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All questions of Polynomial Equations and Inequalities for Grade 12 Exam
For real number x and y, we writeis an irrationalnumber. Then the relation R is:a)Reflexiveb)Symmetricc)Transitived)EquivalenceCorrect answer is option 'A'. Can you explain this answer?
|
Avantika Dasgupta answered |
xRy => x - y + √2 is an irrational number.
Let R is a binary relation on real numbers x and y.
Now, R is transitive iff for all (x, y) ∈ R and (y, z) ∈ R implies (x, z) ∈ R
Given, xRy => x - y + √2 is irrational ............1
and yRz => y - z + √2 is irrational ............2
Add equation 1 and 2, we get
(x - y + √2) + (y - z + √2) is irrational
= x - z + √2 is irrational
= xRz is irrational
So, the relation R is transitive.
If A = { (1, 2, 3}, then the relation R = {(2, 3)} in A isa)symmetric and transitive
|
Pallavi Ahuja answered |
Symmetric
The relation R is said to be symmetric if for every element (a, b) in R, the element (b, a) is also in R. In other words, if (a, b) ∈ R, then (b, a) ∈ R.
In this case, the relation R = {(2, 3)} has only one element. To check if it is symmetric, we need to check if (3, 2) is also in R. Since (3, 2) is not in R, the relation R is not symmetric.
Transitive
The relation R is said to be transitive if for every elements (a, b) and (b, c) in R, the element (a, c) is also in R. In other words, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.
In this case, we need to check if (2, 3) ∈ R and (3, 2) ∈ R imply that (2, 2) ∈ R. Since (2, 3) ∈ R and (3, 2) is not in R, we cannot determine if (2, 2) ∈ R or not. Therefore, the relation R is not transitive.
Summary
- The relation R = {(2, 3)} is not symmetric because it does not contain the element (3, 2).
- The relation R is not transitive because we cannot determine if (2, 2) ∈ R or not based on the given relation R.
Therefore, the relation R = {(2, 3)} is neither symmetric nor transitive.
The relation R is said to be symmetric if for every element (a, b) in R, the element (b, a) is also in R. In other words, if (a, b) ∈ R, then (b, a) ∈ R.
In this case, the relation R = {(2, 3)} has only one element. To check if it is symmetric, we need to check if (3, 2) is also in R. Since (3, 2) is not in R, the relation R is not symmetric.
Transitive
The relation R is said to be transitive if for every elements (a, b) and (b, c) in R, the element (a, c) is also in R. In other words, if (a, b) ∈ R and (b, c) ∈ R, then (a, c) ∈ R.
In this case, we need to check if (2, 3) ∈ R and (3, 2) ∈ R imply that (2, 2) ∈ R. Since (2, 3) ∈ R and (3, 2) is not in R, we cannot determine if (2, 2) ∈ R or not. Therefore, the relation R is not transitive.
Summary
- The relation R = {(2, 3)} is not symmetric because it does not contain the element (3, 2).
- The relation R is not transitive because we cannot determine if (2, 2) ∈ R or not based on the given relation R.
Therefore, the relation R = {(2, 3)} is neither symmetric nor transitive.
Can you explain the answer of this question below:Let f: Q→Q be a function given by f(x) = x2,then f -1(9) =
- A:
{3}
- B:
f
- C:
{-3, 3}
- D:
{-3}
The answer is c.
Let f: Q→Q be a function given by f(x) = x2,then f -1(9) =
{3}
f
{-3, 3}
{-3}
|
Manish Suthar answered |
Function has only one image not 2 . is it right So the answer should be 3
Number of binary sets on the set {p, q,r} is:- a)39
- b)16
- c)18
- d)36
Correct answer is option 'A'. Can you explain this answer?
Number of binary sets on the set {p, q,r} is:
a)
39
b)
16
c)
18
d)
36
|
Guru Randhawa answered |
It os must to remember these basic results the answer is 1 at 8 corners 1/8 sphere is present :8*1/8=1 that means overall( effective) one atom is present in ssc
Let R be a relation on N (set of natural numbers) such that (m, n) R (p, q)mq(n + p) = np(m + q). Then, R is- a)An Equivalence Relation
- b)Only Reflexive
- c)Symmetric and reflexive.
- d)Only Transitive
Correct answer is option 'C'. Can you explain this answer?
Let R be a relation on N (set of natural numbers) such that (m, n) R (p, q)mq(n + p) = np(m + q). Then, R is
a)
An Equivalence Relation
b)
Only Reflexive
c)
Symmetric and reflexive.
d)
Only Transitive
|
Anaya Patel answered |
(m, n) R (p, q) <=> mq(n + p) = np(m + q)
For all m,n,p,q € N
Reflexive:
(m, n) R (m, n) <=> mn(n + m) = nm(m + n)
⇒ mn2 + m2n = nm2 + n2m
⇒ mn2 + m2n = mn2 + m2n
⇒ LHS = RHS
So, (m, n) R (m, n) exists.
