Sets and Relations — Page 7 | JEE Mathematics

Q61

Consider the following relations

R=\{(x, y) \mid x, y

are real numbers and

x=w y

for some rational number

w\}

;

S=\left\{\left(\dfrac{m}{n}, \dfrac{p}{q}\right) \mid m, n, p\right.

and

q

are integers such that

n, q \neq 0

and

q m=p m\}

. Then

A

R

is an equivalence relation but

S

is not an equivalence relation

B Neither

R

nor

S

is an equivalence relation

C

S

is an equivalence relation but

R

is not an equivalence relation

D

R

and

S

both are equivalence relations

Correct Answer

Option C

Solution

Let's evaluate each relation for the properties of an equivalence relation: reflexivity, symmetry, and transitivity.

Relation R : $R=(x, y) \mid x, y$ are real numbers and $x=w y$ for some rational number $w$ .

Reflexivity : For all $x$ in $R$ , $x = 1x$ .

Since 1 is a rational number, every element is related to itself.

Symmetry : For all $x, y$ in $R$ , if $x = w y$ for some rational $w$ , then $y = \dfrac{1}{w}x$ .

However, if $w = 0$ , then $\dfrac{1}{w}$ is undefined, and therefore, $R$ doesn't satisfy symmetry.

Transitivity : If $x = w y$ and $y = v z$ for some rational numbers $w$ and $v$ , then $x = (w v)z$ .

Since the product of rational numbers is rational, if $x$ is related to $y$ and $y$ is related to $z$ , then $x$ is related to $z$ .

Therefore, $R$ is not an equivalence relation on $R$ since it does not satisfy the symmetry property.

Relation S : $S=\left\{\left(\dfrac{m}{n}, \dfrac{p}{q}\right) \mid m, n, p\right.$ and $q$ are integers such that $n, q \neq 0$ and $q m=p m\}$ Reflexivity : For all $\dfrac{m}{n}$ , $\dfrac{m}{n} = \dfrac{m}{n}$ .

Since $n \neq 0$ and $m = m$ , every element is related to itself.

Symmetry : For all $\dfrac{m}{n}$ , $\dfrac{p}{q}$ , if $q m = p n$ , then $n p = m n$ .

So if $\dfrac{m}{n}$ is related to $\dfrac{p}{q}$ , then $\dfrac{p}{q}$ is related to $\dfrac{m}{n}$ .

Transitivity :

\begin{array}{rlr} \frac{m}{n} R \frac{p}{q} \text{ and } \frac{p}{q} R \frac{r}{s} \\\\ \Rightarrow m q=n p \text{ and } p s=r q \\\\ \Rightarrow m q \cdot p s=n p \cdot r q \\\\ \Rightarrow \quad m s=n r \\\\ \Rightarrow \frac{m}{n}=\frac{r}{s} \Rightarrow \frac{m}{n} R \frac{r}{s} \end{array}

So if $\dfrac{m}{n}$ is related to $\dfrac{p}{q}$ and $\dfrac{p}{q}$ is related to $\dfrac{r}{s}$ , then $\dfrac{m}{n}$ is related to $\dfrac{r}{s}$ .

The relation $S$ is transitive.

Therefore, $S$ is an equivalence relation on the set of all fractions where denominator is not zero.

In conclusion, the correct answer is Option C : $S$ is an equivalence relation but $R$ is not an equivalence relation.

Q62

An organization awarded 48 medals in event 'A', 25 in event 'B' and 18 in event 'C'. If these medals went to total 60 men and only five men got medals in all the three events, then, how many received medals in exactly two of three events?

A 10

B 15

C 21

D 9

Correct Answer

Option C

Solution

We are given the number of medals for events A, B, and C which are 48, 25, and 18 respectively.

We are also given that the total number of unique medal recipients across all events is 60 and that 5 people received a medal in all three events.

