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All questions of Sets, Relations and Functions for JEE Exam
The number of proper subsets of the set { 1, 2 , 3 } is :- a)7
- b)5
- c)6
- d)8
Correct answer is 'A'. Can you explain this answer?
The number of proper subsets of the set { 1, 2 , 3 } is :
a)
7
b)
5
c)
6
d)
8
|
Kritika Bajaj answered |
Number of subsets of a given set = 2n
The Shaded region in the following figure illustrates
- a)A ∩ ( B ∪ C)
- b)A ∩ B ∩ C
- c)A ∪ B ∪ C
- d)(A ∩ B) ∪ (A ∩ C)
Correct answer is option 'D'. Can you explain this answer?
The Shaded region in the following figure illustrates
a)
A ∩ ( B ∪ C)
b)
A ∩ B ∩ C
c)
A ∪ B ∪ C
d)
(A ∩ B) ∪ (A ∩ C)
|
New Words answered |
First which region is over which region Then We will see that A is on the B so A intersection B and after C is on the A so, A intersection C after that we have to take all intersection part so A intersection B is Union with A intersection C.
The shaded region represents (A ∩ B) ∪ (A ∩ C).
If Q={x:x=1/y,where y∈N}, then- a)0∈Q
- b)1∈Q
- c)2∈Q
- d)2/3∈Q
Correct answer is option 'B'. Can you explain this answer?
If Q={x:x=1/y,where y∈N}, then
a)
0∈Q
b)
1∈Q
c)
2∈Q
d)
2/3∈Q
|
Rohit Jain answered |
As y ∈ N, y can be 1, 2, 3, 4...
∴ x will be 1, 1/2, 1/3, 1/4...
⇒ 1 ∈ Q
∴ x will be 1, 1/2, 1/3, 1/4...
⇒ 1 ∈ Q
If A = {1, 2, 3 }, B = {4, 5, 6 }, which of the following is not a relation from A to Ba) R1 = {(1, 4), (1, 5),(1, 6)}b) R2 = {(1, 5), (2, 4), (3, 6)}c) R3 = {(4, 2), (2, 6), (5, 1), (2, 4)}d) R4 = {(1, 4), (1, 5), (3, 6), (2, 6), (3, 4)}Correct answer is option 'C'. Can you explain this answer?
|
Poonam Reddy answered |
A × B = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
Which of the following is not an empty set?- a){x : x is a multiple of 7, x < 7, x ∈ N}
- b)Set of common points of two parallel lines in a plane
- c){x : 6 + 2x > 5x + 3, x ∈ N}
- d)Set of smallest whole number
Correct answer is option 'D'. Can you explain this answer?
Which of the following is not an empty set?
a)
{x : x is a multiple of 7, x < 7, x ∈ N}
b)
Set of common points of two parallel lines in a plane
c)
{x : 6 + 2x > 5x + 3, x ∈ N}
d)
Set of smallest whole number
|
Ayush Joshi answered |
As, the set of Whole numbers is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...};
So, according to the set of whole numbers above, the smallest whole number should be "0" and therefore the set of smallest whole number is not empty.
The set A = {1,4,9,16,25—} in set builder form is written as- a)A = {x:x is a prime number}
- b)A ={x:x is the cube of a natural number}
- c)A = {x:x is the square of a natural number}
- d)A = {x:x is an even natural number}
Correct answer is 'C'. Can you explain this answer?
The set A = {1,4,9,16,25—} in set builder form is written as
a)
A = {x:x is a prime number}
b)
A ={x:x is the cube of a natural number}
c)
A = {x:x is the square of a natural number}
d)
A = {x:x is an even natural number}
|
Krishna Iyer answered |
- We know that, 12 = 1, 22 = 4, 32 = 9, 42 = 16, 52 = 25
- Therefore the set A = {1, 4, 9, 16, 25...} can be written in set builder form as:
A = {x: x is the square of a natural number}
If f(x) = x2 and g(x) = cosx, which of the following is true?- a)f + g is even function
- b)f – g is an odd function
- c)f + g is not defined
- d)f + g is an odd function
Correct answer is option 'A'. Can you explain this answer?
If f(x) = x2 and g(x) = cosx, which of the following is true?
a)
f + g is even function
b)
f – g is an odd function
c)
f + g is not defined
d)
f + g is an odd function
|
Raghav Bansal answered |
if f(x) is an odd function
So, f(−x)=−f(x)
F(−x)=cos(f(−x))
=cos(−f(x))
=cos(f(x))
=F(x)
So cos(f(x)) is an even function
So, f(x) and g(x) is an even function
Can you explain the answer of this question below: Identify the null set from the following- A:Odd natural numbers divisible by 2
- B:The set which contains only zero as its element
- C:The set of even prime numbers
- D:The set of points common to two intersecting lines
The answer is a.
