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All questions of Chapter 1 - Relations and Functions for Commerce 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
|
Rajib Dhar answered |
The relation is transitive for elements in the {a,b,c} if (a,b) belongs to r and again if (b,c) belongs to r and (a,c ) also belongs
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
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
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.
, where a, b are the elements of a set of non-zero rational numbers, is
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
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
|
Mohit Rajpoot answered |
We are given that functions f : A→B and g : B→C are both one-to-one functions.
Suppose a1, a2 ∈ A such that (gof)(a1) = (gof)(a2)
⇒ g(f(a1)) = g(f(a2)) (definition of composition) Since g is one-to-one, therefore,
f(a1) = f(a2)
And since f is one-to-one, therefore, a1 = a2
Thus, we have shown that if (gof)(a1) = (gof)(a2) then a1 = a2
Hence, gof is one-to-one function.
Suppose a1, a2 ∈ A such that (gof)(a1) = (gof)(a2)
⇒ g(f(a1)) = g(f(a2)) (definition of composition) Since g is one-to-one, therefore,
f(a1) = f(a2)
And since f is one-to-one, therefore, a1 = a2
Thus, we have shown that if (gof)(a1) = (gof)(a2) then a1 = a2
Hence, gof is one-to-one function.
The number of binary operations which can be defined on the set P= { p, q} is- a)6
- b)0
- c)1
- d)16
Correct answer is option 'D'. Can you explain this answer?
The number of binary operations which can be defined on the set P= { p, q} is
a)
6
b)
0
c)
1
d)
16
|
Anand Kumar answered |
Total no. of from (pxq) to P is
{n(P)}^n(p×q)=2^4 = 16
Hence total no. of possible binary operation on P = 16
{n(P)}^n(p×q)=2^4 = 16
Hence total no. of possible binary operation on P = 16
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 )
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.
If 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 * 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.
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)
Let 
be defined by 
Then, the function f is- a)bijective
- b)Surjective
- c)neither injective nor surjective
- d)Injective
Correct answer is option 'C'. Can you explain this answer?
Let be defined by
Then, the function f is
be defined by Then, the function f is
a)
bijective
b)
Surjective
c)
neither injective nor surjective
d)
Injective
|
Om Desai answered |
F(n) = n+2 if n is odd
so if n=1, then f(1)= 3
if n=3 , f(3)=5
On the other hand
F(n)= n+3 if n is even
if n= 2,f(2)=5
if n=4 ,f(4)= 7
Clearly f(3) = f(2) = 5
but 3 does not equals to 3, so it is not one one i.e. injective also 1€N
i.e. codomain
but here 1€ range does not equals to codomain so it is not surjective.
so if n=1, then f(1)= 3
if n=3 , f(3)=5
On the other hand
F(n)= n+3 if n is even
if n= 2,f(2)=5
if n=4 ,f(4)= 7
Clearly f(3) = f(2) = 5
but 3 does not equals to 3, so it is not one one i.e. injective also 1€N
i.e. codomain
but here 1€ range does not equals to codomain so it is not surjective.
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.
defined as f(x) = 1/ x where x ∈ N is not an onto function. Which one of the following sets should be replaced by N such that the function f will become onto?
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 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)
|
Shiksha Academy 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)
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}
is defined by f(x) = [x+1], where [x] the greatest integer function, is:
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)
|
Sushil Kumar answered |
f(0)=0×sec0
f(0)=0×1
f(0)=0
f(0)=0×1
f(0)=0
- 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
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.
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.
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.
- 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 A = {1, 2, 3, 4} and B = {1, 3, 5} and R is a relation from A to B defined by(a, b) ∈ element of R ⇔ a < b. Then, R = ?- a){(2, 3), (4, 5), (1, 3), (2, 5)}
- b){(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}
- c){(2, 3), (4, 5), (1, 3), (2, 5), (5, 3)}
- d){(5, 3), (3, 5), (5, 4), (4, 5)}
Correct answer is option 'B'. Can you explain this answer?
If A = {1, 2, 3, 4} and B = {1, 3, 5} and R is a relation from A to B defined by(a, b) ∈ element of R ⇔ a < b. Then, R = ?
a)
{(2, 3), (4, 5), (1, 3), (2, 5)}
b)
{(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}
c)
{(2, 3), (4, 5), (1, 3), (2, 5), (5, 3)}
d)
{(5, 3), (3, 5), (5, 4), (4, 5)}
|
Lavanya Menon answered |
A = {1, 2, 3, 4}
B = {1, 3, 5}
(a, b) ∈ element of R ⇔ a < b for all a ∈ A, b ∈ B
(a, b) pairs satisfying the condition of R are:
(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)
So,
R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}
B = {1, 3, 5}
(a, b) ∈ element of R ⇔ a < b for all a ∈ A, b ∈ B
(a, b) pairs satisfying the condition of R are:
(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)
So,
R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}
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.
Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relations on the set A = {1, 2, 3, 4}. the relation R is- a)Reflexive
- b)transitive
- c)not symmetric
- d)a function
Correct answer is option 'C'. Can you explain this answer?
Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relations on the set A = {1, 2, 3, 4}. the relation R is
a)
Reflexive
b)
transitive
c)
not symmetric
d)
a function
|
Shubham Rajput answered |
Clearly it is not reflexive,not symmetric and not transitive
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"?
