Relations and its properties MCQ Quiz - Objective Question with Answer for Relations and its properties - Download Free PDF

Calculation:

Given, A is the set of even natural numbers less than 8.

⇒ A = {2, 4, 6} 

⇒ m = n(A) = 3

B is the set of prime numbers less than 7.

⇒ B = {2, 3, 5}

⇒ n = n(B) = 3

∴ Number of relations = 2^mn = 2⁹ 

∴ The number of relations from A to B are 2⁹.

The correct answer is Option 4.

Concept 

The total number of possible relations that can be defined over a set with n elements is \(2^{n^2}\)

Calculation 

Given:  A  = {5, 8, 9, 12} 

Number of elements n in set A is 4

From the above concept, the total number of distinct relations that can be defined over A is 

⇒ \(2^{4^2}\) = 2¹⁶

∴ Option 1 is correct

Concept:

If f(x) is a linear function then f(x) = ax + b.

Calculation:

Statement I: Let f = {(1, 1), (2, 3), (0, -1), (-1, -3)} be a linear function from Z to Z. Then f(x) is 2x - 1.

Since f = {(1, 1), (2, 3), (0, -1), (-1, -3)} is a linear function.

Let f(x) = ax + b

But (0,-1) ∈ f

f(0) = a(0) + b = -1 ⇒ b = -1

Similarly, (1, 1) ∈ f 

f(1) = a(1) + b

⇒ 1 = a + b

⇒ a = 2 

So, f(x) = 2x -1

Statement I is correct. 

Statement II: If f(x) =  \(\rm x^3-\frac{1}{x^3}\) then \(\rm f(x)+f\left(\frac{1}{x}\right)\) is equal to 0.

Since f(x) = \(\rm x^3-\frac{1}{x^3}\)

\(\rm f(\frac{1}{x})=\frac{1}{x^3}-\frac{1}{\frac{1}{x^3}}=\frac{1}{x^3}-x^3\)

\(\rm f(x)+f\left(\frac{1}{x}\right)\) = \(\rm x^3-\frac{1}{x^3}+\frac{1}{x^3}-x^3\) = 0

Statement II is correct.

∴ Both statements I and II are correct. 

Concept:

• When any two consecutive numbers in a series have the same difference, they are in Arithmetic Progression or AP. An example of AP is 1 , 3 , 5 ...
• The difference between these two consecutive numbers of the series is called the common difference.
• The reciprocals of the numbers in a series are said to be in Harmonic progression or H.P. An example of Harmonic progression is \(1,\dfrac{1}{3}, \dfrac{1}{5},...\)

Calculation:

Given, f is a function from the set of positive integers N to the set of real numbers R, such that f(1) = 1 and f(1) + 2f(2) + 3f(2) + _____ + nf(n) = n(n + 1)(f(n))

Now, according to the given formula, for n = 2 , \(f(1)+2f(2)=2(2+1)f(2)\Rightarrow 1+2f(2)=6f(2)\Rightarrow f(2)=\dfrac{1}{4}\)

Similarly, for n = 3 , \(f(1)+2f(2)+3f(3)=3(3+1)f(3)\Rightarrow 1+\dfrac{1}{2}+3f(3)=12f(3)\Rightarrow f(3)=\dfrac{1}{6}\)

And from similar methods, \(f(4)=\dfrac{1}{8}\) and \(f(5)=\dfrac{1}{10}\) and so on.

