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The set S = [0, 1] has- a)maximal element only
- b)minimal element only
- c)Both maximal and minimal element
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
The set S = [0, 1] has
a)
maximal element only
b)
minimal element only
c)
Both maximal and minimal element
d)
None of the above
|
Mahira Patel answered |
Question Analysis:
The question asks us to determine the elements of the set S = [0, 1] that can be classified as maximal, minimal, or both. We need to understand the definitions of maximal and minimal elements to answer this question correctly.
Definitions:
- A maximal element in a set is an element that is greater than or equal to all other elements in the set.
- A minimal element in a set is an element that is less than or equal to all other elements in the set.
Answer Explanation:
To determine the maximal and minimal elements in the set S = [0, 1], we need to compare each element to all other elements in the set.
Minimal Element:
- The element 0 is the smallest element in the set S = [0, 1].
- It is less than or equal to all other elements in the set, as 0 <=>
- Therefore, 0 is a minimal element of the set S.
Maximal Element:
- The element 1 is the largest element in the set S = [0, 1].
- It is greater than or equal to all other elements in the set, as 1 >= 0.
- Therefore, 1 is a maximal element of the set S.
Conclusion:
The set S = [0, 1] has both a minimal element (0) and a maximal element (1). Therefore, the correct answer is option 'C' - Both maximal and minimal element.
The question asks us to determine the elements of the set S = [0, 1] that can be classified as maximal, minimal, or both. We need to understand the definitions of maximal and minimal elements to answer this question correctly.
Definitions:
- A maximal element in a set is an element that is greater than or equal to all other elements in the set.
- A minimal element in a set is an element that is less than or equal to all other elements in the set.
Answer Explanation:
To determine the maximal and minimal elements in the set S = [0, 1], we need to compare each element to all other elements in the set.
Minimal Element:
- The element 0 is the smallest element in the set S = [0, 1].
- It is less than or equal to all other elements in the set, as 0 <=>
- Therefore, 0 is a minimal element of the set S.
Maximal Element:
- The element 1 is the largest element in the set S = [0, 1].
- It is greater than or equal to all other elements in the set, as 1 >= 0.
- Therefore, 1 is a maximal element of the set S.
Conclusion:
The set S = [0, 1] has both a minimal element (0) and a maximal element (1). Therefore, the correct answer is option 'C' - Both maximal and minimal element.
If A = {(x, y ) : x2 + y2 = 25} and B = {(x, y ) : x2 + 9y2 = 144}, then A ∩ B contains- a)one point
- b)three points
- c)two points
- d)four points
Correct answer is option 'D'. Can you explain this answer?
If A = {(x, y ) : x2 + y2 = 25} and B = {(x, y ) : x2 + 9y2 = 144}, then A ∩ B contains
a)
one point
b)
three points
c)
two points
d)
four points
|
Chirag Verma answered |
A = Set of all values (x, y) :x2 + y2 = 25 = 52
Clearly, A ∩ B consists of four points.
Clearly, A ∩ B consists of four points.
Then sequence <sn> defined as
- a)diverges
- b)Converges
- c)oscillates
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
Then sequence <sn> defined as
a)
diverges
b)
Converges
c)
oscillates
d)
None of these
|
|
Chirag Verma answered |
Here, we have <sn> such that
Let <sn> & <s2 n–1> be the two subsequences of the sequence <sn> such that
Both the sequences <sn> & <s2n+1> converges to he same limit 0. Hence the sequences <sn> Converges to 0.
Let <sn> & <s2 n–1> be the two subsequences of the sequence <sn> such that
Both the sequences <sn> & <s2n+1> converges to he same limit 0. Hence the sequences <sn> Converges to 0.
Which one of the following is not correct?- a)The set I has no limit point.
- b)The set hi has no limit point.
- c)No point of Q is a limit point.
- d)Every point of the set M is a limit point.
Correct answer is option 'C'. Can you explain this answer?
Which one of the following is not correct?
a)
The set I has no limit point.
b)
The set hi has no limit point.
c)
No point of Q is a limit point.
d)
Every point of the set M is a limit point.
|
|
Chirag Verma answered |
The set l has no limit point, for a nbd 
of m ε l, contains no point of l other than m. Thus the derived set of l is the null set φ. The set N has no limit point for a nbd 
of m ε N, contains no points of N other than m. Thus, the derived set of N is null set φ. Every point of the set Q of rationals is a limit point, for between any two rationals there exist an infinity of rational.
Further every point of R is a limit point, for every nbd of any of its points contains an infinite member of R.
of m ε l, contains no point of l other than m. Thus the derived set of l is the null set φ. The set N has no limit point for a nbd of m ε N, contains no points of N other than m. Thus, the derived set of N is null set φ. Every point of the set Q of rationals is a limit point, for between any two rationals there exist an infinity of rational.Further every point of R is a limit point, for every nbd of any of its points contains an infinite member of R.
Let A = {-1, 2, 5, 8}, B = {0, 1, 3, 6, 7} and R be the relation ‘is one less than’ from A to B, then how many elements will R contain?- a)2
- b)3
- c)5
- d)9
Correct answer is option 'B'. Can you explain this answer?
Let A = {-1, 2, 5, 8}, B = {0, 1, 3, 6, 7} and R be the relation ‘is one less than’ from A to B, then how many elements will R contain?
a)
2
b)
3
c)
5
d)
9
|
Veda Institute answered |
Given,
A = { - 1 , 2 , 5, 8} and
B = {0 ,1 ,3 ,6 ,7 }
R = { ( - 1 , 0 ) , (2,3), (5,6)}
(Since , R = A is one less than from B) Hence, total number of elements in R is 3.
A = { - 1 , 2 , 5, 8} and
B = {0 ,1 ,3 ,6 ,7 }
R = { ( - 1 , 0 ) , (2,3), (5,6)}
(Since , R = A is one less than from B) Hence, total number of elements in R is 3.
Consider the following relations
I. A - B = A - ( A ∩ B )
II. A = (A ∩ B ) ∪ (A -B )
III. A - (B ∪ C) = (A - B) ∪ (A - C)
Q. Which of these is/are correct?- a)I and III
- b)I only
- c)II and III
- d)I and II
Correct answer is option 'D'. 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)
Q. 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)
Q. Which of these is/are correct?
a)
I and III
b)
I only
c)
II and III
d)
I and II
|
|
Chirag Verma answered |
A - B = A - (A ∩ B) is correct.
A = (A ∩ B) ∩ (A - B) is correct.
III is false
Therefore, I and II are True.
A = (A ∩ B) ∩ (A - B) is correct.
III is false
Therefore, I and II are True.
If A = {4n + 2 | n is a natural number} and B = .{3n | n is a natural number}, then what is (A ∩ B) equal to?- a){12n2 + 6n | n is a natural number}
- b){24n -12 | n is a natural number}
- c){60n + 30 | n is a natural number}
- d){12n - 6 | n is a natural number}
Correct answer is option 'D'. Can you explain this answer?