Identify the null set from the following
A:
Odd natural numbers divisible by 2
B:
The set which contains only zero as its element
C:
The set of even prime numbers
D:
The set of points common to two intersecting lines
|
Geetika Shah answered |
Since none of the odd number is divisible by 2, so it would be a null set.
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.
If n(A) = p and n(B) = q, then how many relations are there from A to B.- a)pq
- b)3pq
- c)2pq
- d)(pq)2
Correct answer is option 'C'. Can you explain this answer?
If n(A) = p and n(B) = q, then how many relations are there from A to B.
a)
pq
b)
3pq
c)
2pq
d)
(pq)2
|
|
Preeti Khanna answered |
Number of relation from A to B = 2n(A)×n(B)
n(A) = p, n(B) = q
⇒ Number of relation from A to B = 2pq
⇒ Number of relation from A to B = 2pq
Which of the following statement is false?- a)If R is a relation from A to B, then domain of R is a subset of A and range of R is a subset of B.
- b)If A = {1, 2, 3, 4} and B = {5, 6}, then the number of relations from A to B is 256.
- c)A subset of A × B is called a relation from the set A to set B.
- d)If A = {x, y, z} and B = {1, 2} then the number of relations from A to B is 22.
Correct answer is option 'D'. Can you explain this answer?
Which of the following statement is false?
a)
If R is a relation from A to B, then domain of R is a subset of A and range of R is a subset of B.
b)
If A = {1, 2, 3, 4} and B = {5, 6}, then the number of relations from A to B is 256.
c)
A subset of A × B is called a relation from the set A to set B.
d)
If A = {x, y, z} and B = {1, 2} then the number of relations from A to B is 22.
|
|
Riya Banerjee answered |
Option D: The number of relations from A to B is 23*2= 26
From the sets given below, select equal sets :A = { 2, 4, 8, 12}, B = { 1, 2, 3, 4},
C = { 4, 8, 12, 14}, D = { 3, 1, 4, 2}- a)A and C
- b)A and B
- c)B and D
- d)B and C
Correct answer is option 'C'. Can you explain this answer?
From the sets given below, select equal sets :
A = { 2, 4, 8, 12}, B = { 1, 2, 3, 4},
C = { 4, 8, 12, 14}, D = { 3, 1, 4, 2}
C = { 4, 8, 12, 14}, D = { 3, 1, 4, 2}
a)
A and C
b)
A and B
c)
B and D
d)
B and C
|
Mansi Chopra answered |
The sets are equal, if they have the exact same elements in them. Since option B & D have exactly same number of elements in them So, B & D are equal sets.
Let R be a relation defined on the set of N natural numbers as R = {(x, y): y is a factor of x, x, y∈ N} then,- a)(2, 4) ∈ R
- b)(9, 5) ∈ R
- c)(4, 2) ∈ R
- d)(3, 9) ∈ R
Correct answer is option 'C'. Can you explain this answer?
Let R be a relation defined on the set of N natural numbers as R = {(x, y): y is a factor of x, x, y∈ N} then,
a)
(2, 4) ∈ R
b)
(9, 5) ∈ R
c)
(4, 2) ∈ R
d)
(3, 9) ∈ R
|
Om Desai answered |
R = {(x, y): y is a factor of x, x, y∈ N}
As we know that 2 is a factor of 4
So, according to the options (4, 2) ϵ R
As we know that 2 is a factor of 4
So, according to the options (4, 2) ϵ R
Which of the following is a finite set?- a)Set of roots of the equation x2 – 25 = 0
- b)Set of natural numbers
- c)The set of lines parallel to y – axis
- d)The set of numbers which are multiples of 5
Correct answer is 'A'. Can you explain this answer?
Which of the following is a finite set?
a)
Set of roots of the equation x2 – 25 = 0
b)
Set of natural numbers
c)
The set of lines parallel to y – axis
d)
The set of numbers which are multiples of 5
|
Hansa Sharma answered |
- A finite set is a set that has a finite number of elements.
- Since x2 – 25 = 0 has finite number of elements. So it is a finite set.
If A = {1, 2, 3}, and B = {3, 6} then the number of relations from A to B is
- a)32
- b)23
- c)23
- d)26
Correct answer is option 'D'. Can you explain this answer?
