Test: Subsets & Supersets - JEE MCQ

Null set is the subset of every set so Y ⊂ X and Y ⊂ Z.
Since set X is equal to set Z so, Z = X and X = Z.

Set of rational numbers { x: x=p/q where p and q are integers and q≠0}.
Set of real numbers is not a subset of X. Set of natural numbers, whole numbers, integers are subset of X.

Number of elements in a power set = 2ⁿ, where n = number of elements in the set P.
Hence, 2⁴ = 16.

• A subset is a set whose elements are all members of another set.
• Since all the elements of set A are the members of set E. So A is a subset of E.  

(2,4) written in set builder form is:
{x : 2 < x < 4}

If A = Ф that means A does not contain any element i.e., n = 0.
Now, number of elements in a power set is 2ⁿ.
∴, n[P(A)] = 2⁰ = 1
Therefore P(A) contains 1 element.

• The set containing all objects or elements and of which all other sets are subsets is a Universal set.
• Here, option A has all the elements of set P, Q and R. So {1, 2, 3, 4, 5, 6, 7, 9} may be considered as universal set for all the three sets P, Q and R.

Square brackets [a, b] implies that values of a and b should be included in the range of x i.e., a ≤ x ≤ b.

• We have, A^c in X = The set of elements in X which are not in A = 0,1
• {0} ∈ A^c w.r.t. X is false, because {0} is not an element of A^c in X.
• ϕ ∈ A^c in X is false because ϕ is not an element of A^c in X.
• {0} ⊂ A^c in X is correct because the only element of {0} namely 0 also belong to A^c in X.
• 0 ⊂ A^c in X is false because 0 is not a set.

All the elements in set-A are presented in set-B. So "A" is a subset of "B".

• A set A is a proper subset of a set B if A is a subset of B and there is at least one element of B that's not an element of A.
• Thus, the void set is a subset of all sets, and it's a proper subset of every set except itself.

• P contains {0, 1, 2, 3, 4} and Q contain natural numbers which start with 1 and P = {0, 1, 2, 3, 4} Q = {1, 2, 3}
• Here P isn't a subset of Q because all the elements of P are not in Q.

|x| < 1 ⇒ -1 < x < 1
∴ No natural number exists between (-1, 1).

L.C.M of 4 and 6 is 12.

Given, A = {x:x is a multiple of 4}
              = {4,8,12,16,20,…}

and     B = {x:x is a multiple of 6}
              = {6,12,18,24,…}

∴A ∩ B = {12,24,…}
             = {x:x is amultiple of 12}

{1, 2, 3, 4, 5, 6} is a set of 6 elements, so it has 2⁶ = 64 subsets

• Number of proper subsets of a given set = 2^m - 1, where m is the number of elements.
• Here the number of elements is 3. So the number of proper subsets of A = 2³ - 1 = 7.

Integers contain all the natural numbers. So it can be a universal set for natural numbers. In other options, there are only some of the elements of natural numbers.

Number of elements in power set = 2⁷ = 128

Here, A = {1, 2, (3, 4}, 5}

• 1 is an element of the set. Hence 1 ∈ A is a correct statement.
• In this set {3, 4} is treated as a single element of set A. That means {3,4} belongs to the set. Hence the statement given in option C, {3, 4} ∈ A is correct. 
• 3, 4 are not the separate elements of set A. So {{3, 4}} ⊂ A is correct and {3, 4} ⊂ A is the wrong representation.

Hence the correct option is option A.