If A = {4n + 2 | n is a natural number} and B = .{3n | n is a natural number}, then what is (A ∩ B) equal to?
a)
{12n2 + 6n | n is a natural number}
b)
{24n -12 | n is a natural number}
c)
{60n + 30 | n is a natural number}
d)
{12n - 6 | n is a natural number}
|
|
Chirag Verma answered |
Since, A = {An + 2 | n e N}
= {6,10,14,18,22,26,30,.,.}
and B = {3n | n ∈ N}
= {3,6,9,12,15,18,21,24,...}
So, A ∩ B = { 6 ,18,30,...}
= {6 + (n - 1)12 | n ∈ N}
= {12n - 6 | n ∈ N}
(c) x = 3(mod 7) => x - 3 = Ip, (p ∈ z) implies x= 7p + 3,p ∈ z i.e., {7p + 3 :p ∈ z}
= {6,10,14,18,22,26,30,.,.}
and B = {3n | n ∈ N}
= {3,6,9,12,15,18,21,24,...}
So, A ∩ B = { 6 ,18,30,...}
= {6 + (n - 1)12 | n ∈ N}
= {12n - 6 | n ∈ N}
(c) x = 3(mod 7) => x - 3 = Ip, (p ∈ z) implies x= 7p + 3,p ∈ z i.e., {7p + 3 :p ∈ z}
If X and Y are any two non-empty sets, then what is (X - Y)' equal to?- a)X' - Y'
- b)X' ∩ Y
- c)X' ∪ Y
- d)X - Y'
Correct answer is option 'C'. Can you explain this answer?
If X and Y are any two non-empty sets, then what is (X - Y)' equal to?
a)
X' - Y'
b)
X' ∩ Y
c)
X' ∪ Y
d)
X - Y'
|
|
Chirag Verma answered |
From the figure, it is clear that
Let A = {x : x ∈ R, | x | < 1};
B = { x : x ∈ R, |x — 1| ≥ 1} and A ∪ B = R - D , then the set D is- a){x : 1 < x ≤ 2}
- b){x : 1 ≤ x < 2}
- c){x : 1 ≤ x ≤ 2}
- d)None o f these
Correct answer is option 'B'. Can you explain this answer?
Let A = {x : x ∈ R, | x | < 1};
B = { x : x ∈ R, |x — 1| ≥ 1} and A ∪ B = R - D , then the set D is
B = { x : x ∈ R, |x — 1| ≥ 1} and A ∪ B = R - D , then the set D is
a)
{x : 1 < x ≤ 2}
b)
{x : 1 ≤ x < 2}
c)
{x : 1 ≤ x ≤ 2}
d)
None o f these
|
|
Chirag Verma answered |
A =[x : x ∈ R, -1 < x < 1]
B =[x : x ∈ R : x - 1 ≤ - 1 or x - 1 ≥ 1] =[x :x ∈ R : x ≤ 0 o r x ≥ 2]
Hence, A ∪ B = R - D , where
D = [x : x ∈ R, 1 ≤ x < 2]
B =[x : x ∈ R : x - 1 ≤ - 1 or x - 1 ≥ 1] =[x :x ∈ R : x ≤ 0 o r x ≥ 2]
Hence, A ∪ B = R - D , where
D = [x : x ∈ R, 1 ≤ x < 2]
Out of 800 boys in a school, 224 played cricket, 240 played hockey and 336 played basketball. Of the total, 64 played both basketball and hockey; 80 played cricket and basketball and 40 played cricket and hockey; 24 played all the three games. The number of boys who did not play any game is- a)128
- b)216
- c)240
- d)160
Correct answer is option 'D'. Can you explain this answer?
Out of 800 boys in a school, 224 played cricket, 240 played hockey and 336 played basketball. Of the total, 64 played both basketball and hockey; 80 played cricket and basketball and 40 played cricket and hockey; 24 played all the three games. The number of boys who did not play any game is
a)
128
b)
216
c)
240
d)
160
|
|
Chirag Verma answered |
Here,
Out of 32 persons, 30 invest in National savings Certificates and 17 invest in shares. What is the number of persons who invest in both?- a)13
- b)15
- c)17
- d)19
Correct answer is option 'B'. Can you explain this answer?
Out of 32 persons, 30 invest in National savings Certificates and 17 invest in shares. What is the number of persons who invest in both?
a)
13
b)
15
c)
17
d)
19
|
|
Chirag Verma answered |
n(N ∪ S) = 32, n(N) = 30, n(S) = 17 Since, We know that,
n( N ∪ S) = n(N) + n(S) - n(N ∩ 5)
or 32= 30+ 17 - n(N ∩ S)
or n(N ∩ S ) = 47 - 32= 15
Since, R is an equivalence relation on set A, Therefore, (a, a) ∈ R for all a ∈ A, Hence, R has atleast n ordered pairs.
n( N ∪ S) = n(N) + n(S) - n(N ∩ 5)
or 32= 30+ 17 - n(N ∩ S)
or n(N ∩ S ) = 47 - 32= 15
Since, R is an equivalence relation on set A, Therefore, (a, a) ∈ R for all a ∈ A, Hence, R has atleast n ordered pairs.
Assertion (A): 
is not a set. Here, R is the set of real numbers.
Reason (R): For every real number x, x2 ≥ 0.- a)A and B are both correct, also R is the correct explanation of A.
- b)A and R are both correct, but R is not the correct explanation of A..
- c)A is correct but R is wrong.
- d)A is wrong but R is correct.
Correct answer is option 'A'. Can you explain this answer?
Assertion (A): is not a set. Here, R is the set of real numbers.
Reason (R): For every real number x, x2 ≥ 0.
is not a set. Here, R is the set of real numbers.Reason (R): For every real number x, x2 ≥ 0.
a)
A and B are both correct, also R is the correct explanation of A.
b)
A and R are both correct, but R is not the correct explanation of A..
c)
A is correct but R is wrong.
d)
A is wrong but R is correct.
|
|
Chirag Verma answered |
Both (A) and (R) are true and (R) is the correct explanation of (A).
Since, x2 is never negative but it can be positive or zero.
Since, x2 is never negative but it can be positive or zero.
Which one of the following is the empty set?- a){x : x is a real number and x2 - 1 = 0}
- b){x : x is a real number and x2 + 1 = 0 }
- c){x : x is a real number and x2- 9 = 0}
- d){x : x is a real number and x2 = x + 2}
Correct answer is option 'B'. Can you explain this answer?
Which one of the following is the empty set?
a)
{x : x is a real number and x2 - 1 = 0}
b)
{x : x is a real number and x2 + 1 = 0 }
c)
{x : x is a real number and x
2
- 9 = 0}d)
{x : x is a real number and x2 = x + 2}
|
|
Madhav Mehra answered |
The empty set is a set that does not contain any elements. In other words, it is a set with no members.
To determine which of the given options represents the empty set, we need to analyze the conditions specified in each option and check if there are any real numbers that satisfy those conditions.
Option a: {x : x is a real number and x^2 - 1 = 0}
- This option represents the set of real numbers that satisfy the equation x^2 - 1 = 0.
- Solving this equation, we get x = ±1.
