Option 2 : 8

**Concept:**

The **power set** (or **powerset**) of a **Set** A is defined as the **set** of all subsets of the **Set** A including the **Set** itself and the null or empty **set**.

**Calculations:**

Given, A = {x, y, z}.

The **power set** (or **powerset**) of a **Set** A is defined as the **set** of all subsets of the **Set** A including the **Set** itself and the null or empty **set**.

Powerset of A = {\(\rm \phi\),x, y, z, {x, y}, {y, z}, {x, z},{x, y, z}}.

Hence, the number of subsets in powerset of A is 8.

Suppose A and B are two sets in the same broad set. Then, A - B =

Option 3 : Only A

The given two sets - A and B, can be shown as follows:

In the Venn Diagram:

x = Elements which are in set A only,

y = Elements which are in both - set A and set B,

z = Elements which are in set B only.

We need to obtain A – B

From the Venn Diagram, it is clear that:

⇒ A – B = (x + y) – y = x

**∴ (A – B) = only A **

Option 2 : 23

__Calculation:__

Given: We have to form all distinct numbers of the form \(\frac{{\rm{p}}}{{\rm{q}}}\)

As 1/2, 2/4, 3/6 are same and 1/1, 2/2, 3/3...6/6 are same.

We will find only co-prime numbers.

p |
q |
Total elements in S |

1 |
{1, 2, 3, 4, 5, 6} |
6 |

2 |
{1, 3, 5} |
3 |

3 |
{1, 2, 4, 5} |
4 |

4 |
{1, 3, 5} |
3 |

5 |
{1, 2, 3, 4, 6} |
5 |

6 |
{1, 5} |
2 |

∴ Cardinality of set S = 6 + 3 + 4 + 3 + 5 + 2 = 23

In class of 105 students out of three subjects Maths, Physics, Chemistry each student studies at least one subject. In Maths 47, in Physics 50, and in Chemistry 52 students studies, 16 in Maths and Physics, 17 in Maths and Chemistry and 16 in Physics and Chemistry students both subjects.

What will be the number of those students who study only two subjects?

Option 4 : 34

**Given:**

In a class of 105 students out of three subjects Maths, Physics, Chemistry each student studies at least one subject. In Maths 47, in Physics 50, and in Chemistry 52 students studies, 16 in Maths and Physics, 17 in Maths and Chemistry and 16 in Physics and Chemistry students both subjects.

**Calculations:**

Total number of students M ⋃ P ⋃ C = 105

Only mathematics students M = 47

Only physics students P = 50

Only chemistry students C = 52

Mathematics and physic students M ⋂ P = 16

Mathematics and chemistry students M ⋂ C = 17

Physics and chemistry students P ⋂ C = 16

As we know that, M ⋃ P ⋃ C = M + P + C – (M ⋂ P) – (M ⋂ C) – (C ⋂ P) + (M ⋂ P ⋂ C)

105 = 47 + 50 + 52 – 16 – 17 – 16 + (M ⋂ P ⋂ C)

(M ⋂ P ⋂ C) = 105 – 100

(M ⋂ P ⋂ C) = 5

Total students learning two subjects = 11 + 12 + 11 = 34

**∴ The correct choice will be option 4.**

Consider the following statements:

1. The null set is a subset of every set.

2. Every set is a subset of itself.

3. If a set has 10 elements, then its power set will have 1024 elements.

Which of the above statements are correct?

Option 4 : 1, 2 and 3

**Concept:**

1. The null set is a subset of every set. (ϕ ⊆ A)

2. Every set is a subset of itself. (A ⊆ A)

3. The number of subsets of a set with n elements is 2^{n}.

**Explanation:**

1. The null set is a subset of every set - The intersection of two sets is a subset of each of the original sets.

So if {} is the empty set and A is any set then {} intersect A is {} which means {} is a subset of A and {} is a subset of {}.

You can prove it by contradiction. Let's say that you have the empty set {} and a set A.

2. Every set is a subset of itself. (A ⊆ A)

3. If n = 10 then 2^{10} = 1024

So, all three statements are true.

Option 2 : 4

**Concept:**

**The number of elements in the power set** of any set A is **2**^{n} where n is the number of elements of the set A.

**Calculation:**

Given, set S = {2, {1, 4}},

Number of elements in set S = 2

\(\therefore \)The number of element in the power set P(S) = 2^{2} = 4

Option 3 : 45

__Concept:__

**Combination:** Selecting r objects from given n objects.

- The number of selections of r objects from the given n objects is denoted by \({{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}}{\rm{\;}}\)
- \({{\rm{\;}}^{\rm{n}}}{{\rm{C}}_{\rm{r}}}{\rm{}} = {\rm{}}\frac{{{\rm{n}}!}}{{{\rm{r}}!\left( {{\rm{n\;}} - {\rm{\;r}}} \right)!}}\)

**Note:** Use combinations if a problem calls for the number of ways of selecting objects.

__Calculation:__

Number of elements in A = 10

Number of subsets of A containing exactly two elements = Number of ways we can select 2 elements from 10 elements

⇒ Number of ways we can select 2 elements from 10 elements = ^{10}C_{2} = 45

∴ Number of subsets of A containing exactly two elements = 45

Option 4 : (A ∩ B) \ C

__Concept:__

**Null set**is the set that does not contain anything.- In mathematical sets, the null set, also called the empty set,
- It is symbolized or { }.