Hence, it is Reflexive
Symmetric:
Let (m, n) R (p, q) exists
mq(n + p) = np(m + q) --- (eqn1)
(p, q) R (m, n) <=> pn(q + m) = qm (p + n)
⇒ np(m + q) = mq(n + p)
⇒ mq(n + p) = np(m + q)
This equation is true by (eqn1).
So, (p, q) R (m, n) exists
Hence, it is not symmetric.
Transitive:
Let (m, n) R (p, q) and (p, q) R (r, s) exists.
Therefore,
mq(n + p) = np(m + q) --- (eqn1)
ps(q + r) = qr (p + s) --- (eqn2)
We cannot obtain ms(n+r) = nr(m+s) using eqn1 and eqn2.
So, ms(n + r) ≠ nr(m + s)
Therefore, (m, n) R (r, s) doesn’t exist.
Hence, it is transitive.
For all m,n,p,q € N
Reflexive:
(m, n) R (m, n) <=> mn(n + m) = nm(m + n)
⇒ mn2 + m2n = nm2 + n2m
⇒ mn2 + m2n = mn2 + m2n
⇒ LHS = RHS
So, (m, n) R (m, n) exists.
Hence, it is Reflexive
Symmetric:
Let (m, n) R (p, q) exists
mq(n + p) = np(m + q) --- (eqn1)
(p, q) R (m, n) <=> pn(q + m) = qm (p + n)
⇒ np(m + q) = mq(n + p)
⇒ mq(n + p) = np(m + q)
This equation is true by (eqn1).
So, (p, q) R (m, n) exists
Hence, it is not symmetric.
Transitive:
Let (m, n) R (p, q) and (p, q) R (r, s) exists.
Therefore,
mq(n + p) = np(m + q) --- (eqn1)
ps(q + r) = qr (p + s) --- (eqn2)
We cannot obtain ms(n+r) = nr(m+s) using eqn1 and eqn2.
So, ms(n + r) ≠ nr(m + s)
Therefore, (m, n) R (r, s) doesn’t exist.
Hence, it is transitive.
Let * be any binary operation on the set R defined by a * b = a + b – ab, then the binary operation * is- a)Associative
- b)Commutative but not associative
- c)Commutative and associative
- d)Commutative
Correct answer is option 'C'. Can you explain this answer?
Let * be any binary operation on the set R defined by a * b = a + b – ab, then the binary operation * is
a)
Associative
b)
Commutative but not associative
c)
Commutative and associative
d)
Commutative
|
|
Rajat Patel answered |
The answer is b.
Given that,
Given that,
a * b = 1 + ab, a, b ∈ R.
If 
, then inverse of f is:- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
If , then inverse of f is:
, then inverse of f is:a)
b)
c)
d)
|
Kaia Patil answered |
F(×)=4x+3/6x-4
y=4x+3/6x-4
inter change x and y
x=4y+3/6y-4
find y
6xy-4x=4y+3
y(6x-4)=4x+3
y=4x+3/6x-4
y is the inverse of function
y=4x+3/6x-4
inter change x and y
x=4y+3/6y-4
find y
6xy-4x=4y+3
y(6x-4)=4x+3
y=4x+3/6x-4
y is the inverse of function
, where a, b are the elements of a set of non-zero rational numbers, isIf f(x) = |x| and g(x) = | 5x – 2 |. Then, fog =- a)|g|
- b)| f |
- c)f
- d)f2
Correct answer is option 'A'. Can you explain this answer?
If f(x) = |x| and g(x) = | 5x – 2 |. Then, fog =
a)
|g|
b)
| f |
c)
f
d)
f2
|
|
Aryan Khanna answered |
f(x) = |x| g(x) = |5x - 2|
fog = f(g(x) = g(x) = |5x - 2|
fog = f(g(x) = g(x) = |5x - 2|
Let C = {(a, b): a2 + b2 = 1; a, b ∈ R} a relation on R, set of real numbers. Then C isa)Equivalence relationb)Reflexivec)Transitived)SymmetricCorrect answer is option 'A'. Can you explain this answer?
|
|
Krishna Iyer answered |
Correct Answer :- D
Explanation:- Check for reflexive
Consider (a,a)
∴ a2+a2
=1 which is not always true.
If a=2
∴ 22+22
=1⇒4+4=1 which is false.
∴ R is not reflexive ---- ( 1 )
Check for symmetric
aRb⇒a2+b2=1
bRa⇒b2+a2 =1
Both the equation are the same and therefore will always be true.
∴ R is symmetric ---- ( 2 )
Check for transitive
aRb⇒a2+b2=1
bRc⇒b2+c2=1
∴ a2+c2=1 will not always be true.
Let a=−1,b=0 and c=1
∴ (−1)2+02=1,
= 02+12
=1 are true.
But (−1)2+12
=1 is false.
∴ R is not transitive ---- ( 3 )
If A = {1,3,5,7} and we define a relation
Then the number of elements in the relation R is- a)2
- b)1
- c)3
- d)0
Correct answer is option 'D'. Can you explain this answer?
