Subset MCQ Quiz in मल्याळम - Objective Question with Answer for Subset - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Formula Used:

\(\: ^{2n+1}C_0 + \: ^{2n+1}C_1 + \dots \: ^{2n+1}C_{n+1} + \dots+ \: ^{2n+1}C_{2n+1} = 2^{2n+1}\)

\(\: ^{n}C_r = \: ^{n}C_{n-r}\)

Calculation:

There are 4096 subsets of S which contain at most n elements.

By using the above formula,

\(^{2n+1}C_{n+1} = ^{2n+1}C_{n}\) &  \(^{2n+1}C_{n+2} = ^{2n+1}C_{n-1}\)

According to the question,

\(\: ^{2n+1}C_0 + \: ^{2n+1}C_1 + \: ^{2n+1}C_2 .......+\: ^{2n+1}C_n\) = 4096    ---(1)

\(\: ^{2n+1}C_0 + \: ^{2n+1}C_1 + \dots \: ^{2n+1}C_{n+1} + \dots+ \: ^{2n+1}C_{2n+1} = 2^{2n+1}\)

\(2(\: ^{2n+1}C_0 + \: ^{2n+1}C_1 + .......+\: ^{2n+1}C_n + \: ^{2n+1}C_{n}) = 2^{2n+1}\)

\(\: ^{2n+1}C_0 + \: ^{2n+1}C_1 + .......+\: ^{2n+1}C_n + \: ^{2n+1}C_{n} = 2^{2n}\)    ----(2)

From equation (1) & (2),

2²ⁿ = 4096

\(2^{2n} = 2^{12}\)

∴ n = 6

Concept:

• A set is a collection of objects. As an example, the set of even numbers between 1 and 9 is {2, 4, 6, 8}.
• If A is a set, then n(A) denotes the number of objects in the set A.
• For two sets A and B \(n(A\cup B)+n(A\cap B)=n(A)+n(B)\)

​Calculation:

Given,  \(n(A\cup B) = 85\) and \(n(A\cap B)=25\) and n(A) : n(B) : n(C) = 5 : 6 : 7

Now, let us assume that n(A) = 5x and n(B) = 6x

Also, according to the formula, 

\(n(A\cup B)+n(A\cap B)=n(A)+n(B)\Rightarrow 85+25=5x+6x\Rightarrow 11x=110\Rightarrow x=10\)

Therefore, n(A) = 50

Concept:

If A, B and C are subsets of a set X. Then

I. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

II. A ∪ A = A, A ∩ (A ∪ B) = A, A ∪ (A ∩ B) = A and A ∩ A = A

III. (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C)

IV. (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)

Calculation:

⇒ A ∪ (A ∩ B) = (A ∪ A) ∩ (A ∪ B) = A ∩ (A ∪ B) = A       --- (Using property I and II)

So, option 1 is not correct.

⇒ A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B) = A ∪ (A ∩ B) = A       --- (Using property I and II)

So, option 2 is correct.

⇒ (A ∩ B) ∪ C = (A ∪ C) ∩ (B ∪ C)       --- (Using property III)

So, option 3 is correct.

⇒ (A ∪ B) ∩ C = (A ∩ C) ∪ (B ∩ C)       --- (Using property IV)

Concept:

If Set A contains m elements then the number of subsets of A containing exactly n elements = the number of ways we can select n elements from m elements, i.e. ^mC_n.

Calculation:

Statement I: If A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} then number of sub sets of A containing 2 or 3 elements is 45.

Given, A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Set A has 10 elements.

The number of subsets of A containing 2 or 3 elements is ¹⁰C₂ + ¹⁰C₃

 10C2 + 10C3 = \(\frac{10!}{2!8!}+\frac{10!}{3!7!}\)

 = \(\frac{10× 9}{2}+\frac{10× 9× 8}{3× 2} \)

 = 45 + 120 = 165

Statement I is incorrect.

Statement II: If f (x) = cos (log x), then the value of \(f (x) × f (4) − \frac{1}{2} × [f (\frac{x}{4}) + f (4x)] \) is 0.

f (x) = cos (log x)

Let y = f (x) × f (4) − \(\frac{1}{2}\) × [f (x / 4) + f (4x)]

⇒ y = cos (log x) × cos (log 4) − \(\frac{1}{2}\) × [cos log (x/4) + cos (log 4x)]

⇒ y = cos (log x) cos (log 4) − \(\frac{1}{2}\) × [cos (log x − log 4) + cos (log x + log 4)]

⇒ y = cos (log x) cos (log 4) − \(\frac{1}{2}\) × [2 cos (log x) cos (log 4)]

⇒ y = 0

Statement II is correct.

∴ Only Statement II is correct.

Concept:

Number of subsets of a set A having n elements is equal to 2n

Calculation:

Statement I: Number of subsets of a set A having n elements is equal to 2n

By mathematical Induction,

Let P(n): Number of subset of a set containing n distinct elements is 2ⁿ, for all n N.

For n = 1, consider set A = {1}.

So, set of subsets is {{1}, ∅}, which contains 2¹ elements.

So, P(1) is true.

Let us assume that P(n) is true, for some natural number n = k.

P(k): Number of subsets of a set containing k distinct elements is 2^k 

To prove that P(k + 1) is true, we have to show that P(k + 1): The number of subsets of a set containing (k + 1) distinct elements is 2^k+1.

We know that, with the addition of one element in the set, the number of subsets become double.

Number of subsets of a set containing (k+ 1) distinct elements = 2 × 2^k = 2^k+1

So, P(k + 1) is true.

Hence, P(n) is true.

