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Sets, Relations and Functions Test - 6

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Sets, Relations and Functions Test - 6
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  • Question 1
    2 / -0.83

    R = {(1, 1), (2, 2), (1, 2), (2, 1), (2, 3)} be a relation on A = {1, 2, 3}, then R is

    Solution

    A relation R on a non empty set A is said to be reflexive if fx Rx for all x  ∈ R, Therefore , R is not reflexive.
    A relation R on a non empty set A is said to be symmetric if fx Ry ⇔yRx, for all x, y  ∈R Therefore, R is not symmetric.
    A relation R on a non empty set A is said to be antisymmetric if fx Ry and y Rx ⇒x = y, for all x, y  ∈R. Therefore, R is not antisymmetric.

  • Question 2
    2 / -0.83

    If R is a relation from a set A to a set B and S is a relation from B to C, then the relation  S   R.

    Solution

    If R is a relation from A →B and S is a relation fromB →C, then  S  ∘  R is a composite function from A to C.

  • Question 3
    2 / -0.83

    Let A be the set of all students of a boys school. The relation R in A given by R = {(a, b) : a is sister of b} is

    Solution

    Because, the relation is defined over the set A which is  the set of all students of a boys school.

  • Question 4
    2 / -0.83

    The domain of definition of the function  

    Solution

    y is defined  If -x  ≥ 0, i.e.

  • Question 5
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    Solution

    By definition, The Signum function =  

  • Question 6
    2 / -0.83

    Let A = {1, 2, 3}. Which of the following is not an equivalence relation on A ?

    Solution

    A relation R on a non empty set A is said to be reflexive iff xRx for all x  ∈ R . A relation R on a non empty set A is said to be symmetric iff xRy ⇔yRx, for all x , y  ∈R .
    A relation R on a non empty set A is said to be transitive iff xRy and yRz ⇒xRz, for all x  ∈ R. An equivalence relation satisfies all these three properties.
    None of the given relations satisfies all three properties of equivalence relation.

  • Question 7
    2 / -0.83

    The binary operation * defined on the set of integers as  a ∗b = |a −b|-1 is:

    Solution

    Here * is commutative as b*a = |b −a|−1 = |a −b|−1 = a ∗b.
    Because ,|−x| = |x| for  all  x ∈R.

  • Question 8
    2 / -0.83

    Let T be the set of all triangles in a plane with R a relation in T given by  R  = {(T1 , T2 ) : T1  is congruent to  T2 }. Then R is

    Solution

    Let T be the set of all triangles in a plane with R a relation in T given by R = {(T1 , T2 ): T1 is congruent to T2 }. (T1 T2 ) ∈ R iff  T1 is congruent to  T2
    Reflexivity :T1 ≅T1 ⇒(T1 T1 ) ∈R . Symmetry :(T1 , T2 )∈R  ⇒T1 ≅T2 ⇒T2 ≅T1 ⇒(T2 , T1 ) ∈R
    Transitivity :(T1 ,T1 ) ∈R and (T2 , T3 ) ∈R  ⇒T1 ≅T2 and T2 ≅T3 ⇒T1 ≅T3 ⇒(T1 , T3 ) ∈R .
    Therefore, R is an equivalence relation on T.

  • Question 9
    2 / -0.83

    The range of the function   

    Solution

    The given function is defined as Signum function. i.e.


    domain is R and Range is {-1 , 0 , 1}.

  • Question 10
    2 / -0.83

    The function  f  (x) = x2 + sin  x is

    Solution

    For even function : f(-x) = f(x) , therefore, f(- x) = (− x)2 + sin  (− x)
    = x2 −sin  x  ≠ f(x).
    For an odd function : f(-x) = - f(x) , therefore, f(-x) = (− x)2 + sin  (− x)
    = x2 − sin  x
     ≠− f(x).
    Therefore f(x) is neither even nor odd.

  • Question 11
    2 / -0.83

    If A is a finite set containing n distinct elements, then the number of relations on A is equal to

    Solution

  • Question 12
    2 / -0.83

    R is a relation from { 11, 12, 13} to {8, 10, 12} defined by y = x –3. The relation  R−1

    Solution

    R is a relation from {11, 12, 13} to {8, 10, 12} defined by y = x –3. The relation is given by x = y + 3,from{8, 10, 12} to {11, 12, 13} ⇒ relation = {(8,11),(10,13)}.

  • Question 13
    2 / -0.83

     If A = {(1, 2, 3}, then the relation R = {(2, 3)} in A is

    Solution

    A = {1,2,3}
    B = {(2,3)} is not reflexive or symmetric on A but it is transitive
     ∵if (a,b) exists but (b,c) does not exist then (a,c) does not need to exist and the relation is still transitive.

  • Question 14
    2 / -0.83

    The range of the function f(x) = [sin x] is

    Solution

    The only possible integral values of sin x are {-1 ,0, 1}.

  • Question 15
    2 / -0.83

    The period of the function  f(x) = sin2 x  + tan  x is

    Solution

    We have the function f(x) = sin2 x  + tan x, therefore, f(π+x)
    = {sin2 (π+x)+tan(π+x)}
    = {sin2 x+tanx} = f(x).
    This implies that period of the function f(x) is π.

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