- So, this set contains two elements: {1, -1}. Therefore, it is not the empty set.
Option b: {x : x is a real number and x^2 + 1 = 0}
- This option represents the set of real numbers that satisfy the equation x^2 + 1 = 0.
- However, this equation has no real solutions since the square of any real number is always non-negative.
- Therefore, there are no real numbers that satisfy this equation, and the set is empty.
Option c: {x : x is a real number and x^2 - 9 = 0}
- This option represents the set of real numbers that satisfy the equation x^2 - 9 = 0.
- Solving this equation, we get x = ±3.
- So, this set contains two elements: {3, -3}. Therefore, it is not the empty set.
Option d: {x : x is a real number and x^2 = x + 2}
- This option represents the set of real numbers that satisfy the equation x^2 = x + 2.
- Solving this equation, we get x = -1, x = 2.
- So, this set contains two elements: {-1, 2}. Therefore, it is not the empty set.
Therefore, option b is the correct answer as it represents the empty set.
To determine which of the given options represents the empty set, we need to analyze the conditions specified in each option and check if there are any real numbers that satisfy those conditions.
Option a: {x : x is a real number and x^2 - 1 = 0}
- This option represents the set of real numbers that satisfy the equation x^2 - 1 = 0.
- Solving this equation, we get x = ±1.
- So, this set contains two elements: {1, -1}. Therefore, it is not the empty set.
Option b: {x : x is a real number and x^2 + 1 = 0}
- This option represents the set of real numbers that satisfy the equation x^2 + 1 = 0.
- However, this equation has no real solutions since the square of any real number is always non-negative.
- Therefore, there are no real numbers that satisfy this equation, and the set is empty.
Option c: {x : x is a real number and x^2 - 9 = 0}
- This option represents the set of real numbers that satisfy the equation x^2 - 9 = 0.
- Solving this equation, we get x = ±3.
- So, this set contains two elements: {3, -3}. Therefore, it is not the empty set.
Option d: {x : x is a real number and x^2 = x + 2}
- This option represents the set of real numbers that satisfy the equation x^2 = x + 2.
- Solving this equation, we get x = -1, x = 2.
- So, this set contains two elements: {-1, 2}. Therefore, it is not the empty set.
Therefore, option b is the correct answer as it represents the empty set.
If A, B and C are three sets and U is the universal set such that n(U) = 700, n(A) = 200, n(B) = 300 and h (A ∩ B) = 100, then what is the value of (A' ∩ B')?- a)100
- b)200
- c)300
- d)400
Correct answer is option 'C'. Can you explain this answer?
If A, B and C are three sets and U is the universal set such that n(U) = 700, n(A) = 200, n(B) = 300 and h (A ∩ B) = 100, then what is the value of (A' ∩ B')?
a)
100
b)
200
c)
300
d)
400
|
|
Chirag Verma answered |
Given that, n(U) = 700, n(A) = 200, n(B) = 300 and n(A ∩ B)=100
We know that,
= 700-400 = 300
We know that,
= 700-400 = 300
Let R be the relation defined on the set of natural number N as aRb; a, b ∈ N, if a divides b.Then, which one of the following is correct?- a)R is reflexive only
- b)R is symmetric only
- c)R is transitive only
- d)R is reflexive and transitive
Correct answer is option 'D'. Can you explain this answer?
Let R be the relation defined on the set of natural number N as aRb; a, b ∈ N, if a divides b.Then, which one of the following is correct?
a)
R is reflexive only
b)
R is symmetric only
c)
R is transitive only
d)
R is reflexive and transitive
|
|
Chirag Verma answered |
For reflexive aRa implies a divides a
So, R is reflexive.
For symmetric
aRb impliesa divides b
bRa impliesa divides a
which may not be possible.
So, R is not symmetric.
For transitive
aRb impliesa divides b implies b = ka bRc impliesb divides c implies c = lb Now, c = Ika
or a divides c implies aRc
or bRc implies cRa
Thus, R is transitive.
So, R is reflexive.
For symmetric
aRb impliesa divides b
bRa impliesa divides a
which may not be possible.
So, R is not symmetric.
For transitive
aRb impliesa divides b implies b = ka bRc impliesb divides c implies c = lb Now, c = Ika
or a divides c implies aRc
or bRc implies cRa
Thus, R is transitive.
If A, B, C are three sets, then what is A - (B - C) equal to- a)A - (B ∩ C )
- b)(A - B ) ∪ C
- c)( A - B ) ∪ ( A ∩ C)
- d)(A - B) ∪ (A - C)
Correct answer is option 'C'. Can you explain this answer?
If A, B, C are three sets, then what is A - (B - C) equal to
a)
A - (B ∩ C )
b)
(A - B ) ∪ C
c)
( A - B ) ∪ ( A ∩ C)
d)
(A - B) ∪ (A - C)
|
|
Veda Institute answered |
In the above Venn diagram, shaded area shows A - (B - C)
In the above Venn diagram, horizontal lines mean (A - B) and vertical lines mean (A ∩ C).
Total shaded portion =
Therefore,
In the above Venn diagram, horizontal lines mean (A - B) and vertical lines mean (A ∩ C).
Total shaded portion =
Therefore,
Let M be the set of men and R is a relation ‘is son of’ defined on M. Then, R is- a)an equivalence relation
- b)a symmetric relation only
- c)a transitive relation only
- d)None Of the above
Correct answer is option 'D'. Can you explain this answer?
Let M be the set of men and R is a relation ‘is son of’ defined on M. Then, R is
a)
an equivalence relation
b)
a symmetric relation only
c)
a transitive relation only
d)
None Of the above
|
|
Chirag Verma answered |
M = set of men and R is a relation ‘is son of ’ defined on M.
Reflexive relation aRa.
Since, a cannot be a son o f a.
Symmetric relation aRb implies bR
Transitive relation aRb, bRc implies cRa which is not possible.
Reflexive relation aRa.
Since, a cannot be a son o f a.
Symmetric relation aRb implies bR
a
which is also not possible.Transitive relation aRb, bRc implies cRa which is not possible.
Let A = {x | x ≤ 9 , x ε N}. Let B = {a, b, c} be the subset of A where (a + b + c) is a multiple of 3. What is the largest possible number of subsets like B?- a)12
- b)21
- c)27
- d)30
Correct answer is option 'D'. Can you explain this answer?
Let A = {x | x ≤ 9 , x ε N}. Let B = {a, b, c} be the subset of A where (a + b + c) is a multiple of 3. What is the largest possible number of subsets like B?
a)
12
b)
21
c)
27
d)
30
|
|
Veda Institute answered |
Here,
A -
= {1,2,3,4,5,6,7,8,9} ;
Total possible multiple of 3 are
= 3,6,9,12,15,18,21,24,27
But 3 and 27 are not possible.