Statement II:  The power set of set A contains 128 elements then number of elements in set A is 7.

If A = {1, 2, 3, ... n}

Number of subsets of A = 2ⁿ

2ⁿ = 128 ( ∵ Power set is the set of all subsets of A)

⇒ 2ⁿ = 2⁷

⇒ n =7

Number of elements in sets A = 7

∴ Both Statements I and II are correct.

Concept:

• For a given set A, a set B is a subset of set A if all elements of set B are also elements of set A. Set A is called the super-set of set B.
• Null set "{}" or "ϕ" is a subset of all sets.
• The number of ways in which r objects can be selected from a group of n distinct objects is given by ⁿC_r = \(\rm \dfrac{n!}{r!(n-r)!}\).
• (a + b)ⁿ = C₀ aⁿ + C₁ a^n–1b + C₂ a^n–2 b² + … C_r a^x–r b^r + … + C_n–1 a b^n–1 + C_n bⁿ, where C₀, C₁ … C_n are the Binomial Coefficients with coefficient of the (r + 1)^th term C_r = ⁿC_r.
• 2n = (1 + 1)ⁿ = C₀ + C₁ + C₂ + C₃ + … + C_n–1 + C_n. (by replacing a = 1 and b = 1 in the above Binomial Expansion).
• 0! = 1.

Calculation:

For a given set of n elements:

The number of subsets with no elements = ⁿC₀ = C₀.

The number of subsets with 1 element = The number of ways in which 1 element can be selected from n elements = ⁿC₁ = C₁.

The number of subsets with 2 elements = The number of ways in which 2 elements can be selected from n elements = nC2 = C₂.

And so on.

∴ The total number of subsets = C0 + C1 + C2 + C3 + … + Cn–1 + C_n = 2ⁿ.

Concept:

We say, set ‘A’ is a subset of a set ‘B’, if all elements of A are also elements of B. It is denoted by A ⊂ B.

The implication p ⇒ q (read: p implies q, or if p then q) is the statement which asserts that if p is true, then q is also true. We agree that p → q is true when p is false. The statement p is called the hypothesis of the implication, and the statement q is called the conclusion of the implication.

The biconditional or double implication p ⇔ q (read: p if and only if q) is the statement which asserts that p and q if p is true, then q is true, and if q is true then p is true. Put differently, p ⇔ q asserts that p and q have the same truth value.

Calculation:

Statement P: A ∩ (B U C) = (A ∩ B) U C

LHS = (a, b, c, d) ∩ (b, c, d, e, f, g) = b, c, d

RHS = (c, d) ∪ (b, c, f, g) = b, c, d, f, g

⇒ b, c, d = b, c, d, f, g

∴ f, g = 0,

So, set C has region b, c only.

Set A has region a, b, c, d, which implies C is the subset of set A

Thus P ⇒ Q     ---(1)

Now, in statement Q, it is given that C is a subset of A

From statement P again:

LHS = a, b, c, d ∩ (b, c, d, e) = b, c, d

RHS = (c, d) ∪ (b, c) = b, c, d

⇒ b, c, d = b, c, d

i.e., LHS = RHS shows Q ⇒ P     ---(2)

∴ P ⇔ Q     ---(From (1) and (2))

Hence, option (2) is correct.

Concept:

• A set is a collection of objects. As an example, the set of even numbers between 1 and 9 is {2, 4, 6, 8}.
• If A is a set, then n(A) denotes the number of objects in the set A.
• \(n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\)

Calculation:

Given, \(n(A\cap C)=20\) and \(n(A\cap B\cap C) = 10\).

Now, we know that n(A) = 50, n(B) = 60 and n(C) = 70. We also know that \(n(A\cap C)=20\), \(n(A\cap B)=25\) and \(n(B\cap C)=30\)

Now, according to the formula, 

\(n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\)

\(=50+60+70-20-25-30+10\)

\(n(A\cup B\cup C)=115\)

Concept:

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

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

The number of subsets of a set with n elements is 2ⁿ.

The number of proper subsets of a given set is 2n - 1

Calculation:

Statement I: If A = {x: x is an even natural number} and B = {y: y is a natural number}, A subset B.

A = {2, 4, 6, 8, 10, 12, ...} and A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, ...}.

It is clear that all the elements of set A are included in set B.

So, set A is the subset of set B.

Statement I is correct.

Statement II:  Number of subsets for the given set A = {5, 6, 7, 8) is 15.

Given: A = {5, 6, 7, 8}

The number of elements in the set is 4

We know that,

The formula to calculate the number of subsets of a given set is 2ⁿ

 = 2⁴ = 16

Number of subsets is 16

Statement II is incorrect.

Statement III: Number of proper subsets for the given set A = {5, 6, 7, 8) is 15.

The formula to calculate the number of proper subsets of a given set is 2ⁿ - 1

 = 2⁴ - 1

 = 16 - 1 = 15

The number of proper subsets is 15.

Statement III is correct.

∴ Statements I and III are correct.

Concept:

The number of subsets of any set A having 'n' elements is given by 2ⁿ. 

Calculation:

Let set A have m elements and set B have n elements

According to the question, 2^m = 2ⁿ + 56

⇒ 2^m - 2ⁿ = 56

⇒ 2ⁿ(2^{m - n} - 1) = 2³×7

On comparing both sides, we get the following:

2ⁿ = 2³ ⇒ n = 3

and 2^{m - n} - 1 = 7

⇒ 2^{m - n} = 8 = 2³

⇒ m - n = 3

⇒ m = n + 3 = 3 + 3 = 6

∴ The values of m and n, respectively, are (6, 3).