Using the Principle of Inclusion and Exclusion (PIE), we know that the total number of unique medal recipients can be calculated by adding the number of medal recipients in each event, subtracting the number of people who received a medal in any two events (to correct for double counting), and then adding back the number of people who received a medal in all three events (since we subtracted these people too much).

Mathematically, this can be represented as : |A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |B ∩ C| - |C ∩ A| + |A ∩ B ∩ C| However, we want to find the total number of people who received a medal in any two events (which is represented by |A ∩ B| + |B ∩ C| + |C ∩ A| in the equation).

To find this, we rearrange the PIE formula to solve for |A ∩ B| + |B ∩ C| + |C ∩ A| : |A ∩ B| + |B ∩ C| + |C ∩ A| = |A| + |B| + |C| + |A ∩ B ∩ C| - |A ∪ B ∪ C| Substituting the given values, we find that the total number of people who received a medal in any two events is 48 + 25 + 18 + 5 - 60 = 36.

However, this includes people who received a medal in all three events, and we want to find the number of people who received a medal in exactly two events.

Therefore, we need to subtract the people who received a medal in all three events from our calculated value.

Since each person who received a medal in all three events is counted three times in |A ∩ B| + |B ∩ C| + |C ∩ A| (once for each pair of events), we subtract three times the number of people who received a medal in all three events from our calculated value: Number of people who received a medal in exactly two events = |A ∩ B| + |B ∩ C| + |C ∩ A| - 3 $\times$ |A ∩ B ∩ C| Substituting the values we know, we find that the number of people who received a medal in exactly two events is 36 - 3 $\times$ 5 = 21.

Therefore, 21 people received a medal in exactly two of the three events.

Q63

Two sets A and B are as under : A = {(

a

, b)

\in

R

\times

R : |

a

- 5| < 1 and |b - 5| < 1}; B = {(

a

, b)

\in

R

\times

R : 4(

a

- 6)2 + 9(b - 5)2

\le

36 }; Then

A neither A

\subset

B nor B

\subset

A

B B

\subset

A

C A

\subset

B

D A

\cap

B =

\phi

( an empty set )

Correct Answer

Option C

Solution

Given,

4{\left( {a - 6} \right)^2} + 9{\left( {b - 5} \right)^2} \le 36

Let

a - 6 = x

and

b - 5 = y

\therefore\,\,\,\,

4{x^2} + 9{y^2} \le 36

\Rightarrow \,\,\,\,{{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1

This is a equation of ellipse. This ellipse will look like this, According to set A,

\left| {a - 5} \right| < 1

as

a - 6 = x

then

a - 5 = x + 1

\therefore\,\,\,

\left| {x + 1} \right| < 1

\Rightarrow \,\,\, - 1 < x + 1 < 1

\Rightarrow \,\,\, - 2 < x < 0

\left| {b - 5} \right| < 1

as

b - 5 = y

\therefore\,\,\,

\left| y \right| < 1

\Rightarrow \,\,\,\, - 1 < y < 1

This will look like this, By combining both those graphs it will look like this, To check entire set A is inside of B or not put any paint of set A on the equation

{{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1

and if this inequality satisfies, then it is inside B.

By looking at the graph you can surely say (0, 1) or (0, $-$ 1) inside the graph.

But to check point (

-2

, 1) or ( $-$ 2, $-$ 1) is inside of the ellipse or not, put on the

{{{x^2}} \over 9} + {{{y^2}} \over 4} \le 1.

Putting ( $-$ 2, 1) on the inequality, LHS

= {{\left( { - 2} \right){}^2} \over 9} + {{{1^2}} \over 4}

= {4 \over 9} + {1 \over 4}

= {{25} \over {36}} < 1

\therefore\,\,\,

Inequality holds. So,

\left( { - 2,1} \right)

is inside the ellipse. Similarly by checking we can see

\left( { - 2, - 1} \right)

is also inside the ellipse. Hence, we can say entire set A is inside of the set B.