6 → 1 + 2 + 3
9 → 2 + 3 + 4 , 5 + 3 + 1,6 + 2 + 1
12 → 9 + 2 + 1,8 + 3 + 1,7 + 1 + 4,
7 + 2 + 3 ,( 5+ 4 + 2 , 6 + 5 + 1, 5 + 4 + 3
15 → 9 + 4 + 2,9 + 5 + l, 8 + 6 + 1 , 8 + 5 + 2 , 8 + 4 + 3 , 7 + 6 + 2 , 7 + 5 +3 , 6 + 5 + 4
18 → 9 + 8+ 1,9 + 7 + 2,9 + 6 + 3,9 + 5 + 4, 8 + 7 + 3, 8 + 6 + 4,7 + 6 + 5
21→9 + 8 +4, 9 + 7 + 5, 8 + 7 + 6
24 → 9 + 8 + 7
Hence, total number of largest possible subsets are 30.
A -
= {1,2,3,4,5,6,7,8,9} ;
Total possible multiple of 3 are
= 3,6,9,12,15,18,21,24,27
But 3 and 27 are not possible.
6 → 1 + 2 + 3
9 → 2 + 3 + 4 , 5 + 3 + 1,6 + 2 + 1
12 → 9 + 2 + 1,8 + 3 + 1,7 + 1 + 4,
7 + 2 + 3 ,( 5+ 4 + 2 , 6 + 5 + 1, 5 + 4 + 3
15 → 9 + 4 + 2,9 + 5 + l, 8 + 6 + 1 , 8 + 5 + 2 , 8 + 4 + 3 , 7 + 6 + 2 , 7 + 5 +3 , 6 + 5 + 4
18 → 9 + 8+ 1,9 + 7 + 2,9 + 6 + 3,9 + 5 + 4, 8 + 7 + 3, 8 + 6 + 4,7 + 6 + 5
21→9 + 8 +4, 9 + 7 + 5, 8 + 7 + 6
24 → 9 + 8 + 7
Hence, total number of largest possible subsets are 30.
Given a non-empty subset S of R (real number) on the interval [0, 5]. Then, any numbers greater than 5 is an upper bound of S since it is greater than all of the numbers contained in S. Therefore, we can say that 5.01, 5.1, 6 and 7 are all upper bounds of S. Among all these upper bound, the one with the smallest value is known as the _______ of S. - a)Supremum
- b)Minimum
- c)Infimum
- d)Maximum
Correct answer is option 'A'. Can you explain this answer?
Given a non-empty subset S of R (real number) on the interval [0, 5]. Then, any numbers greater than 5 is an upper bound of S since it is greater than all of the numbers contained in S. Therefore, we can say that 5.01, 5.1, 6 and 7 are all upper bounds of S. Among all these upper bound, the one with the smallest value is known as the _______ of S.
a)
Supremum
b)
Minimum
c)
Infimum
d)
Maximum
|
|
Chirag Verma answered |
The supremum of a set of numbers defined on an interval is also known as the least upper bound, lub. On the contrary, the infimum of a set of numbers is also known as the greatest lower bound, glb.
What is the number of proper subsets of a given finite set with n elements?- a)2n - l
- b)2n - 2
- c)2n - 1
- d)2n - 2
Correct answer is option 'C'. Can you explain this answer?
What is the number of proper subsets of a given finite set with n elements?
a)
2n - l
b)
2n - 2
c)
2
n
- 1d)
2n - 2
|
|
Chirag Verma answered |
If a set contains ‘n’ elements, then the number of proper subsets of the set is 2n - 1.
If A = {p, q} the proper subsets of A are [{ }, {p}, {q}]
⇒ Number of proper subsets of A are 3 = 22 - 1 = 4 - 1
In general, number of proper subsets of a given set = 2m - 1, where m is the number of elements.
If A = {p, q} the proper subsets of A are [{ }, {p}, {q}]
⇒ Number of proper subsets of A are 3 = 22 - 1 = 4 - 1
In general, number of proper subsets of a given set = 2m - 1, where m is the number of elements.
If, A = {1,2,3,4} and
R = {(1, 1), (1, 3), (2, 2), (3, 1) (3, 4), (4, 3) (4, 4) is a relation on A x A, then which one of the following is correct?
- a)R is reflexive
- b)R is symmetric and transitive
- c)R is reflexive and symmetric
- d)R is neither reflexive nor transitive
Correct answer is option 'C'. Can you explain this answer?
If, A = {1,2,3,4} and
R = {(1, 1), (1, 3), (2, 2), (3, 1) (3, 4), (4, 3) (4, 4) is a relation on A x A, then which one of the following is correct?
R = {(1, 1), (1, 3), (2, 2), (3, 1) (3, 4), (4, 3) (4, 4) is a relation on A x A, then which one of the following is correct?
a)
R is reflexive
b)
R is symmetric and transitive
c)
R is reflexive and symmetric
d)
R is neither reflexive nor transitive
|
|
Chirag Verma answered |
Correct Answer :- c
Explanation : Let A = {1,2,3,4} and R be a relation on A given by R={(1,1),(2,2),(3,3),(4,4),(1,2),(2,1),(3,1),(1,3)}.
Now for all 1,2,3,4 ∈ A, (1,1),(2,2),(3,3),(4,4) ∈ R, this gives the relation R is reflexive.
Again for (1,2) ∈ R
⇒(2,1)∈R and (1,3)∈R
⇒(3,1)∈R for 1,2∈A.
This gives the relation R is symmetric.
But the relation is not transitive as (2,1),(1,3)∈R but (2,3)does not ∈ R.
If the sequence is increasing, then it ____- a)converges to its supremum.
- b)diverges.
- c)may converges to its supremum.
- d)is bounded
Correct answer is option 'A'. Can you explain this answer?
If the sequence is increasing, then it ____
a)
converges to its supremum.
b)
diverges.
c)
may converges to its supremum.
d)
is bounded
|
|
Chirag Verma answered |
Informally, the theorems state that if a sequence is increasing and bounded above by a supremum, then the sequence will converge to the supremum; in the same way, if a sequence is decreasing and is bounded below by an infimum, it will converge to the infimum.
Let U= { 1 , 2 , 3 , ...,20}. Let A, B, C be the subsets of U.
Let A be the set of all numbers, which are prefect squares, B be the set of all numbers which are multiples of 5 and C be the set of all numbers, which are divisible by 2 and 3.
Q. Consider the following statements.
I. A, B, C are mutually exclusive.
II. A, B, C are mutually exhaustive.
III. The number of elements in the complement set of A∪ B is 12.
Q. Which of the statements given above the correct?
- a)I and II
- b)I and III
- c)II and III
- d)I, II and III
Correct answer is option 'B'. Can you explain this answer?
Let U= { 1 , 2 , 3 , ...,20}. Let A, B, C be the subsets of U.
Let A be the set of all numbers, which are prefect squares, B be the set of all numbers which are multiples of 5 and C be the set of all numbers, which are divisible by 2 and 3.
Q. Consider the following statements.
I. A, B, C are mutually exclusive.
II. A, B, C are mutually exhaustive.
III. The number of elements in the complement set of A∪ B is 12.
Q. Which of the statements given above the correct?
Let A be the set of all numbers, which are prefect squares, B be the set of all numbers which are multiples of 5 and C be the set of all numbers, which are divisible by 2 and 3.
