Sets and Relations — Page 6 | JEE Mathematics

Q51

Let

\mathrm{X}=\mathbf{R} \times \mathbf{R}

. Define a relation R on X as :

\left(a_1, b_1\right) R\left(a_2, b_2\right) \Leftrightarrow b_1=b_2

Statement I:

\quad \mathrm{R}

is an equivalence relation. Statement II : For some

(\mathrm{a}, \mathrm{b}) \in \mathrm{X}

, the

\operatorname{set} \mathrm{S}=\{(x, y) \in \mathrm{X}:(x, y) \mathrm{R}(\mathrm{a}, \mathrm{b})\}

represents a line parallel to

y=x

. In the light of the above statements, choose the correct answer from the options given below :

A Both Statement I and Statement II are true

B Statement I is true but Statement II is false

C Both Statement I and Statement II are false

D Statement I is false but Statement II is true

Correct Answer

Option B

Solution

Statement - I : Reflexive : $\left(a_1, b_1\right) R\left(a_1, b_1\right) \Rightarrow b_1=b_1 \quad$ True $\left.\begin{array}{rl}\text{ Symmetric : } & \left(\mathrm{a}_1, \mathrm{~b}_1\right) \mathrm{R}\left(\mathrm{a}_2, \mathrm{~b}_2\right) \Rightarrow \mathrm{b}_1=\mathrm{b}_2 \\ & \left(\mathrm{a}_2, \mathrm{~b}_2\right) \mathrm{R}\left(\mathrm{a}_1, \mathrm{~b}_1\right) \Rightarrow \mathrm{b}_2=\mathrm{b}_1\end{array}\right\}$ True $\left.\begin{array}{ll}\text{ Transitive: } & \left(a_1, b_1\right) R\left(a_2, b_2\right) \Rightarrow b_1=b_2 \\ & \&\left(a_2, b_2\right) R\left(a_3, b_3\right) b_2=b_3\end{array}\right\} b_1=b_3$

\Rightarrow\left(\mathrm{a}_1, \mathrm{~b}_1\right) \mathrm{R}\left(\mathrm{a}_3 \cdot \mathrm{~b}_3\right) \Rightarrow \text{ True }

Hence Relation $R$ is an equivence relation Statement-I is true.

For statement - II $\Rightarrow \mathrm{y}=\mathrm{b}$ so False

Q52

The number of non-empty equivalence relations on the set

\{1,2,3\}

is :

A 7

B 4

C 5

D 6

Correct Answer

Option C

Solution

An equivalence relation on a finite set is uniquely determined by its partition into equivalence classes.

Hence, counting the number of equivalence relations on a set is equivalent to counting the number of ways to partition that set.

Step: Counting partitions of $\{1,2,3\}$ We want all possible ways to split the set $\{1,2,3\}$ into nonempty subsets (its “blocks”). 3 blocks (each element in its own block) $\{\{1\}, \{2\}, \{3\}\}.$ 2 blocks $\{\{1,2\}, \{3\}\}$ $\{\{1,3\}, \{2\}\}$ $\{\{2,3\}, \{1\}\}$ 1 block (all elements together) $\{\{1,2,3\}\}.$ Counting these, there are a total of 5 distinct partitions, and thus 5 equivalence relations on the set $\{1,2,3\}$ .

All equivalence relations are automatically nonempty (they include at least $(1,1), (2,2), (3,3)$ because they are reflexive), so the answer to “the number of nonempty equivalence relations” is also 5.

Answer: Option C (5)

Q53

Let

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

. Let R be a relation on

A

defined by

x \mathrm{R} y

if and only if

y=\max \{x, 1\}

. Let

l

be the number of elements in R . Let

m

and

n

be the minimum number of elements required to be added in R to make it reflexive and symmetric relations, respectively. Then

l+m+n

is equal to

A 11

B 12

C 14

D 13

Correct Answer

Option B

Solution

To solve the problem, we start by defining the set $A = \{-2, -1, 0, 1, 2, 3\}$ and the relation $R$ on set $A$ , where an element $x$ is related to $y$ (written as $x \, R \, y$ ) if and only if $y = \max\{x, 1\}$ .

This leads us to the following pairs in the relation $R$ : For $x = -2$ , $y = \max\{-2, 1\} = 1$ , so $(-2, 1)$ is in $R$ .

For $x = -1$ , $y = \max\{-1, 1\} = 1$ , so $(-1, 1)$ is in $R$ .

For $x = 0$ , $y = \max\{0, 1\} = 1$ , so $(0, 1)$ is in $R$ .

For $x = 1$ , $y = \max\{1, 1\} = 1$ , so $(1, 1)$ is in $R$ .

For $x = 2$ , $y = \max\{2, 1\} = 2$ , so $(2, 2)$ is in $R$ .

For $x = 3$ , $y = \max\{3, 1\} = 3$ , so $(3, 3)$ is in $R$ .

Thus, the relation $R$ consists of the pairs: $\{(-2, 1), (-1, 1), (0, 1), (1, 1), (2, 2), (3, 3)\}$ , and there are $l = 6$ elements in $R$ .

Making the Relation Reflexive A relation is reflexive if every element in the set $A$ relates to itself.

Therefore, the missing reflexive pairs are: $(-2, -2)$ $(-1, -1)$ $(0, 0)$ Adding these three pairs will make the relation reflexive, so $m = 3$ .

Making the Relation Symmetric A relation is symmetric if whenever $(x, y)$ is in $R$ , $(y, x)$ must also be in $R$ .

Therefore, the missing symmetric pairs are: $(1, -2)$ $(1, -1)$ $(1, 0)$ Thus, we need to add these three pairs for symmetry, so $n = 3$ .