\therefore\,\,\,\,

A $\subset$ B

Q64

Let

A=\{n \in[100,700] \cap \mathrm{N}: n

is neither a multiple of 3 nor a multiple of 4

\}

. Then the number of elements in

A

is

A 300

B 310

C 290

D 280

Correct Answer

Option A

Solution

n \in[100,700]

n(A)=

Total $-$ (multiple of

3

+ multiple of 4) + (multiple of 12) Total

=601

Multiple of

3=102,105, \ldots, 699

\begin{aligned} & n=699=102+(n-1) 3 \\ & \Rightarrow n=200 \end{aligned}

Multiple of

4=100,104 \ldots ., 700

\begin{aligned} & n=700=100+(n-1) 4 \\ & \frac{600}{4}+1=n \\ & \Rightarrow n=151 \end{aligned}

Multiple of

12=108,120 \ldots .696

\begin{aligned} & n=696=108+(n-1) 12 \\ & n=50 \\ & \therefore n(A)=601-(200+151)+50 \\ & =300 \end{aligned}

Q65

Let a set A = A1

\cup

A2

\cup

.....

\cup

Ak, where Ai

\cap

Aj =

\phi

for i

\ne

j, 1

\le

j, j

\le

k. Define the relation R from A to A by R = {(x, y) : y

\in

Ai if and only if x

\in

Ai, 1

\le

i

\le

k}. Then, R is :

A reflexive, symmetric but not transitive.

B reflexive, transitive but not symmetric.

C reflexive but not symmetric and transitive.

D an equivalence relation.

Correct Answer

Option D

Solution

R = \{ (x,y):y \in {A_i},\,iff\,x \in {A_i}\,1 \le i \ge k\}

(1) Reflexive (a, a) $\Rightarrow$

a \in {A_i}

iff

a \in {A_i}

(2) Symmetric (a, b) $\Rightarrow$

a \in {A_i}

iff

b \in {A_i}

(b, a)

\in

R as

b \in {A_i}

iff

a \in {A_i}

(3) Transitive (a, b)

\in

R & (b, c)

\in

R. $\Rightarrow$

a \in {A_i}

iff

b \in {A_i}

&

b \in {A_i}

iff

c \in {A_i}

$\Rightarrow$

a \in {A_i}

iff

c \in {A_i}

$\Rightarrow$ (a, c)

\in

R. $\Rightarrow$ RElation is equivalnece.

Q66

Consider the two sets : A = {m

\in

R : both the roots of x2 – (m + 1)x + m + 4 = 0 are real} and B = [–3, 5). Which of the following is not true?

A A

\cap

B = {–3}

B B – A = (–3, 5)

C A

\cup

B = R

D A - B = (

-

\propto

,

-

3)

\cup

(5,

\propto

)

Correct Answer

Option D

Solution

As roots are real so,

D \ge 0

{(m + 1)^2} - 4(m + 4) \ge 0

\Rightarrow {m^2} - 2m - 15 \ge 0

$\Rightarrow$

(m - 5)(m + 3) \ge 0

m\, \in \,

( $-$ $\propto$ , $-$ 3] $\cup$ [5, $\propto$ )

A= ( -

$\propto$ , $-$ 3] $\cup$ [5, $\propto$ ) Given B = [ $-$ 3, 5) Now, let's examine the options.

Option A : A ∩ B = {–3} The intersection of sets A and B would be the set of elements common to both sets.

In this case, the only common element is -3.

So, option A is true.

Option B : B – A = (–3, 5) The subtraction (or difference) of sets A from B is the set of elements that are in B but not in A.

B is [–3, 5), and A is (-∞, -3] U [5, ∞).

Subtracting A from B would leave an open interval (-3, 5), not including -3 and 5.

So, option B is also true.

Option C : A ∪ B = R The union of sets A and B is the set of elements that are in A, or B, or both.

Here, A U B would cover all real numbers.

So, option C is true.

Option D : A - B = (-∞, -3) ∪ (5, ∞) The subtraction (or difference) of set B from A is the set of elements that are in A but not in B.