Q. Consider the following statements.
I. A, B, C are mutually exclusive.
II. A, B, C are mutually exhaustive.
III. The number of elements in the complement set of A∪ B is 12.
Q. Which of the statements given above the correct?
a)
I and II
b)
I and III
c)
II and III
d)
I, II and III
|
|
Veda Institute answered |
U = { 1 ,2 ,3 , ...,20 } A = Set of all natural numbers which are perfect square
= {1,4 ,9 ,16 }
B = Set of all natural numbers which are multiple of 5
= {5,10,15,20}
C = Set of all natural numbers which are divisible by 2 and 3
= { 6 ,12 ,18 }
Since, A ∩ B ∩ C = φ
and A ∪ B = {1 ,4 ,9 ,1 6 ,5 ,10,15 ,20}
implies n ( A ∪ B ) = 8
n(A ∪ B)' = 20 - 8 = 12
hence A, B, C are mutually exclusive and the number of elements in the complement set of A ∪ B is 12 .
= {1,4 ,9 ,16 }
B = Set of all natural numbers which are multiple of 5
= {5,10,15,20}
C = Set of all natural numbers which are divisible by 2 and 3
= { 6 ,12 ,18 }
Since, A ∩ B ∩ C = φ
and A ∪ B = {1 ,4 ,9 ,1 6 ,5 ,10,15 ,20}
implies n ( A ∪ B ) = 8
n(A ∪ B)' = 20 - 8 = 12
hence A, B, C are mutually exclusive and the number of elements in the complement set of A ∪ B is 12 .
Let S be the set of all real numbers. Then, relation R = {(a, b ) : 1 + ab > 0} on S is- a)reflexive and symmetric but not transitive
- b)reflexive and transitive but not symmetric
- c)symmetric, transitive but not reflexive
- d)None of the above is true
Correct answer is option 'A'. Can you explain this answer?
Let S be the set of all real numbers. Then, relation R = {(a, b ) : 1 + ab > 0} on S is
a)
reflexive and symmetric but not transitive
b)
reflexive and transitive but not symmetric
c)
symmetric, transitive but not reflexive
d)
None of the above is true
|
Aarav Singh answered |
Explanation:
Reflexive:
- For a relation to be reflexive, (a, a) must be in the relation for all a in S.
- In this case, let's consider a = 0. When a = 0, we have 1 + 0*b = 1, which is greater than 0.
- Therefore, (0, 0) is in the relation R.
- Hence, the relation R is reflexive.
Symmetric:
- For a relation to be symmetric, if (a, b) is in the relation, then (b, a) must also be in the relation.
- Let's consider (1, 2) in R, since 1 + 1*2 = 3 > 0. Hence, (1, 2) is in R.
- However, 1 + 1*2 = 3 > 0, but 1 + 2*1 = 3 > 0, so (2, 1) is not in R.
- Therefore, the relation R is not symmetric.
Transitive:
- For a relation to be transitive, if (a, b) and (b, c) are in the relation, then (a, c) must also be in the relation.
- Let's consider (1, 2) and (2, 3) in R. We have 1 + 1*2 = 3 > 0 and 1 + 2*3 = 7 > 0.
- However, 1 + 1*3 = 4 > 0, but 4 is not greater than 0. Therefore, (1, 3) is not in R.
- Hence, the relation R is not transitive.
Therefore, the relation R = {(a, b) : 1 + ab > 0} on set S is reflexive and symmetric but not transitive, making option 'A' the correct answer.
Reflexive:
- For a relation to be reflexive, (a, a) must be in the relation for all a in S.
- In this case, let's consider a = 0. When a = 0, we have 1 + 0*b = 1, which is greater than 0.
- Therefore, (0, 0) is in the relation R.
- Hence, the relation R is reflexive.
Symmetric:
- For a relation to be symmetric, if (a, b) is in the relation, then (b, a) must also be in the relation.
- Let's consider (1, 2) in R, since 1 + 1*2 = 3 > 0. Hence, (1, 2) is in R.
- However, 1 + 1*2 = 3 > 0, but 1 + 2*1 = 3 > 0, so (2, 1) is not in R.
- Therefore, the relation R is not symmetric.
Transitive:
- For a relation to be transitive, if (a, b) and (b, c) are in the relation, then (a, c) must also be in the relation.
- Let's consider (1, 2) and (2, 3) in R. We have 1 + 1*2 = 3 > 0 and 1 + 2*3 = 7 > 0.
- However, 1 + 1*3 = 4 > 0, but 4 is not greater than 0. Therefore, (1, 3) is not in R.
- Hence, the relation R is not transitive.
Therefore, the relation R = {(a, b) : 1 + ab > 0} on set S is reflexive and symmetric but not transitive, making option 'A' the correct answer.
For a set A, consider the following statements
1. A ∪ P(A) = P(A)
2. {A} ∩ P(A)=A
3. P(A) - {A} = P{A}
where P denotes power set.
Which of the statements given above is/are correct?- a)I only
- b)II only
- c)III only
- d)I, II and III
Correct answer is option 'B'. Can you explain this answer?
For a set A, consider the following statements
1. A ∪ P(A) = P(A)
2. {A} ∩ P(A)=A
3. P(A) - {A} = P{A}
where P denotes power set.
Which of the statements given above is/are correct?
1. A ∪ P(A) = P(A)
2. {A} ∩ P(A)=A
3. P(A) - {A} = P{A}
where P denotes power set.
Which of the statements given above is/are correct?
a)
I only
b)
II only
c)
III only
d)
I, II and III
|
|
Veda Institute answered |
Let A = {1,2} and {A} = {(1,2)} implies,P(A) = {{1}, {2},.{1,2}, φ}
Then, A ∪ P(A) ≠ P(A)
and{A} ∩ P(A) = {{ l , 2}} ∩ {{1}, {2}, {1,2},φ}
= {1,2}=A
Then, A ∪ P(A) ≠ P(A)
and{A} ∩ P(A) = {{ l , 2}} ∩ {{1}, {2}, {1,2},φ}
= {1,2}=A
Let R1 be a relation defined by R1 = {(a, b ) | a ≥ b, a, b ∈ R}. Then R1 is- a)an equivalence relation on R
- b)reflexive, transitive but not symmetric
- c)symmetric, transitive but not reflexive
- d)neither transitive nor reflexive but symmetric
Correct answer is option 'B'. Can you explain this answer?
Let R1 be a relation defined by R1 = {(a, b ) | a ≥ b, a, b ∈ R}. Then R1 is
a)
an equivalence relation on R
b)
reflexive, transitive but not symmetric
c)
symmetric, transitive but not reflexive
d)
neither transitive nor reflexive but symmetric
|
|
Chirag Verma answered |
For any a ∈ R, we haye a ≥ a. Therefore, the relation R is reflexive but it is hot symmetric as (2, 1) ∈ R but (1, 2) ∉ R. The relation R is transitive also, because (a, b) ∈ R, (b, c) ∈ R imply that a ≥ b and b ≥. c which is turn imply that a ≥ c.