If A = {1, 2, 3}, and B = {3, 6} then the number of relations from A to B is
a)
32
b)
23
c)
23
d)
26
|
Preeti Iyer answered |
The number of relations between sets can be calculated using 2mn where m and n represent the number of members in each set.
So, number of relations from A to B is 26.
So, number of relations from A to B is 26.
The relation shown in the arrow diagram in set builder form is
- a){(x, y): x is the square of y, x ∈ P, y ∈ Q}
- b){(x, y), x = 4, 1, y = 2, 1, -1, -2}
- c){(x, y): y is the square of x, x ∈ P, y ∈ Q }
- d){(4, 2), (4, -2), (1, 1), (1, -1)}
Correct answer is option 'A'. Can you explain this answer?
The relation shown in the arrow diagram in set builder form is
a)
{(x, y): x is the square of y, x ∈ P, y ∈ Q}
b)
{(x, y), x = 4, 1, y = 2, 1, -1, -2}
c)
{(x, y): y is the square of x, x ∈ P, y ∈ Q }
d)
{(4, 2), (4, -2), (1, 1), (1, -1)}
|
Lohit Matani answered |
Set P is square of the element of set Q.
{(x, y): x is the square of y, x ∈ P, y ∈ Q}
{(x, y): x is the square of y, x ∈ P, y ∈ Q}
Let A = {1, 2, 3, …, 50} and R be the relation “is square of” on A. Then R as a subset of A x A is written as:
- a){(1, 1), (2, 2), (3, 3), (4, 4) (5, 5), (6, 6), (7, 7)}
- b){(1, 1), (2, 5), (3, 9), (4, 15) (5, 25), (6, 36), (7, 49)}
- c){(1, 1), (4, 4), (9, 9), (16, 16) (25, 25), (36, 36), (49, 49)}
- d){(1, 1), (4, 2), (9, 3), (16, 4) (25, 5), (36, 6), (49, 7)}
Correct answer is option 'D'. Can you explain this answer?
Let A = {1, 2, 3, …, 50} and R be the relation “is square of” on A. Then R as a subset of A x A is written as:
a)
{(1, 1), (2, 2), (3, 3), (4, 4) (5, 5), (6, 6), (7, 7)}
b)
{(1, 1), (2, 5), (3, 9), (4, 15) (5, 25), (6, 36), (7, 49)}
c)
{(1, 1), (4, 4), (9, 9), (16, 16) (25, 25), (36, 36), (49, 49)}
d)
{(1, 1), (4, 2), (9, 3), (16, 4) (25, 5), (36, 6), (49, 7)}
|
Tejas Verma answered |
Since x is the square of y in relation R.
i.e. option D is correct.
i.e. option D is correct.
In rule method the null set is represented by- a){}
- b){x: x ≠ x}
- c){x : x = x}
- d)φ
Correct answer is option 'B'. Can you explain this answer?
In rule method the null set is represented by
a)
{}
b)
{x: x ≠ x}
c)
{x : x = x}
d)
φ
|
Aarav Sharma answered |
It is fundamental concept.
Let R be a relation defined as
R = {(x, y): y = 2x, x is natural number < 5} then Range of R is given as ,- a){2, 4, 6, 8}
- b){2, 4, 6, 8, 10}
- c){1, 2, 3, 4}
- d)N
Correct answer is option 'A'. Can you explain this answer?
Let R be a relation defined as
R = {(x, y): y = 2x, x is natural number < 5} then Range of R is given as ,
R = {(x, y): y = 2x, x is natural number < 5} then Range of R is given as ,
a)
{2, 4, 6, 8}
b)
{2, 4, 6, 8, 10}
c)
{1, 2, 3, 4}
d)
N
|
|
Gaurav Kumar answered |
X is a natural number and x < 5, the number is 1, 2, 3, 4
y = 2x
y = 2x
► x(1), y = 2 × 1 = 2
► x(2), y = 2 × 2 = 4
► x(3), y = 2 × 3 = 6
► x(4), y = 2 × 4 = 8
► x(2), y = 2 × 2 = 4
► x(3), y = 2 × 3 = 6
► x(4), y = 2 × 4 = 8
Range = {2, 4, 6, 8}
, then inverse of f is:
Let U = {1,2,3,4,5,6,7,8,9,10} , A = {1,2,5} , B = {6,7}. Then A∩B’ is :- a)A
- b)B
- c)B’
- d)none
Correct answer is option 'A'. Can you explain this answer?