B is [–3, 5), and A is (-∞, -3] U [5, ∞).

Subtracting B from A would leave (-∞, -3) U [5, ∞), not including -3 and 5.

But according to the convention for writing intervals, it should be (-∞, -3) U (5, ∞).

So, option D is not true.

Q67

Let a relation

\mathrm{R}

on

\mathrm{N} \times \mathbb{N}

be defined as:

\left(x_1, y_1\right) \mathrm{R}\left(x_2, y_2\right)

if and only if

x_1 \leq x_2

or

y_1 \leq y_2

. Consider the two statements: (I)

\mathrm{R}

is reflexive but not symmetric. (II)

\mathrm{R}

is transitive Then which one of the following is true?

A Only (II) is correct.

B Both (I) and (II) are correct.

C Neither (I) nor (II) is correct.

D Only (I) is correct.

Correct Answer

Option D

Solution

\begin{aligned} & \left(x_1, y_1\right) R\left(x_2, y_2\right) \\ & \text{ If } x_1 \leq x_2 \text{ or } y_1 \leq y_2 \end{aligned}

For reflexive;

\begin{aligned} & \left(x_1, y_1\right) R\left(x_1, y_1\right) \\ & \Rightarrow x_1 \leq x_1 \text{ or } y_1 \leq y_1 \end{aligned}

So,

R

is reflexive For symmetric When

\left(x_1, y_1\right) R\left(x_2, y_2\right)

\Rightarrow x_1 \leq x_2 \text{ or } y_1 \leq y_2

For

\left(x_2, y_2\right) R\left(x_1, y_1\right)

\Rightarrow x_2 \leq x_1 \text{ or } y_2 \leq y_1

Not true for

(1,2)

and

(3,4)

For transitive Take pairs as

(3,9),(4,6),(2,7)

(3,9) R(4,6)

as

4 \geq 3

(4,6) R(2,7)

As

7 \geq 6

But

(3,9) R(2,7)

As neither

2 \geq 3

nor

7 \geq 9

So not transitive

Q68

Let

\mathrm{A}=\{-3,-2,-1,0,1,2,3\}

. Let R be a relation on A defined by

x \mathrm{R} y

if and only if

0 \leq x^2+2 y \leq 4

. Let

l

be the number of elements in R and

m

be the minimum number of elements required to be added in R to make it a reflexive relation. Then

l+m

is equal to

A 18

B 20

C 17

D 19

Correct Answer

Option A

Solution

The relation $R$ is defined for the set $\mathrm{A} = \{-3, -2, -1, 0, 1, 2, 3\}$ with the condition $0 \leq x^2 + 2y \leq 4$ .

Let's determine the pairs $(x, y)$ that satisfy this condition.

For $y = -3$ : Solving $x^2 + 2(-3) \leq 4$ , we find $x = \{3, -3\}$ .

For $y = -2$ : Solving $x^2 + 2(-2) \leq 4$ , we find $x = \{-2, 2\}$ .

For $y = -1$ : Solving $x^2 + 2(-1) \leq 4$ , we find $x = \{-2, 2\}$ .

For $y = 0$ : Solving $x^2 + 0 \leq 4$ , we find $x = \{-2, -1, 0, 1, 2\}$ .

For $y = 1$ : Solving $x^2 + 2(1) \leq 4$ , we find $x = \{-1, 0, 1\}$ .

For $y = 2$ : Solving $x^2 + 2(2) \leq 4$ , we find $x = \{0\}$ .

The relation $R$ consists of the following pairs: $R = \{(3, -3), (-3, -3), (-2, -2), (2, -2), (-2, -1), (2, -1), (-2, 0), (-1, 0), (0, 0), (1, 0), (2, 0), (-1, 1), (0, 1), (1, 1), (0, 2)\}$ Currently, $R$ has $l = 15$ elements.

To make $R$ reflexive, it must include all pairs $(x, x)$ for every $x \in \mathrm{A}$ .