__Calculation:__

Given:

A = {x : x is a multiple of 3}

∴ A = {3, 6, 9, 12, 15, 18, 24…}

B = {x : x is a multiple of 4}

∴ B = {4, 8, 12, 16, 20, 24, 28, 32 ...}

C = {x : x is a multiple of 12}

∴ C = {12, 24, 36, 48, 60, 72, 84, 96 ...}

Now,

A∩B = {12, 24 ...} = C

∴ (A∩B) \ C = { } = Null set

Option 3 : A

**Calculation:**

According to the question,

A is a non-empty subset = A ⊆ C

B is a non-empty subset = B ⊆ C

By Distributive property,

A ∪ (A ∩ B) = (A ⋃ A) ∩ (A ⋃ B) = A

Option 4 : P = Q.

__Concept:__

- \(\rm \tan\theta=\frac{\sin\theta}{\cos\theta}\).

__Calculation:__

Consider the relation in the set P = {θ∶ sin θ - cos θ = √2 cos θ}.

sin θ - cos θ = √2 cos θ

Dividing both sides by cos θ, we get:

tan θ - 1 = √2

⇒ tan θ = √2 + 1

∴ P = {θ: tan θ = √2 + 1} ... (1)

Consider the relation in the set Q = {θ∶ sin θ + cos θ = √2 sin θ}.

sin θ + cos θ = √2 sin θ

Dividing both sides by cos θ, we get:

tan θ + 1 = √2 tan θ

⇒ \(\rm \tan\theta = \frac{1}{\sqrt2 - 1}=\frac{1}{\sqrt2 - 1}\times\frac{\sqrt2+1}{\sqrt2 + 1}=\frac{\sqrt2+1}{2 - 1}=\sqrt2+1\)

∴ Q = {θ: tan θ = √2 + 1} ... (2)

Comparing equations (1) and (2), we have:

**P = Q**.

Option 1 : There exist at least two natural numbers which are prime to each other

__ Concept__:

**The Pigeonhole Principle**: Let there be n boxes and (n + 1) objects. Then, under any assignment of objects to the boxes, there will always be a box with more than one object in it. This can be reworded as, if m pigeons occupy n pigeonholes, where m > n, then there is at least one pigeonhole with two or more pigeons in it.

__ Calculation__:

We divide the set into n classes {1, 2}, {3, 4},......{2n - 1, 2n}.

By the pigeonhole principle, given n +1 elements at least two of them will be in the same case {2k - 1, 2k} (1 ≤ k ≤ n). But 2k - 1 and 2k are relatively prime because their difference is 1.

Option 2 : 8

**Concept:**

The **power set** (or **powerset**) of a **Set** A is defined as the **set** of all subsets of the **Set** A including the **Set** itself and the null or empty **set**.

**Calculations:**

Given, A = {x, y, z}.

**power set** (or **powerset**) of a **Set** A is defined as the **set** of all subsets of the **Set** A including the **Set** itself and the null or empty **set**.

Powerset of A = {\(\rm \phi\),x, y, z, {x, y}, {y, z}, {x, z},{x, y, z}}.

Hence, the number of subsets in powerset of A is 8.

Option 4 : -5 < x < 3/2

**GIVEN:**

x = {n; 2n^{2} + 7n - 15 < 0}

**CONCEPT:**

Concept of sets and relation.

**FORMULA USED:**

No formula

**CALCULATION:**

x = {n; 2n^{2} + 7n - 15 < 0}

⇒ 2n^{2} + 7n - 15 < 0

⇒ 2n^{2} + 10n - 3n - 15 < 0

⇒ 2n(n + 5) - 3(n + 5) < 0

⇒ (n + 5)(2n - 3) < 0

⇒ n = (-5, 3/2)

-5 < x < 3/2Option 3 : A

**Given:**

The given set is A.

**Formula:**

If A is a set and U is a universal set.

Then complement of set A = U – A.

**Calculation:**

Let us consider a set A and a universal set U.