Let R and S be two non-void relations on a set A.
Q. Which of the following statements is false?- a)R and S are transitive ⇒ R ∪ S is transitive
- b)R and S are transitive ⇒ R ∩ S is transitive
- c)R and S are symmetric ⇒ R ∪ S is symmetric
- d)R and S are reflexive ⇒ R ∩ S is reflexive
Correct answer is option 'A'. Can you explain this answer?
Let R and S be two non-void relations on a set A.
Q. Which of the following statements is false?
Q. Which of the following statements is false?
a)
R and S are transitive ⇒ R ∪ S is transitive
b)
R and S are transitive ⇒ R ∩ S is transitive
c)
R and S are symmetric ⇒ R ∪ S is symmetric
d)
R and S are reflexive ⇒ R ∩ S is reflexive
|
|
Chirag Verma answered |
Let A = {1, 2, 3} and R = {(1, 1), (1, 2)}
S = {(2, 2), (2, 3)} be transitive relations on A.
Then, R ∪ S= {(1,1); (1,2); (2,2); (2, 3)} Obviously, R ∪ S is transitive.
Since, (1, 2) ∈ R ∪ S and (2, 3 ) ∈ R ∪ S but (l , 3) ∉ R ∪ S.
S = {(2, 2), (2, 3)} be transitive relations on A.
Then, R ∪ S= {(1,1); (1,2); (2,2); (2, 3)} Obviously, R ∪ S is transitive.
Since, (1, 2) ∈ R ∪ S and (2, 3 ) ∈ R ∪ S but (l , 3) ∉ R ∪ S.
Let N be the set of integers. A relation R on N is defined as R = {(x, y) | xy > 0, x, y ∈ N}. Then, which one of the following is correct?- a)R is symmetric but not reflexive
- b)R is reflexive but not symmetric
- c)R is symmetric and reflexive but not transitive
- d)R is an equivalence relation
Correct answer is option 'D'. Can you explain this answer?
Let N be the set of integers. A relation R on N is defined as R = {(x, y) | xy > 0, x, y ∈ N}. Then, which one of the following is correct?
a)
R is symmetric but not reflexive
b)
R is reflexive but not symmetric
c)
R is symmetric and reflexive but not transitive
d)
R is an equivalence relation
|
|
Chirag Verma answered |
Since, R = {(x, y) | xy > 0, x, y ∈ N}
Reflexive
Since, x, y ∈ N
So, x, x ∈ N
implies x2 > 0
Hence, R is reflexive.
Symmetric
Since, x, y ∈ N and xy > 0 implies yx > 0 Hence, R is also symmetric.
Transitive
Since, x,y, z, ∈ N
implies xy > 0, yz > 0
implies xz > 0
So, R is also transitive
Thus, R is an equivalence relation.
Reflexive
Since, x, y ∈ N
So, x, x ∈ N
implies x2 > 0
Hence, R is reflexive.
Symmetric
Since, x, y ∈ N and xy > 0 implies yx > 0 Hence, R is also symmetric.
Transitive
Since, x,y, z, ∈ N
implies xy > 0, yz > 0
implies xz > 0
So, R is also transitive
Thus, R is an equivalence relation.
Let N denote the set of natural numbers and A = {n2 : n ∈ N} and B = {n3/2 : n ∈ N}. Which one of the following is correct?- a)A ∪ B = N
- b)The complement of (A ∪ B) is an infinite set
- c)A ∩ B must be a finite set
- d)A ∩ B must be a proper subset o f {m6 : m ∈ N)
Correct answer is option 'D'. Can you explain this answer?
Let N denote the set of natural numbers and A = {n2 : n ∈ N} and B = {n3/2 : n ∈ N}. Which one of the following is correct?
a)
A ∪ B = N
b)
The complement of (A ∪ B) is an infinite set
c)
A ∩ B must be a finite set
d)
A ∩ B must be a proper subset o f {m6 : m ∈ N)
|
|
Chirag Verma answered |
Since, A = { n2 : n ∈ N } and
B = {n3 : n ∈ N}
A = { 1 ,4 ,9 ,1 6 ,2 5 ,4 9 ,6 4 ,8 1 ,...}
B = {1 ,8 ,2 7 ,6 4 ,1 2 5 ,...}
A ∩ B = { 1 , 6 4 , . .....}
A ∩ B must be a proper subset of {m6 : m ∈ N}
B = {n3 : n ∈ N}
A = { 1 ,4 ,9 ,1 6 ,2 5 ,4 9 ,6 4 ,8 1 ,...}
B = {1 ,8 ,2 7 ,6 4 ,1 2 5 ,...}
A ∩ B = { 1 , 6 4 , . .....}
A ∩ B must be a proper subset of {m6 : m ∈ N}
If A = P {1, 2} where P denotes the power set, then which one of the following is correct?- a){1,2} ⊂ A
- b)1 ε A
- c)φ ∉ A
- d){1,2} ε A
Correct answer is option 'D'. Can you explain this answer?
If A = P {1, 2} where P denotes the power set, then which one of the following is correct?
a)
{1,2} ⊂ A
b)
1 ε A
c)
φ ∉ A
d)
{1,2} ε A
|
|
Charvi Verma answered |
Understanding the Power Set
The power set P{1, 2} includes all possible subsets of the set {1, 2}. Let's first enumerate the elements of this power set:
- The subsets of {1, 2} are:
- The empty set: {}
- The set containing just the first element: {1}
- The set containing just the second element: {2}
- The set containing both elements: {1, 2}
Thus, the power set P{1, 2} is:
P{1, 2} = { {}, {1}, {2}, {1, 2} }
This is the set A.
Analyzing the Options
Now, let's evaluate each of the provided options:
- Option a: {1, 2} ⊆ A
- This is incorrect because {1, 2} is not a subset of A; rather, it is an element of A.
- Option b: 1 ∈ A
- This is also incorrect. The number 1 is not an element of A; the elements of A are sets, not individual numbers.
- Option c: ∅ ∉ A
- This statement is incorrect as well. The empty set ∅ is indeed an element of A.
- Option d: {1, 2} ∈ A
- This is correct. The set {1, 2} is one of the elements in the power set P{1, 2}.
Conclusion
Thus, the correct answer is option 'D': {1, 2} ∈ A. The confusion often arises when distinguishing between elements and subsets within the context of power sets.
The power set P{1, 2} includes all possible subsets of the set {1, 2}. Let's first enumerate the elements of this power set:
- The subsets of {1, 2} are:
- The empty set: {}
- The set containing just the first element: {1}
- The set containing just the second element: {2}
- The set containing both elements: {1, 2}
Thus, the power set P{1, 2} is:
P{1, 2} = { {}, {1}, {2}, {1, 2} }
This is the set A.
Analyzing the Options
Now, let's evaluate each of the provided options:
- Option a: {1, 2} ⊆ A
- This is incorrect because {1, 2} is not a subset of A; rather, it is an element of A.
- Option b: 1 ∈ A
- This is also incorrect. The number 1 is not an element of A; the elements of A are sets, not individual numbers.