If A = {1,3,5,7} and we define a relationThen the number of elements in the relation R is
Then the number of elements in the relation R isa)
2
b)
1
c)
3
d)
0
|
|
Genius answered |
Ya.. it will Nile.. means 0 ... coz there is no any element of set A ... which follow set R elements !!!
as it is said.. element a nd b belongs to set A... bt follow a-b|8
as it is said.. element a nd b belongs to set A... bt follow a-b|8
In the set N x N, the relation R is defined by (a, b) R (c, d) ⇔ a+d = b+c. Then R is
a)symmetric and transitive but not reflexiveb)reflexive and transitive but not symmetricc)Equivalence relationd)Partial order relationCorrect answer is option 'C'. Can you explain this answer?
|
|
Rahul Bansal answered |
The correct answer is c
If f(x) = ax + b and g(x) = cx + d, then f[g(x)] – g[f(x)] is equivalent to- a)f(c) + g(a)
- b)f(a) – g(c)
- c)f(d) – g(b)
- d)f(b) – g(b)
Correct answer is option 'C'. Can you explain this answer?
If f(x) = ax + b and g(x) = cx + d, then f[g(x)] – g[f(x)] is equivalent to
a)
f(c) + g(a)
b)
f(a) – g(c)
c)
f(d) – g(b)
d)
f(b) – g(b)
|
|
Leelu Bhai answered |
F(g(x)) - g(f(x)) = f(cx + d) - g(ax + b)= a(cx + d) + b - c(ax + b) - d= acx + ad + b - acx - bc - d= ad + b - bc - dnow by putting each
How many onto functions from set A to set A can be formed for the set A = { 1,2,3,4,5,……n} ?- a)n2
- b)n
- c)n!
- d)2n
Correct answer is option 'C'. Can you explain this answer?
How many onto functions from set A to set A can be formed for the set A = { 1,2,3,4,5,……n} ?
a)
n2
b)
n
c)
n!
d)
2n
|
Mehakbir Singh answered |
The correct answer is n! because it is a formula.
The law a + b = b + a is called- a)Associative law
- b)Distributive law
- c)Closure law
- d)Commutative law
Correct answer is option 'D'. Can you explain this answer?
The law a + b = b + a is called
a)
Associative law
b)
Distributive law
c)
Closure law
d)
Commutative law
|
Naina Bansal answered |
Commutative law is the Law that says you can swap numbers around and still get the same answer when you add or when you multiply i.e a + b = b + a
Example:
You can swap when you add: 6 + 3 = 3 + 6 (= 9)
- a)8
- b)16
- c)1
- d)4
Correct answer is 'D'. Can you explain this answer?
|
Janhavi Kaur answered |
Number of reationsl would be 4 as.. 1+7,7+1,3+5,5+3 all are equal to 8
If R is a relation from a non – empty set A to a non – empty set B, then- a)R⊂A×B.
- b)R=A∪B
- c)R=A∩B
- d)R=A×B
Correct answer is option 'A'. Can you explain this answer?
If R is a relation from a non – empty set A to a non – empty set B, then
a)
R⊂A×B.
b)
R=A∪B
c)
R=A∩B
d)
R=A×B
|
|
Anaya Patel answered |
Let A and B be two sets. Then a relation R from set A to set B is a subset of A × B. Thus, R is a relation from A to B ⇔ R ⊆ A × B.
If f be a mapping defined by f(x) = x2 + 5, then 
is:- a)(x - 5)1/2
- b)(x2 + 5)1/2
- c)(x + 5)1/2
- d)
Correct answer is option 'A'. Can you explain this answer?
If f be a mapping defined by f(x) = x2 + 5, then is:
is:a)
(x - 5)1/2
b)
(x2 + 5)1/2
c)
(x + 5)1/2
d)
|
|
Kaia Patil answered |
F(x)=x^2+5
y=x^2+5
interchange x and y
x=y^2+5
y=(x-5)^1/2
y is the inverse of f(x)
y=x^2+5
interchange x and y
x=y^2+5
y=(x-5)^1/2
y is the inverse of f(x)
Can you explain the answer of this question below:Let R = {(P,Q) : OP = OQ , O being the origin} be an equivalence relation on A. The equivalence class [(1,2)] is
- A:
{(x,y) : x2+y2 = 5}
- B:
{(x,y) : x2 = y2}
- C:
{(x,y) : x2+y2 = 1}
- D:
{(x,y) : x2+y2 = 4}
The answer is a.
Let R = {(P,Q) : OP = OQ , O being the origin} be an equivalence relation on A. The equivalence class [(1,2)] is
{(x,y) : x2+y2 = 5}
{(x,y) : x2 = y2}
{(x,y) : x2+y2 = 1}
{(x,y) : x2+y2 = 4}
|
Om Desai answered |
Correct Answer : a
Explanation : A = {(x,y) : x2 + y2 = 5}
=> (1,2) {(1)2 + (4)2 = 5}
=> {1 + 4 = 5}
=> {5 = 5}
If ƒ(x) = xsecx, then ƒ(0) =- a)−1
- b)0
- c)1
- d)√(2)
Correct answer is option 'B'. Can you explain this answer?