Hence, from this it is easy to understand that in general case, \(f(n)=\dfrac{1}{2n}\) 

Now, for n = 2006 , \(f(2006)=\dfrac{1}{2\times 2006}=\dfrac{1}{4012}\)

Concept:

• When any two consecutive numbers in a series have the same difference, they are in Arithmetic Progression or AP. An example of AP is 1 , 3 , 5 ...
• The difference between these two consecutive numbers of the series is called the common difference.
• The reciprocals of the numbers in a series are said to be in Harmonic progression or H.P. An example of Harmonic progression is \(1,\dfrac{1}{3}, \dfrac{1}{5},...\)

Calculation:

Given, f is a function from the set of positive integers N to the set of real numbers R, such that f(1) = 1 and f(1) + 2f(2) + 3f(2) + _____ + nf(n) = n(n + 1)(f(n))

Now, according to the given formula, for n = 2 , \(f(1)+2f(2)=2(2+1)f(2)\Rightarrow 1+2f(2)=6f(2)\Rightarrow f(2)=\dfrac{1}{4}\)

Similarly, for n=3 , \(f(1)+2f(2)+3f(3)=3(3+1)f(3)\Rightarrow 1+\dfrac{1}{2}+3f(3)=12f(3)\Rightarrow f(3)=\dfrac{1}{6}\)

And from similar methods, \(f(4)=\dfrac{1}{8}\) and \(f(5)=\dfrac{1}{10}\) and so on.

Hence, from this, it is easy to understand that in the general case, \(f(x)=\dfrac{1}{2x}\) 

Concepts:

If  A and B are two non-empty sets such that n(A) = p and n(B) = q then number of relations that can be defined from A to B = 2^pq 

Calculation:

Given: A = {x, y, z} and B = {1, 2}

⇒ n(A) = 3 and n(B) = 2

As we know that, if A and B are two non-empty sets such that n(A) = p and n(B) = q then number of relations that can be defined from A to B = 2pq 

Here, p = 3 and q = 2

So, the number of relations from A to B = 2⁶ 

Hence, option 2 is the correct answer.

Calculation:

Given, f(x) = 2x, g(x) = x² + 2

then, (f + g)(2) = f(2) + g(2)

= (2 × 2) + (2² + 2)

= 4 + 6

= 10

Calculation:

Given, f(x) = 3x, g(x) = 3x2 + 9

then, (f + g)(2) = f(2) + g(2)

= (3 × 2) + (3 × 22 + 9)

= 6 + 21

= 27

 ∴ The value of (f + g) (2) is 27

Concept:

1). Reflexive: Each element is related to itself.

• R is reflexive if for all x ∈ A, xRx.

2). Symmetric: If any one element is related to any other element, then the second element is related to the first.

• R is Symmetric if for all x, y ∈ A, if xRy, then yRx.

3). Transitive: If any one element is related to a second and that second element is related to a third, then the first

    element is related to the third.

• R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

4). R is an equivalence relation if A is nonempty and R is reflexive, symmetric, and transitive.

5). Let R be a relation from a set A to another set B. Then R is of the form {(x, y): x ∈ A and y ∈ B}. The inverse

    relationship of R is denoted by R^-1 and its formula is R^-1 = {(y, x): y ∈ B and x ∈ A}.

Calculation:

Statement I: If R is reflexive, then R-1 is also reflexive

R is reflexive

⇒ (a,a) ∈ R,  a ∈ A

⇒ (a,a) ∈ R⁻¹      [by the definition of R^-1]

⇒ R⁻¹ is also reflexive relation.
Statement II: If R is symmetric, then R-1 is also symmetric

Let (b,a) ∈ R−1

⇒ (a,b) ∈ R, a,b ∈ A      [by the definition of R]

⇒ (b,a) ∈ R         [R is symmetric]

⇒ (a,b) ∈ R⁻¹        [by the definition of R-1]

If (b,a) ∈ R−1 then (a,b) ∈ R−1

⇒ R⁻¹ is also symmetric relation.

Statement III: If R is transitive, then R-1 is also transitive

Let (b,a), (a,c) ∈ R−1

⇒ (a,b), (c,a) ∈ R       [by the definition of R-1]

⇒ (c,a),(a,b) ∈ R

⇒ (c,b) ∈ R        [R is transitive]

⇒ (b,c) ∈ R^-1       [by the definition of R-1]

If (b,a), (a,c) ∈ R-1 then (b,c) ∈ R-1

⇒ R⁻¹ is also transitive relation.

∴ R⁻¹ is reflexive, symmetric and transitive.

