# Sets And Functions Multiple Choice Questions & Answers

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• 1.

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

• A.

F is one- to- one and onto

• B.

F is one- to- one but not onto

• C.

F is onto but not one-to -one

• D.

F need not be either one- to -one or onto

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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• 2.

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

• A.

The closure of the interior of A

• B.

A countable set

• C.

A compact set

• D.

Not open

D. 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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• 3.

### Which of the following is/are true?

• A.

(1+ 1 /n) n +1 → e   as n →∞

• B.

(1+ 1 /n+1 )n →e  as n→∞

• C.

(1+ 1 /n )n 2 →e  as n→∞

• D.

(1+ 1 /n 2) n →e  as n→∞

B. (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 X⊂R  be an infinite countable bounded subset  of R  which of the statements is true

• A.

X cannot be compact

• B.

X contains an interior point

• C.

X may be closed

• D.

Closure of X is countable

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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• 5.

### Which is compact in Rn ?

• A.

{x1,x2,x3,……….xn   : xi<1,   1≤i≤n}

• B.

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

• C.

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

• D.

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

A. {x1,x2,x3,……….xn   : xi<1,   1≤i≤n}
Explanation
The correct answer is {x1,x2,x3,...,xn : xi

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• 6.

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

• A.

Discontinuous somewhere

• B.

Continuous on R but differentiable only at x=1

• C.

Continuous on R but differentiable only at x=1,2

• D.

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

D. 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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• 7.

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

• A.

If β=1 then f is differentiable

• B.

If β>0 then f is uniform continuous

• C.

If β>1  then f is constant function

• D.

F must be a polynomial

B. If β>0 then f is uniform continuous
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.

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• 8.

### Which of the following subsets of  R2  is /are convex

• A.

{(x,y):  |x|≤5 , |y|≤10}

• B.

{(x,y) :   x 2 + y 2 =1}

• C.

{(x,y) :  y ≥ x 2 }

• D.

{(x,y) :  y≤ x 2 }

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

• A.

0 is an isolated point.

• B.

(–2, 0] is an open set

• C.

0 is a limit point of the subset {1 /n ,n∈N}

• D.

(–2, 0) is an open set

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

• A.

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

• B.

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

• C.

F (G 1 ∩ G 2) =f (G 1 ) ∩ f ( G 2 )

• D.

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

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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