Test: Real Analysis- 8 - Mathematics MCQ

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.

In the given statements, statement (I) is correct.

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.

An injective function means one-one.
In option (d),f (x) = -x
For every values of x, we get a different values of f.
Hence, it is injective.

A = {2,4,6}, B = {2,3, 5}
So, A x B contains 3 x 3 = 9 elements.
Hence, number of relations from A to B = 2⁹.

The shaded region in the Venn diagram is representes

Since shaded region is a combination of area of C and combined with intersection area of A and B.
So, the shaded region 

C∪(A∩B)

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.

Since, R = {3,6,9,12,15,..., 99}
and S= {5,10,15, ...,95}
Now, (R x S) ∩ ( S x  R)
= (R ∩ S ) x ( S ∩ R ) = (15,30,45,60,75,90) x (15,30, 45,60,75,90)
Hence, number of elements in
(R x S) ∩ (S x R) = 6 x 6 = 36

xRy ⇒ x and y are graduates of the same university.
Reflexive xRx ⇔ x and y are graduates of the same university.
So, relation is reflexive.
Symmetric xRy ⇔ x and y are graduates of the same university.
implies, yRx ⇔ y and x are graduates of the same university.
So, relation is symmetric.
Transitive xRy, yRz ⇔ xRz
It means x and y, y and z are graduates of the same university, then x and z are also graduates of the same university.
So, relation is transitive.
Hence, relation is reflexive, symmetric and transitive.

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

Here,

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.

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.

We have (a,b) R (a, b) for all (a, b) ∈ N x N
Since. a + b = b + a.
Hence, R is reflexive.
R is symmetric for we have (a , b) R (c, d)
implies a + d = b + c
implies d + a = c + b
implies c + b = d + a
implies (c, d) R (e, f)
Then, by definition of R,
we have a + d = b + c and c + f = d + e when by addition, we get
a + d + c + f = b + c + d + e
or a + f = b + e
Hence, (a, b) R (e , f)
Thus, (a, b) R( c, d) and (c, d) R (e , f) implies (a , b) R (e , f)

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}

Since, A ∩ B = φ (given)
So, A - B = A - ( A ∩ B )
(Since, A - B = A)
Now, B - A' = φ
and B ∩ A = φ
or B ‘- A ' = B ∩ A
and (A- B ) ∩ B = A ∩ B
So, option (d) is correct.

M = set of men and R is a relation ‘is son of ’ defined on M.
Reflexive relation _aR_a.
Since, a cannot be a son o f a.
Symmetric relation _aR_b implies _bRa which is also not possible.
Transitive relation _aR_b, bR_c implies _cR_a which is not possible.

Since, f(x) = x |x| 
If f(x₁) = f(x₂)
or x₁ | x₁| = x₂ | x₂ |
or x₁ = x₂
So, f(x) is one-one.
Also, range of f(x) = codomain of f(x)
So, f(x) is onto.
Hence, f(x) is both one-one and onto.

Since, f(x) = cos 2x - sin 2x
[Since, f( x ) = a cos x + b sin x,

Since, on the set of real numbers, R is a relation defined by _XR_y if and only if 3x + 4y = 5 for which


Hence, both the statements 11 and 111 are correct.