If ƒ(x) = xsecx, then ƒ(0) =
a)
−1
b)
0
c)
1
d)
√(2)
|
Nabanita Bajaj answered |
You could have any superpower, what would it be?
A relation R from C to R is defined by x Ry if f |x| = y. Which of the following is correct?- a)iR1
- b)3R(–3)
- c)(2 + 3 i)R13
- d)(1 + i)R2
Correct answer is option 'A'. Can you explain this answer?
A relation R from C to R is defined by x Ry if f |x| = y. Which of the following is correct?
a)
iR1
b)
3R(–3)
c)
(2 + 3 i)R13
d)
(1 + i)R2
|
Preeti Iyer answered |
|2+3i| = sqroot( square(2) + square(3) ) = sqroot(13) which is not equal to 13
so, (2+3i)R 13 is wrong.
|3| = sqroot( square(3) + square(0) ) = sqroot(9) = 3 which is not equal to -3
so, 3R (-3) is wrong.
|1+i| = sqroot( square(1) + square(1) ) = sqroot(2) which is not equal to 2
so, (1+i)R 2 is wrong.
|i| =sqroot( square(0) + square(1) ) = sqroot(1) which is equal to 1
so, iR1 is correct.
A function f: X → Y is injective if and only if- a)
- b)
- c)f(X) = Y
- d)X = Y
Correct answer is option 'A'. Can you explain this answer?
A function f: X → Y is injective if and only if
a)
b)
c)
f(X) = Y
d)
X = Y
|
Gunjan Lakhani answered |
Let X, Y be sets, and let f : X → Y be a function. We say that f is injective (sometimes called one-to-one) if ∀x1, x2 ∈ X, f(x1) = f(x2)
⇒ x1 = x2.
⇒ x1 = x2.
Let f and g be the function from the set of integers to itself, defined by f(x) = 2x + 1 and g(x) = 3x + 4. Then the composition of f and g is ____________- a)6x + 9
- b)6x + 7
- c)6x + 6
- d)6x + 8
Correct answer is option 'A'. Can you explain this answer?
Let f and g be the function from the set of integers to itself, defined by f(x) = 2x + 1 and g(x) = 3x + 4. Then the composition of f and g is ____________
a)
6x + 9
b)
6x + 7
c)
6x + 6
d)
6x + 8
|
|
Arjun Singhania answered |
The composition of f and g is given by f(g(x)) which is equal to 2(3x + 4) + 1.
Let f : A→B and g : B→C be one-one functions. Then, gof: A→C is- a)one-one
- b)does not exist
- c)not onto
- d)onto
Correct answer is option 'A'. Can you explain this answer?
Let f : A→B and g : B→C be one-one functions. Then, gof: A→C is
a)
one-one
b)
does not exist
c)
not onto
d)
onto
|
|
Naina Bansal answered |
f is one-one g is one-oneIf f(x1) = f(x2) If g(x2) = g(x2)then x1 = x2 then x1= x2C
Let A = {1, 2, 3} and B = {5, 6, 7, 8, 9} and let f(x) = {(1, 8), (2, 7), (3, 6)} then f is- a)Surjective
- b)Not a function
- c)Injective
- d)Bijective
Correct answer is option 'C'. Can you explain this answer?
Let A = {1, 2, 3} and B = {5, 6, 7, 8, 9} and let f(x) = {(1, 8), (2, 7), (3, 6)} then f is
a)
Surjective
b)
Not a function
c)
Injective
d)
Bijective
|
|
Sharmila Choudhury answered |
Solution:
To determine whether f is surjective, injective, or bijective, we need to understand what these terms mean.
- Surjective: A function f from set A to set B is surjective (onto) if for every element y in set B, there is an element x in set A such that f(x) = y. In other words, every element in set B has at least one preimage in set A.
- Injective: A function f from set A to set B is injective (one-to-one) if for every pair of distinct elements x₁ and x₂ in set A, f(x₁) and f(x₂) are distinct in set B. In other words, no two distinct elements in set A have the same image in set B.
- Bijective: A function f from set A to set B is bijective if it is both surjective and injective. In other words, every element in set B has exactly one preimage in set A, and no two distinct elements in set A have the same image in set B.
Now, let's look at the given function f(x) = {(1, 8), (2, 7), (3, 6)}.
- f is not surjective because set B has elements 5 and 9 that do not have any preimages in set A.
- f is not a function if there is an element in set A that has more than one image in set B. However, this is not the case for f, so it is a function.
- f is injective because no two distinct elements in set A have the same image in set B. For example, f(1) = 8, f(2) = 7, and f(3) = 6, so no two distinct elements in set A have the same image in set B.
