Suppose f : R→R is a function that satisfies |f(x) -f(y)| ≤ |x-y| β, β>0 then

If β>0 then f is uniform continuous

If β=1 then f is differentiable

If β>1 then f is constant function

Let G 1 and G 2 be two subsets of R 2 and f: R 2 →R 2 be a function, then

f -1 (G 1 ∪ G 2 )= f -1 ( G 1 )∪ f -1 ( G 2 )

f -1 ( G 1) c = (f -1 ( G 1 )) c

f (G 1 ∩ G 2) =f (G 1 ) ∩ f ( G 2 )

If G1 is open and G2 is closed then G1+ G2 = {x+y : x∈ G1,y∈ G2 is neither open nor closed

Which is compact in R n ?

{x1,x2,x3,……….xn : xi<1, 1≤i≤n}

{x 1 , x 2 , x 3 ,………. x n : x 1 + x 2 + x 3 ……. x n =0}

{x 1 , x 2 , x 3 ,………. x n : x i ≥0, 1≤i≤n}

{x 1 , x 2 , x 3 ,………. x n : 1≤x i ≤2, 1≤i≤n }

Sets And Functions Multiple Choice Questions & Answers

2.

Let f : X → X such that f(f (x )= x for all x∈ X then

F is one- to- one and onto
F is one- to- one but not onto
F is onto but not one-to -one
F need not be either one- to -one or onto

Correct Answer

A. F is one- to- one and onto

Explanation

If f(f(x)) = x for all x in X, it means that for every element x in X, applying the function f twice will give back x. This implies that f is both one-to-one and onto.

One-to-one means that every element in the domain corresponds to a unique element in the range, and onto means that every element in the range is mapped to by at least one element in the domain. In this case, since applying f twice gives back the original element, it shows that f is one-to-one. And since f(f(x)) = x for all x, it means that every element in the range is mapped to by at least one element in the domain, making f onto as well.

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3.

Which of the following is/are true?

(1+ 1 /n) ^{n +1} → e as n →∞
(1+ 1 /n+1 )ⁿ →e as n→∞
(1+ 1 /n )ⁿ²→e as n→∞
(1+ 1 /n ²) ^ⁿ →e as n→∞

Correct Answer

A. (1+ 1 /n+1 )ⁿ →e as n→∞

Explanation

As n approaches infinity, the expression (1+ 1/n+1 )n approaches the mathematical constant e. This can be proved using the limit definition of e. By taking the limit as n approaches infinity, we can see that the expression converges to e.

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4.

Let G _₁ and G _₂ be two subsets of R ² and f: R ^² →R ^² be a function, then

F ^-1 (G ₁ ∪ G ₂)= f ^-1 ( G ₁ )∪ f ^-1 ( G ₂)
F -1 ( G 1) c = (f -1 ( G 1 )) c
F (G 1 ∩ G 2) =f (G 1 ) ∩ f ( G 2 )
If G₁is open and G₂is closed then G₁+ G₂= {x+y : x∈ G₁,y∈ G₂is neither open nor closed

Correct Answer

A. F ^-1 (G ₁ ∪ G ₂)= f ^-1 ( G ₁ )∪ f ^-1 ( G ₂)

Explanation

The correct answer states that the preimage of the union of two subsets, G1 and G2, under the function f is equal to the union of the preimages of G1 and G2. This means that if we apply the function f to the elements in the union of G1 and G2, and then take the inverse image of that result, it will be the same as taking the inverse images of G1 and G2 separately and then taking their union. This property holds for functions between sets, and it allows us to analyze the behavior of the function on subsets of the domain.

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5.

Which is compact in R ^ⁿ ?

{x_₁,x_₂,x_₃,……….x_{_n} : x_{_i}<1, 1≤i≤n}
{x _₁ , x _₂ , x _₃ ,………. x _{_n} : x _₁ + x _₂ + x _₃……. x _{_n} =0}
{x _₁ , x _₂ , x _₃ ,………. x _{_n} : x _{_i}≥0, 1≤i≤n}
{x _₁ , x _₂ , x _₃ ,………. x _{_n} : 1≤x _{_i} ≤2, 1≤i≤n }

Correct Answer

A. {x_₁,x_₂,x_₃,……….x_{_n} : x_{_i}<1, 1≤i≤n}

Explanation

The correct answer is {x1,x2,x3,...,xn : xi

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6.

Let X⊂R be an infinite countable bounded subset of R which of the statements is true

X cannot be compact
X contains an interior point
X may be closed
Closure of X is countable

Correct Answer

A. X cannot be compact

Explanation

An infinite countable bounded subset of R cannot be compact because compactness requires that every open cover of the set has a finite subcover. Since X is infinite, it cannot be covered by a finite number of open sets, making it not compact.

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7.