. Then f is
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).
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.
Let R be a relation on set A of triangles in a plane.R = {(T1 , T2) : T1 , T2 element of A and T1 is congruent to T2} Then the relation R is______- a)Equivalence relation
- b)Transitive
- c)Symmetric
- d)Reflexive
Correct answer is option 'A'. Can you explain this answer?
Let R be a relation on set A of triangles in a plane.
R = {(T1 , T2) : T1 , T2 element of A and T1 is congruent to T2} Then the relation R is______
a)
Equivalence relation
b)
Transitive
c)
Symmetric
d)
Reflexive
|
Akash Singh answered |
It is equivalence relation...
as a triangle is congruent to itself that means (T1 , T1) exist in relation...which implies it is reflexive.
And also if T1 is congruent to T2 then T2 is also congruent to T1 as simple...that means (T1,T2) and (T2,T1) both belongs to relation ...which implies it is symmetric.
And is T1 is congruent to T2 and T2 is congruent to T3...then T1 is also congruent to T3.... congruency rule...that means (T1,T2),(T2,T3)and(T1,T3) belongs to relation...which implies it transitive....
this relation is reflexive, symmetric, and transitive...hence it is equivalence relation..
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.
If f: R → R and g: R → R defined by f(x) = 2x + 3 and g(x) = x2 + 7, then the value of x for which f(g(x)) = 25 is- a)±3
- b)±1
- c)±4
- d)±2
Correct answer is option 'D'. Can you explain this answer?
If f: R → R and g: R → R defined by f(x) = 2x + 3 and g(x) = x2 + 7, then the value of x for which f(g(x)) = 25 is
a)
±3
b)
±1
c)
±4
d)
±2
|
|
Devika Choudhury answered |
To find the value of x for which f(g(x)) = 25, we need to substitute g(x) into the function f(x) and solve for x.
Given:
f(x) = 2x - 3
g(x) = x^2 - 7
Substituting g(x) into f(x), we have:
f(g(x)) = 2(g(x)) - 3
= 2(x^2 - 7) - 3
= 2x^2 - 14 - 3
= 2x^2 - 17
Now we can set this expression equal to 25 and solve for x:
2x^2 - 17 = 25
Adding 17 to both sides:
2x^2 = 42
Dividing both sides by 2:
x^2 = 21
Taking the square root of both sides:
x = ±√21
Since we are looking for a real value of x, we can discard the negative square root.
Therefore, the value of x for which f(g(x)) = 25 is x = √21, which is approximately 4.58.
So, the correct answer is option D) 2.
Given:
f(x) = 2x - 3
g(x) = x^2 - 7
Substituting g(x) into f(x), we have:
f(g(x)) = 2(g(x)) - 3
= 2(x^2 - 7) - 3
= 2x^2 - 14 - 3
= 2x^2 - 17
Now we can set this expression equal to 25 and solve for x:
2x^2 - 17 = 25
Adding 17 to both sides:
2x^2 = 42
Dividing both sides by 2:
x^2 = 21
Taking the square root of both sides:
x = ±√21
Since we are looking for a real value of x, we can discard the negative square root.
Therefore, the value of x for which f(g(x)) = 25 is x = √21, which is approximately 4.58.
So, the correct answer is option D) 2.
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
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.
Find the number of bijective functions from set A to itself when A contains 106 elements.- a)106
- b)1062
- c)106!
- d)2106
Correct answer is option 'D'. Can you explain this answer?
Find the number of bijective functions from set A to itself when A contains 106 elements.
a)
106
b)
1062
c)
106!
d)
2106
|
|
Varun Kapoor answered |
For a finite set S, there is a bijection between the set of possible total orderings of the elements and the set of bijections from S to S. That is to say, the number of permutations of elements of S is the same as the number of total orderings of that set, i.e. n!.
We have the set A that contains 106 elements, so the number of bijective functions from set A to itself is 106!.
We have the set A that contains 106 elements, so the number of bijective functions from set A to itself is 106!.
Let R be an equivalence relation on Z, the set of integers.R = {(a,b): a,b∈Z and a – b is a multiple of 3} The Equivalence class of [1] is- a){..-7,-4,2,5,8,.}
- b){.-4,-1,2,5,8,.}
- c){.-6,-1,3,5,9,.}
- d){…..-5,-2,1,4,7,.}
Correct answer is option 'D'. Can you explain this answer?
Let R be an equivalence relation on Z, the set of integers.
R = {(a,b): a,b∈Z and a – b is a multiple of 3} The Equivalence class of [1] is
a)
{..-7,-4,2,5,8,.}
b)
{.-4,-1,2,5,8,.}
c)
{.-6,-1,3,5,9,.}
d)
{…..-5,-2,1,4,7,.}
|
Gaurav Kumar answered |
Correct Answer :- d
Explanation : R = (a,b) : 3 divides (a-b)
⇒(a−b) is a multiple of 3.
To find equivalence class 1, put b=1
So, (a−0) is a multiple of 3
⇒ a is a multiple of 3
So, In set z of integers, all the multiple
of 3 will come in equivalence
class {1}
Hence, equivalence class {1} = {3x+1}
{-5,-2,1,4,7}
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