Real Analysis - 5 - Free MCQ Practice Test with solutions, Maths

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.

since, n(A) = 3 and n(B) = 2 Therefore, Number of relations from

= 64

Since, x² + 1 = 0, given x² = -1 
x = ± i
Therefore, x is not real but x is real (given)
Hence, No value of x is possible.

A-P(A)=A,
which is correct as A and P(A) are disjoint sets.

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]

Since, intelligency is not defined for students in a class i. e., not a well defined collection.

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 .

Since, A and B are two sets satisfying A - B = B - A , which is possible only if A = B.

The total number of elements common in (A x B) and (B x A) is n².(by property)

Here,

or A = φ
Here, B = φ satisfy the given condition (i)

∵ Equation of normal at (x, y) is
x − y = dx/dy (X − x)
Put, y = 0
Then, X = x + y dy/dx
Given, y² = 2x X
⇒ y² = 2x ( x + y dy/dx)
⇒ dy/dx = (y² − 2x²)/(2xy) = ((y/x)² − 2)/ (2y/x)
Put y = vx, we get dy/dx = v + x dv/dx
Then, v + x dv/dx = v²−2/2v
On integrating both sides, we get
ln (2 + v² ) + ln|x| = ln c
⇒ ln (|x|(2 + v² )) = ln c
⇒ |x| ( 2 + y²/x²) = c
∵ It passes through (2, 1), then 2 (2 + 1/4 ) = c
⇒ c = 9/2
Then, |x| ( 2 + y²/x² ) = 9/2
⇒ 2x² + y² = 9/2 |x|
⇒ 4x² + 2y² = 9|x|

Since, y = e^x, y = e^-x will meet,
when e^x = e^-x
implies e^2x = 1,
So, x = 0,y = 1
Hence, A and B meet on (0,1),
Hence, A ∩ B = φ

Given that, n(U) = 700, n(A) = 200, n(B) = 300 and n(A ∩ B)=100
We know that,


= 700-400 = 300