Let I={1}∪{2} for x∈R let ϕ (x) =dist {x,I} =Inf{ |x-y |:y∈I} then ∅ is

Discontinuous somewhere
Continuous on R but differentiable only at x=1
Continuous on R but differentiable only at x=1,2
Continuous on R but not differentiable only at x=1, 3/ 2 ,2

Correct Answer

A. Continuous on R but not differentiable only at x=1, 3/ 2 ,2

Explanation

The function ϕ(x) is defined as the distance between x and the set I={1, 2}. This means that ϕ(x) will be 0 if x is in I, and the absolute difference between x and the closest element in I if x is not in I. Since I={1, 2}, the function ϕ(x) will be 0 when x=1 or x=2. However, when x=3/2, the distance between 3/2 and the set I is 1/2, which is not 0. Therefore, ϕ(x) is not differentiable at x=3/2. Similarly, ϕ(x) is not differentiable at x=1 and x=2 because the distance between x and I is not 0 at these points. However, ϕ(x) is continuous on R because it is defined for all real numbers. Hence, the correct answer is "Continuous on R but not differentiable only at x=1, 3/2, 2."

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8.

Which of the following subsets of R^²is /are convex

{(x,y): |x|≤5 , |y|≤10}
{(x,y) : x ^² + y ^² =1}
{(x,y) : y ≥ x ^² }
{(x,y) : y≤ x ^² }

Correct Answer

A. {(x,y): |x|≤5 , |y|≤10}

Explanation

The subset {(x,y): |x|≤5 , |y|≤10} is convex because it satisfies the definition of convexity. A set is convex if for any two points within the set, the line segment connecting them is also contained within the set. In this case, for any two points (x1, y1) and (x2, y2) where |x1|≤5 , |y1|≤10 and |x2|≤5 , |y2|≤10, the line segment connecting them will also have points with |x|≤5 and |y|≤10. Therefore, the subset is convex.

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9.

Consider the set X={(-∞,0)∪ 1/n, n ∈ N}⊂R with the subspace topology. Then

0 is an isolated point.
(–2, 0] is an open set
0 is a limit point of the subset {1 /n ,n∈N}
(–2, 0) is an open set

Correct Answer

A. 0 is an isolated point.

Explanation

0 is an isolated point in the set X because it has a neighborhood that does not contain any other point of the set. Specifically, the neighborhood (-∞, 0) does not contain any other point besides 0. Therefore, 0 is isolated from the other points in the set.

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10.

Let A be a closed subset of R, A≠∅ and A≠R . Then A is

The closure of the interior of A
A countable set
A compact set
Not open

Correct Answer

A. Not open

Explanation

If A is a closed subset of R and A is not equal to the empty set or R, then A cannot be open. This is because if A is open, then its complement, which is R - A, would be closed. However, since A is not equal to R, its complement is not empty, and therefore A cannot be open.

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1

Sets And Functions Multiple Choice Questions & Answers

Suppose f : R→R is a function that satisfies |f(x) -f(y)| ≤ |x-y| ^{^β}, β>0 then

Quiz Preview

Let f : X → X such that f(f (x )= x for all x∈ X then

Which of the following is/are true?

Let G _₁ and G _₂ be two subsets of R ² and f: R ^² →R ^² be a function, then

Which is compact in R ^ⁿ ?

Let X⊂R be an infinite countable bounded subset of R which of the statements is true

Let I={1}∪{2} for x∈R let ϕ (x) =dist {x,I} =Inf{ |x-y |:y∈I} then ∅ is

Which of the following subsets of R^²is /are convex

Consider the set X={(-∞,0)∪ 1/n, n ∈ N}⊂R with the subspace topology. Then

Let A be a closed subset of R, A≠∅ and A≠R . Then A is

Sets And Functions Multiple Choice Questions & Answers

Suppose f : R→R is a function that satisfies |f(x) -f(y)| ≤ |x-y| β, β>0 then

Quiz Preview

Let f : X → X such that f(f (x )= x for all x∈ X then

Which of the following is/are true?

Let G 1 and G 2 be two subsets of R 2 and f: R 2 →R 2 be a function, then

Which is compact in R n ?

Let X⊂R be an infinite countable bounded subset of R which of the statements is true

Let I={1}∪{2} for x∈R let ϕ (x) =dist {x,I} =Inf{ |x-y |:y∈I} then ∅ is

Which of the following subsets of R2 is /are convex

Consider the set X={(-∞,0)∪ 1/n, n ∈ N}⊂R with the subspace topology. Then

Let A be a closed subset of R, A≠∅ and A≠R . Then A is

Suppose f : R→R is a function that satisfies |f(x) -f(y)| ≤ |x-y| ^{^β}, β>0 then

Let G _₁ and G _₂ be two subsets of R ² and f: R ^² →R ^² be a function, then

Which is compact in R ^ⁿ ?

Which of the following subsets of R^²is /are convex