Let U = {1,2,3,4,5,6,7,8,9,10} , A = {1,2,5} , B = {6,7}. Then A∩B’ is :
a)
A
b)
B
c)
B’
d)
none
|
|
Pooja Shah answered |
- B' gives us all the elements in U other than 6 and 7 i.e., B' = {1, 2, 3, 4, 5, 8, 9, 10}
- The intersection of this set with A will be the common elements in both of these (A and B') i.e., = {1, 2, 5} which is set A itself.
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.
Which of the following is not an example of polynomial function ?- a)g(x) = 8x2 + 5x – 2
- b)p(x) =5x3 + 3x2 + x – 2
- c)h(x) =3 root of x minus 1 over x
- d)f(x) = 3x2 + x – 2
Correct answer is option 'C'. Can you explain this answer?
Which of the following is not an example of polynomial function ?
a)
g(x) = 8x2 + 5x – 2
b)
p(x) =5x3 + 3x2 + x – 2
c)
h(x) =3 root of x minus 1 over x
d)
f(x) = 3x2 + x – 2
|
Lavanya Menon answered |
A polynomial function is a function which involves only non-negative integer powers or only positive integer exponents of a variable in an equation.
In option C, powers of x are negative and fractional.
In option C, powers of x are negative and fractional.
Which of the following has only one subset?- a){ }
- b){5}
- c){4,5}
- d){0}
Correct answer is option 'A'. Can you explain this answer?
Which of the following has only one subset?
a)
{ }
b)
{5}
c)
{4,5}
d)
{0}
|
Vijay Kumar answered |
- Every set has the empty set as a subset. So if a set has 1 element, like {0}, then it will have 2 subsets: itself and the empty set, which is denoted by { }.
- So, if a set has only one subset, then this set must be the empty set.
If f(x + 2) = f(x) + f(x + 1) for all positive integers x, and f(11) = 91, f(15) = 617, then f(10) equals
(2018)
- a)55
- b)54
- c)52
- d)51
Correct answer is option 'B'. Can you explain this answer?
If f(x + 2) = f(x) + f(x + 1) for all positive integers x, and f(11) = 91, f(15) = 617, then f(10) equals
(2018)
a)
55
b)
54
c)
52
d)
51
|
Shalini Patel answered |
Calculation:
⇒ f(x + 2) = f(x) + f(x + 1)
⇒ f(15) = f(13) + f(14)
⇒ f(13) + f(14) = 617 - (1)
⇒ f(12) + f(13) = f(14) - (2)
⇒ f(11) + f(12) = f(13) - (3)
By solving,
⇒ f(12) + f(13) = 617 - f(13) [ using (1) and (2) ]
⇒ 2f(11) + 3f(12) = 617 [ usining (3) ]
⇒ f(12) = 145
⇒ f(12) = f(11) + f(10)
∴ f(10) = 54
If A and B are two sets, then A∩(A∪B)′ is equal to- a)B
- b)A
- c)∅
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
If A and B are two sets, then A∩(A∪B)′ is equal to
a)
B
b)
A
c)
∅
d)
none of these
|
Bibek Desai answered |
Two sets are disjoint if the intersection of two sets is the empty set.
Consider the following relations:
(i) A - B = A - (A ∩ B)
(ii) A = (A ∩ B) ∪ (A - B)
(iii) A - (B ∪ C) = (A - B) ∪ (A - C)
Which of these is/are correct?
- a)(i) and (ii)
- b)(i) and (iii)
- c)(ii) only
- d)(ii) and (iii)
Correct answer is option 'A'. Can you explain this answer?
Consider the following relations:
(i) A - B = A - (A ∩ B)
(ii) A = (A ∩ B) ∪ (A - B)
(iii) A - (B ∪ C) = (A - B) ∪ (A - C)
Which of these is/are correct?
(i) A - B = A - (A ∩ B)
(ii) A = (A ∩ B) ∪ (A - B)
(iii) A - (B ∪ C) = (A - B) ∪ (A - C)
Which of these is/are correct?
a)
(i) and (ii)
b)
(i) and (iii)
c)
(ii) only
d)
(ii) and (iii)
|
|
Hansa Sharma answered |
Using the Venn Diagram,
(i) A - B = A - (A ∩ B) is true
(ii) A = (A ∩ B) ∪ (A - B) is true
(iii) A - (B ∪ C) = (A - B) ∪ (A - C) is false
The set {x: x is an odd number between 10 and 18}:- a){11, 12, 13, 15, 17}
- b){12, 16, 15, 13}
- c){11, 13, 15, 17}
- d){12, 14, 16, 18}
Correct answer is option 'C'. Can you explain this answer?