- Since f is injective but not surjective, it is not bijective.
Therefore, the correct answer is option 'C' - f is injective.
To determine whether f is surjective, injective, or bijective, we need to understand what these terms mean.
- Surjective: A function f from set A to set B is surjective (onto) if for every element y in set B, there is an element x in set A such that f(x) = y. In other words, every element in set B has at least one preimage in set A.
- Injective: A function f from set A to set B is injective (one-to-one) if for every pair of distinct elements x₁ and x₂ in set A, f(x₁) and f(x₂) are distinct in set B. In other words, no two distinct elements in set A have the same image in set B.
- Bijective: A function f from set A to set B is bijective if it is both surjective and injective. In other words, every element in set B has exactly one preimage in set A, and no two distinct elements in set A have the same image in set B.
Now, let's look at the given function f(x) = {(1, 8), (2, 7), (3, 6)}.
- f is not surjective because set B has elements 5 and 9 that do not have any preimages in set A.
- f is not a function if there is an element in set A that has more than one image in set B. However, this is not the case for f, so it is a function.
- f is injective because no two distinct elements in set A have the same image in set B. For example, f(1) = 8, f(2) = 7, and f(3) = 6, so no two distinct elements in set A have the same image in set B.
- Since f is injective but not surjective, it is not bijective.
Therefore, the correct answer is option 'C' - f is injective.
is defined by f(x) = [x+1], where [x] the greatest integer function, is:
Let g(x) = 1 + x – [x] and 
Then f{g(x)} for all x, is equal to:- a)g(x)
- b)x
- c)1
- d)f(x)
Correct answer is option 'C'. Can you explain this answer?
Let g(x) = 1 + x – [x] and
Then f{g(x)} for all x, is equal to:
a)
g(x)
b)
x
c)
1
d)
f(x)
|
|
Om Desai answered |
g(x) = 1 + x - [x]
f(x) = {-1, x<0 0, x=0 1, x>0}
f(g(x)) = f[1 + x - [x]]
= f[1 + x - 1]
= f(x)
f(g(x)) is equal to f(x)
f(x) = {-1, x<0 0, x=0 1, x>0}
f(g(x)) = f[1 + x - [x]]
= f[1 + x - 1]
= f(x)
f(g(x)) is equal to f(x)
be defined by 
Which one of the following relations on set of real numbers is an equivalence relation?- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Which one of the following relations on set of real numbers is an equivalence relation?
a)
b)
c)
d)
|
Kumari Sakshi answered |
Equivalence relation is symmetric reflexive and transitive now see option A a>=b but in this case relation will not be symmetric now see option B if |a|=|b | then |b|=|a| so this is symmetric as well as reflexive and if |a|=|b| and |b|=|c| then|c|=|a | then it is transitive as well so it is an equivalence relation you can check yourself for the rest two
Let A = {1,2,3,4} and B = { x,y,z}. Then R = {(1,x) , ( 2,z), (1,y), (3,x)} is- a)relation from B to A
- b)Is not a relation
- c)relation from A to B
- d)relation from B to B
Correct answer is option 'C'. Can you explain this answer?
Let A = {1,2,3,4} and B = { x,y,z}. Then R = {(1,x) , ( 2,z), (1,y), (3,x)} is
a)
relation from B to A
b)
Is not a relation
c)
relation from A to B
d)
relation from B to B
|
|
Om Desai answered |
Let a set of A = (1, 2, 3, 4) and set B (x, y, z) so. set A of all elements in set B then the relation of A to B
The function, f(x) = 2x - 1 is
- a)It is not surjective
- b)It is surjective
- c)It is surjective in some intervals of the domain
- d)Insufficient information
Correct answer is option 'B'. Can you explain this answer?
The function, f(x) = 2x - 1 is
a)
It is not surjective
b)
It is surjective
c)
It is surjective in some intervals of the domain
d)
Insufficient information
|
Vivek Patel answered |
The function f(x) = 2x - 1 is surjective.
Explanation:
A function is surjective (also called onto) if for every element y in the codomain, there exists an element x in the domain such that f(x) = y.
In this case, the function f(x) = 2x - 1 is a linear function. For any given y value, we can find an x value such that f(x) = y:
y = 2x - 1
x = (y + 1) / 2
So, for any y value, there exists an x value that satisfies the equation. Therefore, the function is surjective.
Explanation:
A function is surjective (also called onto) if for every element y in the codomain, there exists an element x in the domain such that f(x) = y.
In this case, the function f(x) = 2x - 1 is a linear function. For any given y value, we can find an x value such that f(x) = y:
y = 2x - 1
x = (y + 1) / 2
So, for any y value, there exists an x value that satisfies the equation. Therefore, the function is surjective.
. Then f is
Which of the following is not an equivalence relation on I, the set of integers ; x, y- a)x R y x + y is an even integer
- b)x R y x = y
- c)x R y x < y
- d)x R y x – y is an even integer
Correct answer is option 'C'. Can you explain this answer?
