Test: Real Analysis- 1 - Mathematics MCQ

Since, (a, a), (b, b), (c, c) ε R
Therefore, R is a reflexive relation
But, (a,b ) ε R and (b, a) R So, R is not a symmetric relation Also, (a, b), (b, cj 6 R.
Implies (a, c)
Hence, R is not a transitive relation

Since, the sets A and B are not known thus cardinality of the set A Δ B cannot be determined.

Both (A) and (R) are true and (R) is the correct explanation of (A).
Since, (A ∪ B ∪ C)^c = A^c ∩ B^c ∩ C^c = D

Elements of a population are classified according to the presence or absence of each of 3 attributes A, B and C. Then, the smallest number of smallest ultimate classes into which the population is divided, is 2³ = 8

Let sets A and B have m and n elements, respectively,
Then 2^m - 2ⁿ = 56 (According to question)
implies2ⁿ(2^m-n - 1)
= 8 x 7 = (2)³ x 7 = 2³(2³- l )
On comparing implies n = 3
and m-n = 3
implies m = 6 and n = 3
Number of subsets of A = 2³ + 5⁶ = 64 = 2⁶
Hence,Number of elements of in A = 6

 I. Let l, m, n are parallel lines and R is a relation.
Then l|| l, then R is reflexive,
and l || m and m || l, then R is symmetric.
Also, l || m, m || n => l | n, then R is transitive. Hence, R is an equivalence relation.
II. If x is father of y and y is not father of x then relation is not symmetric, thus relation is not equivalence

Since ,f(- 1) = f(1) = 2³⁵
i.e., Two real number 1 and -1 have the same image.
So, the function is not one-one and let y = (x² + l )³⁵ implies Thus, every real number has no pre-image.
So, the function is not onto.
Hence, the function is neither one-one nor onto.

Since, A ∩ B = φ (given)

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 .

It is clear form the graph that f(x) = e^x,  is one- one but not onto. Since, range ≠ codomain, so f(x) is into.

A = P {1 ,2} = {φ, {1}, {2}, {1 ,2 }}
From above, it is clear that {1 , 2} ε A

Given, R and S are relations on set A.
Then R ε A x A and S ≤ A x A
=> R ∩ C ≤ A x A
implies R ∩ S is also a relation on A.
Reflexivity: Let a be an arbitrary element of A. Then, aA implies (a, a ) R and (a, a) S,
[Since, R and S are reflexive]

implies(a, a) ε R ∩ S
Thus, (a , a ) ε R ∩ S for all a ε A.
So, R ∩ S is a reflexive relation on A.
Symmetry: Let a, b ε A such that (a, b) ε R ∩ S.
Then, (a, b) ε R ∩ S
implies (a, b) ε R and (a, b) ε S
implies (b, a) ε R and (b , a ) ε S
[Since R and S are symmetric]
implies (b, a) ε R ∩ S
Thus, (a, b) ε R ∩ S implies
(b, a) ε R ∩ S for all (a, b ) ε R ∩ S .
So, R ∩ S is symmetric on A.

So, R ∩ S is transitive on A.
Hence, R is an equivalence relation on A.

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.

Given , f(x) = cos x
It is clear from the figure that f(x) is neither one-one nor a onto function.
Since, whenever we drawn a line parallel to x-axis, then it intersects at infinite points to the curve. So, f(x) is not one-one.
And, range o f f ( x ) = [-1, 1]
Codomain of f(x) = R
Range of f(x) ≠ codomain of f(x)
Hence, f(x) is not onto.

n(A) = 115, n(B) = 326

n(A−B) = 47

n(A) = n(A−B) + n(A∩B)

n(A∩B) = n(A) − n(A−B)

∴n(A∩B) =115−47 = 68

∴n(A∪B) = n(A) + n(B) − n(A∩B)

⇒n(A∪B) = 115+326−68

⇒n(A∪B) = 373

Here.
n(A)= 40% of 10000 = 4000
n(B)= 20% of 10000 = 2000
n(C)= 10% of 10000 =1000
n(A ∩ B) = 5% of 10000 = 500
n(B ∩ C) = 3% of 10000 = 300
n(C ∩ A) = 4% of 10000 = 400
= 4000 - [500 + 400 - 200]
 = 4000- 700 - 3300

Let A denote the set ot Americans who like cheese and let B denote the set of Americans who like apples.
Let population of American be 100.
Then, n(A) = 63, n(B) =76