Test: Sets- 1 - CA Foundation MCQ

- A set with n elements can have subsets of varying sizes, from the empty set to the full set itself.
- For each element in the set, there are two choices: include it in a subset or not.
- Therefore, each element's inclusion or exclusion doubles the number of possible subsets.
- Mathematically, this results in 2^n  total subsets.
- This is why the correct answer is 2^n, which corresponds to option A.

Since the value of x has to be greater than 0 but less than 5, the correct answer is Option B: {1, 2, 3, 4}.

CALCULATION:

Given: A = {1, 2, 3, 4, 5, 7, 8, 9} and B = {2, 4, 6, 7, 9}

As we know that, A ∩ B = {x : x ∈ A and x ∈ B}

⇒ A ∩ B = {2, 4, 7, 9}

As we can see that,

The number of elements present in A ∩ B = 4

i.e n(A ∩ B) = 4

As we know that;

If A is a non-empty set such that n(A) = m then

The numbers of proper subsets of A are given by 2m - 1.

So, The number of proper subsets of A ∩  B = 2⁴ - 1 = 15

Hence, the correct option is 2

Explanation:

Intersection of Sets:
- The intersection of two sets P and Q, denoted by P ∩ Q, is the set of elements that are common to both sets.
- In this case, P = {1, 2, 3, 5, 7} and Q = {1, 3, 6, 10, 15}.
- To find the intersection, we look for elements that are present in both sets.
- The common elements between P and Q are 1 and 3.

Calculation:
- n(P ∩ Q) = Number of elements in the intersection of sets P and Q.
- n(P ∩ Q) = 2 (as there are 2 common elements: 1 and 3).

Therefore, the correct answer is A: 2.

To find n(S-Q), which is the number of elements in the set difference S-Q:

- Identify Q: Q = {1, 3, 6, 10, 15}
- Identify Universal Set S: S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15} 
- Calculate ( S-Q ): Remove elements of Q from S.
- Result: S-Q = {2, 4, 5, 7, 8, 9, 11, 12, 13, 14}
- Count elements in (S-Q): n(S-Q) = 10

Thus, the correct answer is B: 10.

- The set {2^x | x is any positive rational number} is an infinite set.
- Rational numbers include both whole numbers and fractions.
- As x varies over all positive rationals, 2^x takes infinitely many distinct values.
- Therefore, the set is infinite, corresponding to option A.

- Set E: Contains positive even numbers (e.g., 2, 4, 6, 8,...).
- Set O: Contains positive odd numbers (e.g., 1, 3, 5, 7,...).
- Union E ∪ O: Combines all elements from both sets E and O.

The union E ∪ O includes all positive integers since each integer is either even or odd.
- Option B (N): Represents the set of natural numbers, which are all positive integers (1, 2, 3,...).
- Thus, E ∪ O = N, the set of natural numbers.

To determine the relationship between the sets R and E:

• Set R is the set of positive rational numbers. Rational numbers are numbers that can be expressed as the quotient of two integers, with a non-zero denominator.
• Set E is the set of all real numbers. This includes rational numbers, irrational numbers, integers, etc.
• Since every positive rational number is also a real number, R is a subset of E. Thus, R ⊆ E.
• However, not all real numbers are rational (e.g., √2), so E is not a subset of R.

 

Therefore, the correct answer is B: R ⊆ E.

- Natural Numbers (N): Traditionally, the set of natural numbers can be defined as {1, 2, 3,...}.

- Positive Integers (I): This set is also defined as {1, 2, 3,...}.

- Both sets start from 1 and include all the numbers increasing by 1 indefinitely.

- Since both sets contain the exact same elements, they are equal.

- Therefore, the correct answer is Option A: N = I.

Since every equilateral triangle is also an isoceles triangle then E is a subset of I.

If R is the set of isosceles right angled triangles and l is set of isosceles triangles, then R belongs to l.

 (R is a subset of l) 

- The set {n(n+1)/2 : n is a positive integer} represents triangular numbers.
- Each triangular number is generated by summing the first n natural numbers: (1, 3, 6, 10, 15, ...).
- As n increases, new triangular numbers can be formed indefinitely.
- There is no upper limit to the value of n, meaning the process can continue forever.
- Thus, the set is infinite.
- Correct answer: B, an infinite set.

• The union of a set with itself, A ∪ A, always results in the set A itself.


• This is because the union operation includes all elements from both sets and since both sets are identical, the result is just  A.


• Thus, the expression A ∪ A, simplifies to A.


• Therefore, the correct answer is Option A: A


 

- When a set A is a subset of a set E, it means every element of  A is also an element of E.
- The union of two sets, A ∪ E, includes all elements from both sets.
- Since A is a subset of E, A ∪ E will contain all elements of E.
- Thus, A ∪ E = E.
- Therefore, the correct answer is Option B: E.

P = {1, 2, 3, 5, 7}, Q = {1, 3, 6, 10, 15},
P ∩ Q = { 1, 3 }.
The cardinal number is the number of elements of a set.
So The cardinal number of P ∩ Q is 2.

To determine the nature of the set (x, y)| xy = 5, consider the following:

• The set represents a hyperbola, where any x has a unique corresponding y = {5/x}.


• Each pair (x, y) satisfies the relation xy = 5, implying a one-to-one correspondence between x and y.


• This mapping is both injective and bijective, as each x maps to exactly one y and vice versa

 

Thus, the correct answer is Option C: one-one mapping.

The correct answer is B: a function.

• A function is a relation where each input (x-value) has exactly one output (y-value).


• The given set {(x, y), y = x²} is a function because for every x, there is one unique y-value, which is x².


• This satisfies the definition of a function as no x-value is paired with more than one y-value.


 

- A function is a relation where each input (x) is associated with exactly one output (y).
- The set {(x, y) | x < y} represents all pairs where x is less than y.
- For a given x, there can be many possible values of y (e.g., x = 1 could pair with y = 2, 3, 4, etc.).
- Therefore, this set does not satisfy the definition of a function.
- The correct answer is Option A: not a function.

Explanation:

Domain of a Relation:
- The domain of a relation is the set of all possible input values (first elements in the ordered pairs) in the relation.
- In this case, the ordered pairs are (1,7) and (2,6).
- The possible input values are 1 and 2.

Determining the Domain:
- The domain of the relation {(1,7), (2,6)} is the set of all possible input values, which are 1 and 2.
- Therefore, the domain of the relation is {1, 2}.

Conclusion:
- The correct answer is option C: (1, 2), as it represents the set of possible input values for the given relation.