Which of the following is not an equivalence relation on I, the set of integers ; x, y
a)
x R y x + y is an even integer
b)
x R y x = y
c)
x R y x < y
d)
x R y x – y is an even integer
|
Aravind Nambiar answered |
If R is a relation defined by xRy : ifx ⩽ y,then R is reflexive and transitive .But , it is not symmetric. Hence , R is not an equivalence relation.
The binary relation S = Φ (empty set) on set A = {1, 2,3} is- a)transitive and relexive
- b)symmetric and relexive
- c)transitive and symmetric
- d)neither reflexive nor symmetric
Correct answer is option 'C'. Can you explain this answer?
The binary relation S = Φ (empty set) on set A = {1, 2,3} is
a)
transitive and relexive
b)
symmetric and relexive
c)
transitive and symmetric
d)
neither reflexive nor symmetric
|
|
Rohit Jain answered |
Reflexive : A relation is reflexive if every element of set is paired with itself. Here none of the element of A is paired with themselves, so S is not reflexive.
Symmetric : This property says that if there is a pair (a, b) in S, then there must be a pair (b, a) in S. Since there is no pair here in S, this is trivially true, so S is symmetric.
Transitive : This says that if there are pairs (a, b) and (b, c) in S, then there must be pair (a,c) in S. Again, this condition is trivially true, so S is transitive.
Symmetric : This property says that if there is a pair (a, b) in S, then there must be a pair (b, a) in S. Since there is no pair here in S, this is trivially true, so S is symmetric.
Transitive : This says that if there are pairs (a, b) and (b, c) in S, then there must be pair (a,c) in S. Again, this condition is trivially true, so S is transitive.
Given below is binary composition table a* b = LCM of a and b on S = { 1,2,3,4}. Then, from the table determine which one of these options is correct.
- a)a* b is a distributive binary operation
- b)a* b is an associative binary operation
- c)a* b is not a binary operation
- d)a* b is a commutative binary operation
Correct answer is option 'C'. Can you explain this answer?
Given below is binary composition table a* b = LCM of a and b on S = { 1,2,3,4}. Then, from the table determine which one of these options is correct.
a)
a* b is a distributive binary operation
b)
a* b is an associative binary operation
c)
a* b is not a binary operation
d)
a* b is a commutative binary operation
|
Lavanya Menon answered |
We have ∗ is defined on the set {1,2,3,4,5} by a∗b = LCM of a and b
Now, 3∗4 = LCM of 3 and 4=12
But, 12 does not belong to {1,2,3,4,5}.
Hence, ′∗′ is not a binary operation.
Now, 3∗4 = LCM of 3 and 4=12
But, 12 does not belong to {1,2,3,4,5}.
Hence, ′∗′ is not a binary operation.
Let R be the relation on the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3,3), (3,2)}. then R is- a)R is reflexive and symmetric but not transitive.
- b)R is symmetric and transitive but not reflexive.
- c)R is an equivalence relation.
- d)R is reflexive and transitive but not symmetric.
Correct answer is option 'D'. Can you explain this answer?
Let R be the relation on the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3,3), (3,2)}. then R is
a)
R is reflexive and symmetric but not transitive.
b)
R is symmetric and transitive but not reflexive.
c)
R is an equivalence relation.
d)
R is reflexive and transitive but not symmetric.
|
Pragati Dey answered |
R be the relation in the set {1,2,3,4] given by R ={(1,2),(2,2),(1,1),(4,4),(1,3),(3,3),(3,2)}
it is seen that (a,a)∈ R for every a∈{1,2,3,4}
so,R is feflexive.
it is seen that (a,b) = (b,a) ∈ R
because, (1,2)∈ R but (2,1) ∉R
so, R is not symmetric.
it is seen that (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ {1, 2, 3, 4}.
so, R is transitive.
Hence, R is reflexive and transitive but not symmetric.
A function f: X → Y is surjective if and only if- a)X = Y
- b)f(X) = Y
- c)X is a singleton set
- d)Y is a singleton set
Correct answer is option 'B'. Can you explain this answer?
A function f: X → Y is surjective if and only if
a)
X = Y
b)
f(X) = Y
c)
X is a singleton set
d)
Y is a singleton set
|
|
Aqsa Kibria answered |
Hiii Saumya!!! Surjective function is one which is onto which means there is a value of y corresponding to every x in the domain..and which is option B f(X)=Y Regards
The void relation (a subset of A x A) on a non empty set A is :- a)Anti symmetric
- b)transitive
- c)Reflexive
- d)Transitive and symmetric
Correct answer is option 'D'. Can you explain this answer?
The void relation (a subset of A x A) on a non empty set A is :
a)
Anti symmetric
b)
transitive
c)
Reflexive
d)
Transitive and symmetric
|
Tejas Chawla answered |
The relation { }⊂ A x A on a is surely not reflexive.However ,neither symmetry nor transitivity is contradicted .So { } is a transitive and symmetry relation on A.