The set {x: x is an odd number between 10 and 18}:
a)
{11, 12, 13, 15, 17}
b)
{12, 16, 15, 13}
c)
{11, 13, 15, 17}
d)
{12, 14, 16, 18}
|
Alok Choudhury answered |
**Explanation:**
To find the set of odd numbers between 10 and 18, we need to consider the numbers in that range and check if they are odd or not.
The set of odd numbers between 10 and 18 includes the numbers 11, 13, 15, and 17.
Let's analyze each option to see which one matches this set.
**Option A: {11, 12, 13, 15, 17}**
This option includes all the numbers in the set of odd numbers between 10 and 18. However, it also includes the number 12, which is an even number. Therefore, this option is incorrect.
**Option B: {12, 16, 15, 13}**
This option includes the numbers 12, 16, 15, and 13. It does not include the numbers 11 and 17, which are part of the set of odd numbers between 10 and 18. Therefore, this option is incorrect.
**Option C: {11, 13, 15, 17}**
This option includes all the numbers in the set of odd numbers between 10 and 18. It does not include any even numbers and only includes the odd numbers 11, 13, 15, and 17. Therefore, this option is correct.
**Option D: {12, 14, 16, 18}**
This option includes all even numbers and does not include any odd numbers. Therefore, this option is incorrect.
Therefore, the correct answer is **Option C: {11, 13, 15, 17}** as it includes all the odd numbers between 10 and 18 and does not include any even numbers.
To find the set of odd numbers between 10 and 18, we need to consider the numbers in that range and check if they are odd or not.
The set of odd numbers between 10 and 18 includes the numbers 11, 13, 15, and 17.
Let's analyze each option to see which one matches this set.
**Option A: {11, 12, 13, 15, 17}**
This option includes all the numbers in the set of odd numbers between 10 and 18. However, it also includes the number 12, which is an even number. Therefore, this option is incorrect.
**Option B: {12, 16, 15, 13}**
This option includes the numbers 12, 16, 15, and 13. It does not include the numbers 11 and 17, which are part of the set of odd numbers between 10 and 18. Therefore, this option is incorrect.
**Option C: {11, 13, 15, 17}**
This option includes all the numbers in the set of odd numbers between 10 and 18. It does not include any even numbers and only includes the odd numbers 11, 13, 15, and 17. Therefore, this option is correct.
**Option D: {12, 14, 16, 18}**
This option includes all even numbers and does not include any odd numbers. Therefore, this option is incorrect.
Therefore, the correct answer is **Option C: {11, 13, 15, 17}** as it includes all the odd numbers between 10 and 18 and does not include any even numbers.
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)
|
|
Lavanya Menon answered |
f(x) = x2 + 5 = y(let)
⇒ y = x2+5
⇒ x2 = y-5
⇒ x = (y-5)1/2
⇒ f-1(x) = (x-5)1/2
⇒ y = x2+5
⇒ x2 = y-5
⇒ x = (y-5)1/2
⇒ f-1(x) = (x-5)1/2
If three positive real numbers a, b and c (c > a) are in Harmonic Progression, then log (a + c) + log (a – 2b + c) is equal to:(2010)- a)2 log (c – b)
- b)2 log (a – c)
- c)2 log (c – a)
- d)log a + log b + log c
Correct answer is option 'C'. Can you explain this answer?
If three positive real numbers a, b and c (c > a) are in Harmonic Progression, then log (a + c) + log (a – 2b + c) is equal to:
(2010)
a)
2 log (c – b)
b)
2 log (a – c)
c)
2 log (c – a)
d)
log a + log b + log c
|
|
Shalini Patel answered |
Since a, b, c are in H.P.
Now log (a + c) + log (a – 2b + c)
= log [(a + c){(a + c) – 2b}]
= log [(a + c)2 – 2b (a + c)]
= log [(a + c)2 – 2 × 2ac]
= log (a – c)2 or log (c – a)2
= 2 log (a – c) or 2 log (c – a)
∴ log (a + c) + log (a – 2b + c) = 2 log (c – a)
Now log (a + c) + log (a – 2b + c)
= log [(a + c){(a + c) – 2b}]
= log [(a + c)2 – 2b (a + c)]
= log [(a + c)2 – 2 × 2ac]
= log (a – c)2 or log (c – a)2
= 2 log (a – c) or 2 log (c – a)
∴ log (a + c) + log (a – 2b + c) = 2 log (c – a)
If A={2,4,5}, B={7,8,9}, then n(A×B) is equal to- a)6
- b)3
- c)9
- d)0
Correct answer is option 'C'. Can you explain this answer?
