Real Analysis- 7 - Free MCQ Practice Test with solutions, Maths

Let the original set contains (2n + 1) elements, then subset of this set containing more than n elements, i.e., subsets containing (n + 1) elements, (n + 2) elements,....... (2n + 1) elements.
Hence, Required number of subsets

Since, n | n for all n ∈ N, therefore R is reflexive.
Since, 2 | 6 but 6 | 2 , therefore R is not symmetric. Let _nR_m

Since, the image of the each element of A is belonging to only one element in B. But d has no pre-image in A Therefore, function is one-one but not onto.
So, its inverse does not exist.

We have, n(M alone) = n(M) - n(M ∩ C) -n (M ∩ P ) + n(M ∩ P ∩ C)
=100 - 28 - 30 + 18 = 60

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, 

A - B = A - (A ∩ B) is correct.
A = (A ∩ B) ∩ (A - B) is correct.

III is false
Therefore, I and II are True.

Since, n(E) = 75, n(M)
= 60 and n(E ∩ M) = 45
Hence, n(E∪M) = n(E) + n(M) - n(E ∩M)
= 75 + 60 - 45 = 90
Required number of students = 90 - 45 = 45

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.

μ =the set of all triangles
P =the set of all isosceles triangles
Q =the set of all equilateral triangles
R =the set of all the right angled triangles
Since, P ∩ Q represents the set of isosceles triangles and R - P represents the set of non-isosceles right angled triangles.

So, 4ⁿ - 3k - 1 is a multiples of 9 for all n ∈ N
Hence, X contains elements, which are multiples of 9 and clearly Y contains all multiples of 9.

(a) Since, (1 ,1 ); (2 ,2 ); (3 ,3 ) ∈ R
therefore R is reflexive.
(1, 2) ∈ R but (2, 1) ∉ R, therefore R is not symmetric. It can be easily seen that R is transitive.
(c) Here,R{(l, 3), (2,2); (3,2)}, &= {(2,1); (3,2); (3,2); (2 ,3 )}. Then RoS = {(2,3), (3,2); (2,2)}

Since, 1 + a • a = 1 + a² > 0,, Therefore, (a, a) ∈ R
So, R is reflexive.
Also, (a, b) ∈ R => 1 + ab > 0 => 1 + ba > 0
=> (b, a) ∈ R,
Hence, R is symmetric.
Since, (a, b) ∈ R and (b, c) ∈ R need
not imply (a, c) ∈ R.
Hence, R is not transitive

For reflexive _aR_a implies a divides a
So, R is reflexive.
For symmetric
_aR_b impliesa divides b
_bR_a impliesa divides a
which may not be possible.
So, R is not symmetric.
For transitive
_aR_b impliesa divides b implies b = ka _bR_c impliesb divides c implies c = lb Now, c = Ika
or a divides c implies aRc
or _bR_c implies _cR_a
Thus, R is transitive.

A = Set of all values (x, y) :x² + y² = 25 = 5²

Clearly, A ∩ B consists of four points.

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}

Since, R is defined over the set of non-negative integers, then R = {(6,0), (0,6)} because the option (b) and (c) does not have integers in all pairs. Also in option (a) and (c), the most appropriate option is (c).

 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 

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.

From the figure, it is clear that