only one connected component, hence this relation has rank 1.
R = {(p, p), (p, q), (q, p), (q, q)}
∴ Reflexive
The largest equivalence relation is n2 = 22 = 4
Diagraph of this relation will have only one connected component, hence this relation has rank 1.
The relation R in the set Integers given by R = {(a, b) : a – b is divisible by 3} is- a)reflexive
- b)reflexive but not symmetric
- c)not symmetric and transitive
- d)equivalence relation
Correct answer is option 'D'. Can you explain this answer?
The relation R in the set Integers given by R = {(a, b) : a – b is divisible by 3} is
a)
reflexive
b)
reflexive but not symmetric
c)
not symmetric and transitive
d)
equivalence relation
|
|
Sanvi Kapoor answered |
Concept:
1. Reflexive: Each element is related to itself.
R is reflexive if for all x ∈ A, xRx.
2. Symmetric: If any one element is related to any other element, then the second element is related to the first.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
3. Transitive: If any one element is related to a second and that second element is related to a third, then the first element is related to the third.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.
4. R is an equivalence relation if A is nonempty and R is reflexive, symmetric and transitive.
1. Reflexive: Each element is related to itself.
R is reflexive if for all x ∈ A, xRx.
2. Symmetric: If any one element is related to any other element, then the second element is related to the first.
R is symmetric if for all x, y ∈ A, if xRy, then yRx.
3. Transitive: If any one element is related to a second and that second element is related to a third, then the first element is related to the third.
R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.
4. R is an equivalence relation if A is nonempty and R is reflexive, symmetric and transitive.
Calculation:
For reflexive:
(a, a) = a – a = 0 is divisible by 3.
For symmetric:
If (a - b) is divisible by 3, then (b – a) = - (a - b) is also divisible by 3.
Thus relation is symmetric.
For transitive:
If (a - b) and (b - c) is divisible by 3
Then (a - c) is also divisible by 3
Thus relation is transitive.
∴ R is an equivalence relation
Hence, option (d) is correct.
For reflexive:
(a, a) = a – a = 0 is divisible by 3.
For symmetric:
If (a - b) is divisible by 3, then (b – a) = - (a - b) is also divisible by 3.
Thus relation is symmetric.
For transitive:
If (a - b) and (b - c) is divisible by 3
Then (a - c) is also divisible by 3
Thus relation is transitive.
∴ R is an equivalence relation
Hence, option (d) is correct.
Which of the following two sets are equal?- a)A = {1, 2} and B = {1}
- b)A = {1, 2} and B = {1, 2, 3}
- c)A = {1, 2, 3} and B = {2, 1, 3}
- d)A = {1, 2, 4} and B = {1, 2, 3}
Correct answer is option 'C'. Can you explain this answer?
Which of the following two sets are equal?
a)
A = {1, 2} and B = {1}
b)
A = {1, 2} and B = {1, 2, 3}
c)
A = {1, 2, 3} and B = {2, 1, 3}
d)
A = {1, 2, 4} and B = {1, 2, 3}
|
|
Sanya Agarwal answered |
Two set are equal if and only if they have the same elements.
The Cartesian Product B x A is equal to the Cartesian product A x B.- a)True
- b)False
Correct answer is option 'B'. Can you explain this answer?
The Cartesian Product B x A is equal to the Cartesian product A x B.
a)
True
b)
False
|
|
Sanya Agarwal answered |
Let A = {1, 2} and B = {a, b}. The Cartesian product A x B = {(1, a), (1, b), (2, a), (2, b)} and the Cartesian product B x A = {(a, 1), (a, 2), (b, 1), (b, 2)}. This is not equal to A x B.
The members of the set S = {x | x is the square of an integer and x < 100} is __________.- a){0, 2, 4, 5, 9, 58, 49, 56, 99, 12}
- b){0, 1, 4, 9, 16, 25, 36, 49, 64, 81}
- c){1, 4, 9, 16, 25, 36, 64, 81, 85, 99}
- d){0, 1, 4, 9, 16, 25, 36, 49, 64, 121}
Correct answer is option 'B'. Can you explain this answer?
The members of the set S = {x | x is the square of an integer and x < 100} is __________.
a)
{0, 2, 4, 5, 9, 58, 49, 56, 99, 12}
b)
{0, 1, 4, 9, 16, 25, 36, 49, 64, 81}
c)
{1, 4, 9, 16, 25, 36, 64, 81, 85, 99}
d)
{0, 1, 4, 9, 16, 25, 36, 49, 64, 121}
|
|
Sanya Agarwal answered |
The set S consists of the square of an integer less than 10.
Chapter doubts & questions for Set Theory & Algebra - Engineering Mathematics 2025 is part of Civil Engineering (CE) exam preparation. The chapters have been prepared according to the Civil Engineering (CE) exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Civil Engineering (CE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of Set Theory & Algebra - Engineering Mathematics in English & Hindi are available as part of Civil Engineering (CE) exam.
Download more important topics, notes, lectures and mock test series for Civil Engineering (CE) Exam by signing up for free.
Engineering Mathematics
65 videos|122 docs|94 tests
|
Signup to see your scores go up within 7 days!
Study with 1000+ FREE Docs, Videos & Tests
10M+ students study on EduRev
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up
within 7 days!
within 7 days!
Takes less than 10 seconds to signup