Let A = {1,2,3,4,5,6,7}. P = {1,2}, Q = {3, 7}. Write the elements of the set R so that P, Q and R form a partition that results in equivalence relation.- a){4,5,6}
- b){0}
- c){1,2,3,4,5,6,7}
- d){ }
Correct answer is option 'A'. Can you explain this answer?
Let A = {1,2,3,4,5,6,7}. P = {1,2}, Q = {3, 7}. Write the elements of the set R so that P, Q and R form a partition that results in equivalence relation.
a)
{4,5,6}
b)
{0}
c)
{1,2,3,4,5,6,7}
d)
{ }
|
|
Rajdeep Singh answered |
Solution:
Equivalence relation is a relation that is reflexive, symmetric and transitive. A partition of a set is a collection of non-empty subsets whose union is the original set and whose members are pairwise disjoint. Let's find the set R so that P, Q and R form a partition that results in equivalence relation.
Reflexive property: Every element is related to itself.
Symmetric property: If a is related to b, then b is related to a.
Transitive property: If a is related to b and b is related to c, then a is related to c.
Partition: P, Q and R are pairwise disjoint. Their union is A.
Elements of the set R: {4, 5, 6}
Explanation:
We can choose the set R as {4, 5, 6} because it is disjoint from both P and Q. This means that the sets P, Q and R are pairwise disjoint. Also, their union is A, which means that every element of A belongs to exactly one of the sets P, Q and R.
Now, let's check whether the partition formed by P, Q and R results in an equivalence relation.
Reflexive property: Every element is related to itself.
- For any element x in A, x belongs to exactly one of the sets P, Q and R.
- If x belongs to P or Q, then x is related to itself because P and Q are both reflexive.
- If x belongs to R, then x is related to itself because R is a set of elements.
Symmetric property: If a is related to b, then b is related to a.
- For any elements a and b in A, a belongs to exactly one of the sets P, Q and R, and b belongs to exactly one of the sets P, Q and R.
- If a and b belong to the same set (P, Q or R), then the relation between them is symmetric because P, Q and R are all symmetric.
- If a and b belong to different sets, then they are not related. Therefore, the relation between them is trivially symmetric.
Transitive property: If a is related to b and b is related to c, then a is related to c.
- For any elements a, b and c in A, a belongs to exactly one of the sets P, Q and R, b belongs to exactly one of the sets P, Q and R, and c belongs to exactly one of the sets P, Q and R.
- If a and b belong to the same set and b and c belong to the same set, then a and c belong to the same set by the transitive property of P, Q and R.
- If a and b belong to different sets or b and c belong to different sets, then a and c are not related. Therefore, the relation between them is trivially transitive.
Therefore, the partition formed by P, Q and R results in an equivalence relation.
Equivalence relation is a relation that is reflexive, symmetric and transitive. A partition of a set is a collection of non-empty subsets whose union is the original set and whose members are pairwise disjoint. Let's find the set R so that P, Q and R form a partition that results in equivalence relation.
Reflexive property: Every element is related to itself.
Symmetric property: If a is related to b, then b is related to a.
Transitive property: If a is related to b and b is related to c, then a is related to c.
Partition: P, Q and R are pairwise disjoint. Their union is A.
Elements of the set R: {4, 5, 6}
Explanation:
We can choose the set R as {4, 5, 6} because it is disjoint from both P and Q. This means that the sets P, Q and R are pairwise disjoint. Also, their union is A, which means that every element of A belongs to exactly one of the sets P, Q and R.
Now, let's check whether the partition formed by P, Q and R results in an equivalence relation.
Reflexive property: Every element is related to itself.
- For any element x in A, x belongs to exactly one of the sets P, Q and R.
- If x belongs to P or Q, then x is related to itself because P and Q are both reflexive.
- If x belongs to R, then x is related to itself because R is a set of elements.
Symmetric property: If a is related to b, then b is related to a.
- For any elements a and b in A, a belongs to exactly one of the sets P, Q and R, and b belongs to exactly one of the sets P, Q and R.
- If a and b belong to the same set (P, Q or R), then the relation between them is symmetric because P, Q and R are all symmetric.
- If a and b belong to different sets, then they are not related. Therefore, the relation between them is trivially symmetric.
Transitive property: If a is related to b and b is related to c, then a is related to c.
- For any elements a, b and c in A, a belongs to exactly one of the sets P, Q and R, b belongs to exactly one of the sets P, Q and R, and c belongs to exactly one of the sets P, Q and R.
- If a and b belong to the same set and b and c belong to the same set, then a and c belong to the same set by the transitive property of P, Q and R.
- If a and b belong to different sets or b and c belong to different sets, then a and c are not related. Therefore, the relation between them is trivially transitive.
Therefore, the partition formed by P, Q and R results in an equivalence relation.
Let A = {a, b, c} and R = {(a, a), (b, b), (c, c), (b, c)} be a relation on A. Here, R is- a)Transitive
- b)anti – symmetric
- c)symmetric
- d)reflexive
Correct answer is option 'D'. Can you explain this answer?