- Option c: ∅ ∉ A
- This statement is incorrect as well. The empty set ∅ is indeed an element of A.
- Option d: {1, 2} ∈ A
- This is correct. The set {1, 2} is one of the elements in the power set P{1, 2}.
Conclusion
Thus, the correct answer is option 'D': {1, 2} ∈ A. The confusion often arises when distinguishing between elements and subsets within the context of power sets.
The set of real numbers in the closed interval {0, 1} is- a)countable set
- b)uncountable set
- c)finite set
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
The set of real numbers in the closed interval {0, 1} is
a)
countable set
b)
uncountable set
c)
finite set
d)
None of these
|
Ishan Kapoor answered |
The set of real numbers in the closed interval [0, 1] is an uncountable set.
Explanation:
To understand why the set of real numbers in the closed interval [0, 1] is uncountable, we need to understand what it means for a set to be countable or uncountable.
A set is countable if its elements can be put into a one-to-one correspondence with the natural numbers (1, 2, 3, ...). In other words, a set is countable if we can list its elements in a sequence.
On the other hand, a set is uncountable if its elements cannot be put into a one-to-one correspondence with the natural numbers. In other words, a set is uncountable if we cannot list all its elements in a sequence.
To determine whether the set of real numbers in the closed interval [0, 1] is countable or uncountable, we can use a proof by contradiction known as Cantor's diagonal argument.
Cantor's diagonal argument starts by assuming that the set of real numbers in the closed interval [0, 1] is countable. This means that we can list all the real numbers in the interval in a sequence.
Next, we construct a new number by taking the digits in the diagonal of the sequence and changing each digit to a different digit. This new number will be different from every number in the original sequence because it will differ from each number in at least one digit.
Now, we have a new number that is not in the original sequence, which contradicts the assumption that the set of real numbers in the closed interval [0, 1] is countable. Therefore, the set of real numbers in the closed interval [0, 1] must be uncountable.
In conclusion, the set of real numbers in the closed interval [0, 1] is an uncountable set.
Explanation:
To understand why the set of real numbers in the closed interval [0, 1] is uncountable, we need to understand what it means for a set to be countable or uncountable.
A set is countable if its elements can be put into a one-to-one correspondence with the natural numbers (1, 2, 3, ...). In other words, a set is countable if we can list its elements in a sequence.
On the other hand, a set is uncountable if its elements cannot be put into a one-to-one correspondence with the natural numbers. In other words, a set is uncountable if we cannot list all its elements in a sequence.
To determine whether the set of real numbers in the closed interval [0, 1] is countable or uncountable, we can use a proof by contradiction known as Cantor's diagonal argument.
Cantor's diagonal argument starts by assuming that the set of real numbers in the closed interval [0, 1] is countable. This means that we can list all the real numbers in the interval in a sequence.
Next, we construct a new number by taking the digits in the diagonal of the sequence and changing each digit to a different digit. This new number will be different from every number in the original sequence because it will differ from each number in at least one digit.
Now, we have a new number that is not in the original sequence, which contradicts the assumption that the set of real numbers in the closed interval [0, 1] is countable. Therefore, the set of real numbers in the closed interval [0, 1] must be uncountable.
In conclusion, the set of real numbers in the closed interval [0, 1] is an uncountable set.
The function f(x) = ex, x ε R is- a)onto but not one-one
- b)one-one onto
- c)one-one but not onto
- d)neither one-one nor onto
Correct answer is option 'C'. Can you explain this answer?
The function f(x) = ex, x ε R is
a)
onto but not one-one
b)
one-one onto
c)
one-one but not onto
d)
neither one-one nor onto
|
|
Chirag Verma answered |
It is clear form the graph that f(x) = ex, 
is one- one but not onto. Since, range ≠ codomain, so f(x) is into.
is one- one but not onto. Since, range ≠ codomain, so f(x) is into.An integer m is said to be related to another integer n, if m is a multiple of n. Then, the relation is- a)reflexive and symmetric
- b)reflexive and transitive
- c)symmetric and transitive
- d)equivalence relation
Correct answer is option 'B'. Can you explain this answer?
An integer m is said to be related to another integer n, if m is a multiple of n. Then, the relation is
a)
reflexive and symmetric
b)
reflexive and transitive
c)
symmetric and transitive
d)
equivalence relation
|
|
Veda Institute answered |
For any integer n, we have n | n => nRn
So, nRn for all n ∈ Z implies R is reflexive
Now, 2|6 but 6 + 2,
implies (2 ,6) ∈ R but (6,2) ∉ R
So, R is not symmetric.
Let (m, n) ∈ R and (n , p) ∈ R
Then,
implies m | p => (m,p) ∈ R
So, R is transitive.
Hence, R is reflexive and transitive but it is not symmetric.
So, nRn for all n ∈ Z implies R is reflexive
Now, 2|6 but 6 + 2,
implies (2 ,6) ∈ R but (6,2) ∉ R
So, R is not symmetric.
Let (m, n) ∈ R and (n , p) ∈ R
Then,
implies m | p => (m,p) ∈ R
So, R is transitive.
Hence, R is reflexive and transitive but it is not symmetric.
Let A = {(x, y) : y = ex, x ∈ R}, B = {(x, y) : y = e-x,x ∈ R}. Then- a) A ∩ B = φ
- b)A ∩ B ≠ φ
- c)A ∪ B = R2
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
Let A = {(x, y) : y = ex, x ∈ R}, B = {(x, y) : y = e-x,x ∈ R}. Then
a)
A ∩ B = φ
b)
A ∩ B ≠ φ
c)
A ∪ B = R2
d)
None of these
|
|
Chirag Verma answered |
Since, y = ex, y = e-x will meet,
when ex = e-x
implies e2x = 1,
So, x = 0,y = 1
Hence, A and B meet on (0,1),
Hence, A ∩ B = φ
when ex = e-x
implies e2x = 1,
So, x = 0,y = 1
Hence, A and B meet on (0,1),
Hence, A ∩ B = φ
Consider the function f:R → {0, 1} such that
Q. Which one of the following is correct?- a)The function is one-one into
- b)The function is many One into
- c)The function is one-one onto
- d)The function is many one onto
Correct answer is option 'D'. Can you explain this answer?
Consider the function f:R → {0, 1} such that
Q. Which one of the following is correct?
Q. Which one of the following is correct?
a)
The function is one-one into
b)
The function is many One into
c)
The function is one-one onto
d)
The function is many one onto
|
|
Arghya answered |
Since, on taking a straight line parallel to x-axis, the group of given function intersect it at many points.
So, f(x) is a many one.
And as range of f(x) = codomain
So, f(x) is onto.
Hence, f(x) is many one onto.
So, f(x) is a many one.
And as range of f(x) = codomain
So, f(x) is onto.
Hence, f(x) is many one onto.
In a class of 55 students, the number of students studying different subjects are 23 in Mathematics, 24 in Physics, 19 in Chemistry, 12 in Mathematics and Physics, 9 in Mathematics and Chemistry, 7 in Physics and Chemistry and 4 in all the three subjects. The number of students who have taken exactly one subject is
- a)23
- b)56
- c)22
- d)45
Correct answer is option 'C'. Can you explain this answer?