Calculation:

Set A has n elements.

Relation is simply A × A

So, we can select first element of ordered pair in n ways and second element in n ways.

So, clearly this set of ordered pairs contain n × n = n² pairs.

Now, each of these n² ordered pairs can be present in the relation or can't be. So, there are 2 possibilities for each of the n² ordered pairs.

Thus, the total no. of relations = 2 × 2 × 2 × ......× (n2  times) 

= \({2^{{n^2}}}\)

Hence, option (3) is correct.

Concepts:

If  A and B are two non-empty sets such that n(A) = p and n(B) = q then number of relations that can be defined from A to B = 2^pq 

Calculation:

Given: A = {x, y, z} and B = {1, 2}

⇒ n(A) = 3 and n(B) = 2

As we know that, if A and B are two non-empty sets such that n(A) = p and n(B) = q then number of relations that can be defined from A to B = 2pq 

Here, p = 3 and q = 2

So, the number of relations from A to B = 2⁶ 

Hence, option 2 is the correct answer.

Calculation:

Given, f(x) = 2x, g(x) = x² + 2

then, (f + g)(2) = f(2) + g(2)

= (2 × 2) + (2² + 2)

= 4 + 6

= 10

Calculation:

Given, f(x) = 3x, g(x) = 3x2 + 9

then, (f + g)(2) = f(2) + g(2)

= (3 × 2) + (3 × 22 + 9)

= 6 + 21

= 27

 ∴ The value of (f + g) (2) is 27

Concept:

1). Reflexive: Each element is related to itself.

• R is reflexive if for all x ∈ A, xRx.

2). Symmetric: If any one element is related to any other element, then the second element is related to the first.

• R is Symmetric if for all x, y ∈ A, if xRy, then yRx.

3). Transitive: If any one element is related to a second and that second element is related to a third, then the first

    element is related to the third.

• R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

4). R is an equivalence relation if A is nonempty and R is reflexive, symmetric, and transitive.

5). Let R be a relation from a set A to another set B. Then R is of the form {(x, y): x ∈ A and y ∈ B}. The inverse

    relationship of R is denoted by R^-1 and its formula is R^-1 = {(y, x): y ∈ B and x ∈ A}.

Calculation:

Statement I: If R is reflexive, then R-1 is also reflexive

R is reflexive

⇒ (a,a) ∈ R,  a ∈ A

⇒ (a,a) ∈ R⁻¹      [by the definition of R^-1]

⇒ R⁻¹ is also reflexive relation.
Statement II: If R is symmetric, then R-1 is also symmetric

Let (b,a) ∈ R−1

⇒ (a,b) ∈ R, a,b ∈ A      [by the definition of R]

⇒ (b,a) ∈ R         [R is symmetric]

⇒ (a,b) ∈ R⁻¹        [by the definition of R-1]

If (b,a) ∈ R−1 then (a,b) ∈ R−1

⇒ R⁻¹ is also symmetric relation.

Statement III: If R is transitive, then R-1 is also transitive

Let (b,a), (a,c) ∈ R−1

⇒ (a,b), (c,a) ∈ R       [by the definition of R-1]

⇒ (c,a),(a,b) ∈ R

⇒ (c,b) ∈ R        [R is transitive]

⇒ (b,c) ∈ R^-1       [by the definition of R-1]

If (b,a), (a,c) ∈ R-1 then (b,c) ∈ R-1

⇒ R⁻¹ is also transitive relation.

∴ R⁻¹ is reflexive, symmetric and transitive.

Calculation:

Set A has n elements.

Relation is simply A × A

So, we can select first element of ordered pair in n ways and second element in n ways.

So, clearly this set of ordered pairs contain n × n = n² pairs.

Now, each of these n² ordered pairs can be present in the relation or can't be. So, there are 2 possibilities for each of the n² ordered pairs.

Thus, the total no. of relations = 2 × 2 × 2 × ......× (n2  times) 

= \({2^{{n^2}}}\)

Hence, option (3) is correct.