Let A = {a, b, c} and R = {(a, a), (b, b), (c, c), (b, c)} be a relation on A. Here, R is
a)
Transitive
b)
anti – symmetric
c)
symmetric
d)
reflexive
|
Anjali Deshpande answered |
Any relation R is reflexive if fx Rx for all x ∈ R. Here ,(a, a), (b, b), (c, c) ∈ R. Therefore , R is reflexive.
- a)the greatest integer function
- b)a constant function.
- c)the absolute value function
- d)The signum function
Correct answer is option 'D'. Can you explain this answer?
a)
the greatest integer function
b)
a constant function.
c)
the absolute value function
d)
The signum function
|
|
Om Desai answered |
By definition, The Signum function =
If f : A→B and g : B →C be two functions. Then, composition of f and g, gof : A→C is defined a- a)(gof)(x) = f(g(x)), for all x ∈ A
- b)(gof)(x) = g(f(x)), for all x ∈ A
- c)(gof)(x) = g(x), for all x ∈ A
- d)None of the above
Correct answer is option 'B'. Can you explain this answer?
If f : A→B and g : B →C be two functions. Then, composition of f and g, gof : A→C is defined a
a)
(gof)(x) = f(g(x)), for all x ∈ A
b)
(gof)(x) = g(f(x)), for all x ∈ A
c)
(gof)(x) = g(x), for all x ∈ A
d)
None of the above
|
|
Rashi Nair answered |
Please provide additional details or context for me to assist you better.
f: R → R is defined by 
- a)x4 + 4x3 - 4x2
- b)x4 - 4x3 + 4x2
- c)x4 + 4x3 + 4x2
- d)x4 - 4x3 - 4x2
Correct answer is option 'B'. Can you explain this answer?
f: R → R is defined by
a)
x4 + 4x3 - 4x2
b)
x4 - 4x3 + 4x2
c)
x4 + 4x3 + 4x2
d)
x4 - 4x3 - 4x2
|
|
Sounak Pillai answered |
I'm sorry, could you please provide more context or a specific question related to the variable "f:R"?
Let a relation T on the set R of real numbers be T = {(a,b) : 1 + ab < 0, a,∈ R}. Then from among the ordered pairs (1,1) (1,2)(1,-2)(2,2), the only pair that belongs to T is________.- a)(2,2),
- b)(1,1),
- c)(1,-2)
- d)(1,2)
Correct answer is option 'C'. Can you explain this answer?
Let a relation T on the set R of real numbers be T = {(a,b) : 1 + ab < 0, a,∈ R}. Then from among the ordered pairs (1,1) (1,2)(1,-2)(2,2), the only pair that belongs to T is________.
a)
(2,2),
b)
(1,1),
c)
(1,-2)
d)
(1,2)
|
Keerthana Deshpande answered |
Relation T on R of real numbers
Definition of Relation T: T is a relation on the set R of real numbers and it is defined as T = {(a,b) : 1 ≤ ab ≤ 0, a ε R}
Explanation: This means that T is a set of ordered pairs (a,b) such that a is a real number and b is a real number, and the product of a and b is between 1 and 0, inclusive. In other words, either a and b are both negative or one of them is negative and the other is between 0 and 1.
Given ordered pairs: (1,1), (1,2), (1,-2), (2,2)
Checking which pairs belong to T:
- (1,1): 1 ≤ 1*1 ≤ 0 is false, so (1,1) does not belong to T
- (1,2): 1 ≤ 1*2 ≤ 0 is false, so (1,2) does not belong to T
- (1,-2): 1 ≤ 1*(-2) ≤ 0 is true, so (1,-2) belongs to T
- (2,2): 1 ≤ 2*2 ≤ 0 is false, so (2,2) does not belong to T
Conclusion: The only ordered pair that belongs to T is (1,-2). Therefore, the correct answer is option C, (1,-2).
Definition of Relation T: T is a relation on the set R of real numbers and it is defined as T = {(a,b) : 1 ≤ ab ≤ 0, a ε R}
Explanation: This means that T is a set of ordered pairs (a,b) such that a is a real number and b is a real number, and the product of a and b is between 1 and 0, inclusive. In other words, either a and b are both negative or one of them is negative and the other is between 0 and 1.
Given ordered pairs: (1,1), (1,2), (1,-2), (2,2)
Checking which pairs belong to T:
- (1,1): 1 ≤ 1*1 ≤ 0 is false, so (1,1) does not belong to T
- (1,2): 1 ≤ 1*2 ≤ 0 is false, so (1,2) does not belong to T
- (1,-2): 1 ≤ 1*(-2) ≤ 0 is true, so (1,-2) belongs to T
- (2,2): 1 ≤ 2*2 ≤ 0 is false, so (2,2) does not belong to T
Conclusion: The only ordered pair that belongs to T is (1,-2). Therefore, the correct answer is option C, (1,-2).
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