Sets And Functions Multiple Choice Questions & Answers

Approved & Edited by ProProfs Editorial Team
The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes.
Learn about Our Editorial Process
| By Themes
T
Themes
Community Contributor
Quizzes Created: 416 | Total Attempts: 700,873
Questions: 10 | Attempts: 148

SettingsSettingsSettings
Sets And Functions Multiple Choice Questions & Answers - Quiz

It's time to unleash your mathematical genius? Take our online quiz and solve questions about sets and functions to test yourself and prepare for an upcoming exam.


Questions and Answers
  • 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  

    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.

    Rate this question:

  • 2. 

    Let A be a closed subset of RA≠∅  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

    Correct Answer
    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.

    Rate this question:

  • 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→∞ 

    Correct Answer
    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.

    Rate this question:

  • 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

    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.

    Rate this question:

  • 5. 

    Which is compact in R n ?

    • 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 }

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

    Rate this question:

  • 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

    Correct Answer
    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."

    Rate this question:

  • 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

    Correct Answer
    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.

    Rate this question:

  • 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 }

    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.

    Rate this question:

  • 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

    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.

    Rate this question:

  • 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

    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.

    Rate this question:

Back to Top Back to top
Advertisement
×

Wait!
Here's an interesting quiz for you.

We have other quizzes matching your interest.