In a class of 55 students, the number of students studying different subjects are 23 in Mathematics, 24 in Physics, 19 in Chemistry, 12 in Mathematics and Physics, 9 in Mathematics and Chemistry, 7 in Physics and Chemistry and 4 in all the three subjects. The number of students who have taken exactly one subject is
a)
23
b)
56
c)
22
d)
45
|
|
Chirag Verma answered |
Let n(M)=student who studying mathematics
n(C)=student who studying chemistry
n(P)=student who studying Physics
∴n(M)=23,n(P)=24,n(C)=19,n(M∩P)=12,n(M∩C)=9,n(P∩C)=7,n(M∩P∩C)=4
Number of student who studying mathematics but not physic and chemistry
⇒n(M)−[(n(M∩C)+n(M∩P)]+n(M∩P∩C)
⇒23−[9+12]+4
⇒23−21+4=6
Number of student who studying chemistry but not physic and matehematics
⇒n(C)−[(n(M∩C)+n(P∩C)]+n(M∩P∩C)
⇒19−[9+7]+4
⇒19−16+4=7
Number of student who studying physics but not mathematics and chemistry
⇒n(P)−[(n(M∩P)+n(P∩C)]+n(M∩P∩C)
⇒24−[12+7]+4
⇒24−19+4=9
∴no. of student studying exactly one subject = 6+7+9 = 22.
What is the range of f(x) = cos 2x - sin 2x?- a)[2,4]
- b)[- 1 , 1]
- c)[-√2, √2]
- d)(-√2. 2)
Correct answer is option 'C'. Can you explain this answer?
What is the range of f(x) = cos 2x - sin 2x?
a)
[2,4]
b)
[- 1 , 1]
c)
[-√2, √2]
d)
(-√2. 2)
|
|
Chirag Verma answered |
Since, f(x) = cos 2x - sin 2x
[Since, f( x ) = a cos x + b sin x,
[Since, f( x ) = a cos x + b sin x,
For every real number x, there is a positive integer n such that- a)n > x
- b)n < x
- c)n = x
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
For every real number x, there is a positive integer n such that
a)
n > x
b)
n < x
c)
n = x
d)
None of these
|
|
Akasha Mallick answered |
(x,y)€R and x>o as it is real
but a positive integer contains n which is=0 andx>0
so n>0¥x>0
but a positive integer contains n which is=0 andx>0
so n>0¥x>0
For real numbers x and y, we write xRy <=> x - y + √2 is an irrational number. Then, the relation R is- a)reflexive
- b)symmetric
- c)transitive
- d)None of these
Correct answer is option 'A'. Can you explain this answer?
For real numbers x and y, we write xRy <=> x - y + √2 is an irrational number. Then, the relation R is
a)
reflexive
b)
symmetric
c)
transitive
d)
None of these
|
|
Veda Institute answered |
For any x ∈ R, we have x - x +√2 = √2 an irrational number.
implies xRx for all x. So, R is reflexive.
R is not symmetric, because
and 
R is not transitive also because 
and 
but 
.
implies xRx for all x. So, R is reflexive.
R is not symmetric, because
and R is not transitive also because and but .Which one of the following is correct?
Here, P(A) denotes the power set of a set A.- a)A ∪ P(A) = P(A)
- b)A ∩ P(A) = A
- c)A - P(A )= A
- d)P(A) - {A}=P(A)
Correct answer is option 'C'. Can you explain this answer?
Which one of the following is correct?
Here, P(A) denotes the power set of a set A.
Here, P(A) denotes the power set of a set A.
a)
A ∪ P(A) = P(A)
b)
A ∩ P(A) = A
c)
A - P(A )= A
d)
P(A) - {A}=P(A)
|
|
Veda Institute answered |
A-P(A)=A,
which is correct as A and P(A) are disjoint sets.
which is correct as A and P(A) are disjoint sets.
Solution set of x = 3 (mod 7), p ∈ Z, is given by- a){3}
- b) {7 p - 3 ; p ∈ Z}
- c) {7p + 3 :p ∈ Z}
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
Solution set of x = 3 (mod 7), p ∈ Z, is given by
a)
{3}
b)
{7 p - 3 ; p ∈ Z}
c)
{7p + 3 :p ∈ Z}
d)
None of these
|
|
Veda Institute answered |
We first find R-1 we have
R-1 = {(5,4); (4 ,1 ); (6 ,4 ); (6 ,7 ); (7,3}.
We now obtain the elements of R-1 OR we first pick the elements of R and then of R-1 Since, (4, 5) ∈ R and
R-1 = {(5,4); (4 ,1 ); (6 ,4 ); (6 ,7 ); (7,3}.
We now obtain the elements of R-1 OR we first pick the elements of R and then of R-1 Since, (4, 5) ∈ R and
In a college of 300 students, every student reads 5 newspaper and every newspaper is read by 60 students. The number of newspaper is- a)atleast 30
- b)almost 20
- c)exactly 25
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
In a college of 300 students, every student reads 5 newspaper and every newspaper is read by 60 students. The number of newspaper is
a)
atleast 30
b)
almost 20
c)
exactly 25
d)
None of these
|
Vivek Kumar answered |
Let number of newspapers be x.
If every students reads one newspaper, the number of students would be
x(60) = 60x
Since, every students reads 5 newspapers
Hence, number of students
If every students reads one newspaper, the number of students would be
x(60) = 60x
Since, every students reads 5 newspapers
Hence, number of students
Let X = {1,2,3,4,5} and Y={ 1,3,5,7,9}. Which of the following (s) is not relations from X to Y?- a)R1 = {( x , y ) | y = 2+x, x ε X , y ε Y}
- b)R2 = {(1,1), (2,1), (3,3), (4,3), (5, 5)}
- c)R3 = {(1,1), (1,3), (3,5), (3,7), (5,7)}
- d)R4 = {(1,3), (2,5), (2,4), (7,9)}
Correct answer is option 'D'. Can you explain this answer?
Let X = {1,2,3,4,5} and Y={ 1,3,5,7,9}. Which of the following (s) is not relations from X to Y?
a)
R1 = {( x , y ) | y = 2+x, x ε X , y ε Y}
b)
R2 = {(1,1), (2,1), (3,3), (4,3), (5, 5)}
c)
R3 = {(1,1), (1,3), (3,5), (3,7), (5,7)}
d)
R4 = {(1,3), (2,5), (2,4), (7,9)}
|
|
Chirag Verma answered |
We are given that
X = { 1 ,2 ,3 ,4 ,5 }
and Y= { 1 ,3 ,5 ,7 ,9 }
Since, (7, 9) ε R4 but (7, 9) ∉ X x Y therefore R4 is not a relation from X to Y.
X = { 1 ,2 ,3 ,4 ,5 }
and Y= { 1 ,3 ,5 ,7 ,9 }
Since, (7, 9) ε R4 but (7, 9) ∉ X x Y therefore R4 is not a relation from X to Y.
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