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

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 ?

{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

1) 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

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.

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.

2) 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 )

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.

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.

3) Which is compact in R ^ⁿ ?

{x1,x2,x3,……….xn : xi<1, 1≤i≤n}

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

Explanation

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

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

If β=1 then f is differentiable

If β>0 then f is uniform continuous

If β>1 then f is constant function

F must be a polynomial

The given statement states that if the function f satisfies the condition |f(x) - f(y)| ≤ |x-y|β, where β > 0, then f is uniform continuous. This means that for any two points x and y, the difference between the function values at those points is always less than or equal to the difference between the points themselves, multiplied by β. This condition ensures that as the distance between x and y approaches zero, the difference between f(x) and f(y) also approaches zero. This is a key property of uniform continuity, which guarantees that the function does not have any sudden jumps or discontinuities and is continuous throughout its domain.

Explanation

The given statement states that if the function f satisfies the condition |f(x) - f(y)| ≤ |x-y|β, where β > 0, then f is uniform continuous. This means that for any two points x and y, the difference between the function values at those points is always less than or equal to the difference between the points themselves, multiplied by β. This condition ensures that as the distance between x and y approaches zero, the difference between f(x) and f(y) also approaches zero. This is a key property of uniform continuity, which guarantees that the function does not have any sudden jumps or discontinuities and is continuous throughout its domain.

5) 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

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.

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.

6) 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 )n 2 →e as n→∞

(1+ 1 /n 2) n →e as n→∞

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.

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.

7) 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 }

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.

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.

8) 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

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.

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.

9) 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

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.

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.

10) 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

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."

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."