If A = {1, 4, 7, 9) and U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

Then the complement of set A = U – A

Let A’ represent the complement of set A

A’ = U – A

⇒ A’ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} - {1, 4, 7, 9)

⇒ A’ = {2, 3, 5, 6, 8, 10, 11}

⇒ (A’)’ = U – A’

⇒ (A’)’ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} - {2, 3, 5, 6, 8, 10, 11}

⇒ (A’)’ = {1, 4, 7, 9}

We can say that complement of complement of a set A equals to the set A.

**∴**** Option (3) is the correct answer.**

Option 2 : Closed set

**Concept: **

The open set has no boundary whether the closed set has a boundary.

If A is an open set and B is a closed set then,

**B - A = **closed set - open set** **

**⇒ closed set**

Option 4 : 4

**Given: **

A = {BELOW}, B = {WOOL}

**Concept used:**

If a set has “n” elements, then the number of subset of the given set is 2^{n}

**Calculation:**

⇒ C = A – B

⇒ C = {B, E} taking out the common letters.

The subsets of C = 2^{2} = 4

**∴ The number of subsets of C is 4**

Option 2 : ϕ, {0}

**Concept:**

A set with one element has 2 subsets i.e. The null set and itself

∴ {0} = ϕ, {0}

Option 1 : ϕ

**Given:**

A Δ B = A

**Concept:**

The Δ in set theory is the symmetric difference of two sets.

A Δ B = (A – B) ∪ (B – A)

**Calculation:**

A Δ B = A

\( \Rightarrow \left( {{\rm{A}} - {\rm{B}}} \right) \cup \left( {{\rm{B}} - {\rm{A}}} \right) = {\rm{A}}\)

\( \Rightarrow \left( {{\rm{A}} \cap {\rm{B'}}} \right) \cup \left( {{\rm{B}} \cap {\rm{A'}}} \right) = {\rm{A}}\)

\( \Rightarrow \left( {{\rm{A}} \cup {\rm{B}} \cap {\rm{A'}}} \right) \cap \left( {{\rm{B'}} \cup {\rm{B}} \cap {\rm{A'}}} \right) = {\rm{A}}\)

\( \Rightarrow \left[ {\left( {{\rm{A}} \cap {\rm{A'}}} \right) \cup {\rm{B}}} \right] \cap \left[ {\left( {{\rm{B'}} \cup {\rm{B}}} \right) \cap {\rm{A'}}} \right] = {\rm{A}}\)

\( \Rightarrow \left[ {\emptyset \cup {\rm{B}}} \right] \cap \left[ {{\rm{U}} \cap {\rm{A'}}} \right] = {\rm{A}}\)

\( \Rightarrow {\rm{B}} \cap {\rm{A'}} = {\rm{A}}\)

\( \Rightarrow {\rm{B}} - {\rm{A}} = {\rm{A}}\)

\( \Rightarrow {\rm{A}} \cap {\rm{B}} = \emptyset \)

\(\therefore {\bf{A}} \cap {\bf{B}} = \emptyset \)

Option 1 : Cardinality of C.

__Concept:__

- The
**cardinality**of a set A is the measure of the "number of elements" of the set. It is denoted by n(A).

For example, a set containing 3 elements has a cardinality of 3. - If A ⊂ B, then A ∪ B = B.

__Calculations:__

Since, A ⊂ B and B ⊂ C, therefore A ∪ B ∪ C = C.

⇒ n(A ∪ B ∪ C) = n(C).

∴ The cardinality (number of elements) of A ∪ B ∪ C is equal to **cardinality (number of elements) of C**.

Option 1 : {x : |x| < 1, x ∈ N}

__ CONCEPT__:

**Null Set**: A set which does not contain any element is called a null set. It is denoted by ϕ or {}.

__ CALCULATION__:

**Option A**: {x : |x| < 1, x ∈ N}

As we know that there is no x ∈ N such that |x| < 1

So, {x : |x| < 1, x ∈ N} is a null set

**Option B**: {x : |x| = 5, x ∈ N}

As we know that, |5| = 5 and 5 ∈ N

⇒ 5 ∈ {x : |x| = 5, x ∈ N}

So, {x : |x| = 5, x ∈ N} is not a null set.

**Option C**: {x : x2 + 2x + 1 = 0, x ∈ R}

⇒ x2 + 2x + 1 = 0

⇒ x = - 1 ∈ R

⇒ - 1 ∈ {x : x2 + 2x + 1 = 0, x ∈ R}

So, {x : x2 + 2x + 1 = 0, x ∈ R} is not a null set.

**Option D**: {x : x2 = 1, x ∈ Z}

⇒ x^{2} = 1

⇒ x = ± 1 ∈ Z

⇒ ± 1 ∈ {x : x2 = 1, x ∈ Z}

So, {x : x2 = 1, x ∈ Z} is not a null set.

Hence, **option A** is the correct answer.