We identify the missing reflexive pairs $(-1, -1)$ , $(2, 2)$ , and $(3, 3)$ , which are required to satisfy reflexivity.

Thus, $m = 3$ more elements are needed.

Therefore, the total $l + m = 15 + 3 = 18$ .

Q69

Let

\mathrm{A}=\left\{x \in(0, \pi)-\left\{\dfrac{\pi}{2}\right\}: \log _{(2 /\pi)}|\sin x|+\log _{(2 / \pi)}|\cos x|=2\right\}

and

\mathrm{B}=\{x \geqslant 0: \sqrt{x}(\sqrt{x}-4)-3|\sqrt{x}-2|+6=0\}

. Then

\mathrm{n}(\mathrm{A} \cup \mathrm{B})

is equal to :

A 4

B 8

C 6

D 2

Correct Answer

Option B

Solution

\begin{aligned} & \mathrm{A}: \log _{2 /\pi}|\sin \mathrm{x}|+\log _{2 /\pi}|\cos \mathrm{x}|=2 \\ & \Rightarrow \log _{2 /\pi}(|\sin \mathrm{x} \cdot \cos \mathrm{x}|)=2 \\ & \Rightarrow|\sin 2 \mathrm{x}|=\frac{8}{\pi^2} \end{aligned}

Number of solution 4 B : let $\sqrt{\mathrm{x}}=\mathrm{t}<2$ Then $\sqrt{\mathrm{x}}(\sqrt{\mathrm{x}}-4)+3(\sqrt{\mathrm{x}}-2)+6=0$

\begin{aligned} & \Rightarrow \mathrm{t}^2-4 \mathrm{t}+3 \mathrm{t}-6+6=0 \\ & \Rightarrow \mathrm{t}^2-\mathrm{t}=0, \mathrm{t}=0, \mathrm{t}=1 \\ & \mathrm{x}=0, \mathrm{x}=1 \end{aligned}

again let $\sqrt{x}=t>2$ then $\mathrm{t}^2-4 \mathrm{t}-3 \mathrm{t}+6+6=0$

\begin{aligned} & \Rightarrow t^2-7 \mathrm{t}+12=0 \\ & \Rightarrow \mathrm{t}=3,4 \\ & \mathrm{x}=9,16 \end{aligned}

Total number of solutions

\mathrm{n}(\mathrm{~A} \cup \mathrm{~B})=4+4=8

Q70

Let N denote the set of all natural numbers. Define two binary relations on N as R = {(x, y)

\in

N

\times

N : 2x + y = 10} and R2 = {(x, y)

\in

N

\times

N : x + 2y = 10}. Then :

A Range of R1 is {2, 4, 8).

B Range of R2 is {1, 2, 3, 4}.

C Both R1 and R2 are symmetric relations.

D Both R1 and R2 are transitive relations.

Correct Answer

Option B

Solution

For R1; 2x + y = 10 and x, y

\in

N possible values for x and y are : x = 1, y = 8 i.e. (1, 8); x = 2, y = 6 i.e (2, 6); x = 3, y = 4 i.e (3, 4); x = 4, y = 2 i.e (4, 2)

\therefore\,\,\,

R1 = { (1, 8), (2, 6), (3, 4), (4, 2) }

\therefore\,\,\,

Range of R1 is {2, 4, 6, 8} R1 is not symmetric. R1 is not transitive also as (3, 4), (4, 2)

\in

R , but (3, 2) $\notin$ R1 For R2 : x + 2y = 10 and x, y

\in

N Possible values of x, and y are : x = 8, y= 1 i.e (8, 1) x = 6, y = 2 i.e (6, 2) x = 4, y = 3 i.e (4, 3) and x = 2, y = 4 i.e (2, 4)

\therefore\,\,\,

R2 = {(8, 1) (6, 2) (4, 3) (2, 4)}

\therefore\,\,\,

Range of R2 =

\left\{ {1,2,3,4} \right\}

R2 is not symmetric and transitive