If A={2,4,5}, B={7,8,9}, then n(A×B) is equal to
a)
6
b)
3
c)
9
d)
0
|
Yash Patel answered |
n(A × B) = n(A).n(B)
⇒ 3 × 3 = 9
A × B = {(2, 7), (2, 8), (2, 9), (4, 7), (4,8), (4, 9), (5, 7), (5, 8), (5, 9)}
⇒ 3 × 3 = 9
A × B = {(2, 7), (2, 8), (2, 9), (4, 7), (4,8), (4, 9), (5, 7), (5, 8), (5, 9)}
Which of the following is incorrect?- a)Modulus function, Domain: R; Range: [0, infinity)
- b)Signum function, Domain: R; Range: {-1, 0,1}
- c)Constant function, Domain: R; Range: R
- d)Identity function, Domain: R; Range: R
Correct answer is option 'C'. Can you explain this answer?
Which of the following is incorrect?
a)
Modulus function, Domain: R; Range: [0, infinity)
b)
Signum function, Domain: R; Range: {-1, 0,1}
c)
Constant function, Domain: R; Range: R
d)
Identity function, Domain: R; Range: R
|
|
Preeti Khanna answered |
Constant Function is defined as the real valued function.
f : R→R, y = f(x) = c for each x∈R and c is a constant.
f : R→R, y = f(x) = c for each x∈R and c is a constant.
So, this function basically associate each real number to a constant value.
It is a linear function where f(x1) = f(x2) for all x1,x2 ∈ R
For f : R→R, y = f(x) = c for each x ∈ R
Domain = R
Range = {c}
The value of c can be any real number.
Domain = R
Range = {c}
The value of c can be any real number.
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.
In probability, the event ‘A or B’ can be associated with set :- a)A B
- b)A ∪ B
- c)(A – B)’
- d)A – B
Correct answer is option 'B'. Can you explain this answer?
In probability, the event ‘A or B’ can be associated with set :
a)
A B
b)
A ∪ B
c)
(A – B)’
d)
A – B
|
Mohit Rajpoot answered |
Probability of event A or B
The probability that Events A and B both occur is the probability of the intersection of A and B. The probability of the intersection of Events A and B is denoted by P(A ∩ B). If Events A and B are mutually exclusive, P(A ∩ B) = 0. The probability that Events A or B occur is the probability of the union of A and B.
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.
If f and g are two functions over real numbers defined as f(x) = 3x + 1, g(x) = x2 + 2, then find f-g- a)x2+1
- b)3x + x2
- c)3x – x2-1
- d)x2+1-3x
Correct answer is option 'C'. Can you explain this answer?
If f and g are two functions over real numbers defined as f(x) = 3x + 1, g(x) = x2 + 2, then find f-g
a)
x2+1
b)
3x + x2
c)
3x – x2-1
d)
x2+1-3x
|
Karthikeyeni Baskaran answered |
It is just a simple subtraction...subtract g from f
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:
If f(x) = x2 and g(x) = x are two functions from R to R then f(g(2)) is
- a)8
- b)4
- c)1
- d)2
Correct answer is option 'B'. Can you explain this answer?
If f(x) = x2 and g(x) = x are two functions from R to R then f(g(2)) is
a)
8
b)
4
c)
1
d)
2
|
|
Raghav Bansal answered |
f(g(2)) compare f(gx) x=2
on comparing g(x)=x, g(2)=2
f(g(x)) = f(2) = x2 = 22= 4
on comparing g(x)=x, g(2)=2
f(g(x)) = f(2) = x2 = 22= 4
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
|
Kiran Sanodiya answered |
F(n) = n+2 if n is odd
so if n=1
f(n)= 3
if n=3 , f(n)=5
On the other hand
F(n)= n+3 if n is even
if n= 2,f(n)=5
if n=4 ,f(n)= 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
range does not equals to codomain
so it is not surjective.
option C is correct.
so if n=1
f(n)= 3
if n=3 , f(n)=5
On the other hand
F(n)= n+3 if n is even
if n= 2,f(n)=5
if n=4 ,f(n)= 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
range does not equals to codomain
so it is not surjective.
option C is correct.
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
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.
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