Finally, we calculate the sum $l + m + n = 6 + 3 + 3 = 12$ .

Q54

Consider the sets

A=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: x^2+y^2=25\right\}, B=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: x^2+9 y^2=144\right\}

,

C=\left\{(x, y) \in \mathbb{Z} \times \mathbb{Z}: x^2+y^2 \leq 4\right\}

and

D=A \cap B

. The total number of one-one functions from the set

D

to the set

C

is:

A 15120

B 18290

C 17160

D 19320

Correct Answer

Option C

Solution

\begin{aligned} & A=\left\{(x, y) \in R \times R: x^2+y^2=25\right\}, B=\{(x, y) \in \mathbb{R} \times \\ & \left.\mathbb{R}: x^2+9 y^2=144\right\} \\ & x^2+9 y^2-\left(x^2+y^2\right)=144-25 \\ & \text{ Plug in } y^2=\frac{119}{8} \text{ into either equation to find } x . \\ & x^2=25-\frac{119}{8} \\ & x^2=\frac{200-119}{8} \\ & x^2=\frac{81}{8} \\ & x= \pm \sqrt{\frac{81}{8}}, y= \pm \sqrt{\frac{119}{8}} \end{aligned}

Now, $C=\left\{(x, y) \in \mathbb{Z} \times \mathbb{Z}: x^2+y^2 \leq 4\right\}$ Valid points are $(-2,0),(-1,-1),(-1,0),(-1,1)$ , $(0,-2),(0,-1),(0,0),(0,1),(0,2),(1,-1),(1,0)$ , $(1,1)$ $\therefore$ Total valid points in $C=13$ $\Rightarrow \quad$ There are 4 distinct real points in set $D$ $\therefore \quad$ The number of one-one functions from $D$ to $C$

\Rightarrow \quad 13 P_4 \Rightarrow \frac{13!}{(13-4)!}=\frac{13!}{9!}=17160

Q55

Let

\mathrm{S}=\mathbf{N} \cup\{0\}

. Define a relation R from S to

\mathbf{R}

by :

\mathrm{R}=\left\{(x, y): \log _{\mathrm{e}} y=x \log _{\mathrm{e}}\left(\frac{2}{5}\right), x \in \mathrm{~S}, y \in \mathbf{R}\right\} .

Then, the sum of all the elements in the range of

R

is equal to :

A

\dfrac{3}{2}

B

\dfrac{10}{9}

C

\dfrac{5}{2}

D

\dfrac{5}{3}

Correct Answer

Option D

Solution

\begin{aligned} & S=\{0,1,2,3 \ldots . .\} \\ & \log _{\mathrm{e}} \mathrm{y}=\log _{\mathrm{e}}\left(\frac{2}{5}\right) \\ & \Rightarrow \mathrm{y}=\left(\frac{2}{5}\right)^{\mathrm{x}} \end{aligned}

Required

\text{ Sum }=1+\left(\frac{2}{5}\right)^1+\left(\frac{2}{5}\right)^2+\left(\frac{2}{5}\right)^3+\ldots . .-=\frac{1}{1-\frac{2}{5}}=\frac{5}{3}

Q56

Let

\mathrm{A}=\{1,2,3,4,5\}

. Let

\mathrm{R}

be a relation on

\mathrm{A}

defined by

x \mathrm{R} y

if and only if

4 x \leq 5 \mathrm{y}

. Let

\mathrm{m}

be the number of elements in

\mathrm{R}

and

\mathrm{n}

be the minimum number of elements from

\mathrm{A} \times \mathrm{A}

that are required to be added to R to make it a symmetric relation. Then m + n is equal to :

A 23

B 26

C 25

D 24

Correct Answer

Option C

Solution

\begin{aligned} & A=\{1,2,3,4,5\} \\ & x R y \Leftrightarrow 4 x \leq 5 y \\ & 4 x \leq 5 y \quad \Rightarrow \quad \frac{x}{y} \leq \frac{5}{4} \quad \Rightarrow \frac{x}{y} \leq 1.25 \end{aligned}

\begin{aligned} & R=\{(1,2),(1,3),(1,4),(1,5),(1,1),(2,2),(2,3),(2,4), \\ & (2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,4),(5,5)\} \\ & \therefore \quad n(R)=m=16 \end{aligned}

Elements to be added to

R

to make it symmetric

(1,2) \in R \quad \Rightarrow \quad(2,1)

should be added Similarly,

(3,1),(4,1),(5,1),(3,2),(4,2),(5,2),(4,3), (5,3)

\therefore 9

elements should be added

\begin{array}{ll} \therefore & n=9 \\ \therefore & m+n=25 \end{array}

Q57

If

A, B

and

C

are three sets such that

A \cap B=A \cap C

and

A \cup B=A \cup C

, then :

A

A=C

B

B=C

C

A \cap B=\phi

D

A=B

Correct Answer

Option B

Solution

From the given conditions, we have : A ∩ B = A ∩ C : The intersection of set A with set B is the same as the intersection of set A with set C.

This indicates that all elements common to A and B are also common to A and C, and vice versa.

A ∪ B = A ∪ C : The union of set A with set B is the same as the union of set A with set C.

This indicates that all elements in A, B, and C are the same.

From these two conditions, we can infer that set B is equal to set C because every element of B is also an element of C and vice versa.

Hence, Option B : B = C is the correct answer.

Q58

Let

\mathrm{R}

be a relation on

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

defined by

(a, b) ~\mathrm{R}~(c, d)

if and only if

a d(b-c)=b c(a-d)

. Then

\mathrm{R}

is

A symmetric and transitive but not reflexive

B reflexive and symmetric but not transitive

C transitive but neither reflexive nor symmetric

D symmetric but neither reflexive nor transitive

Correct Answer

Option D

Solution

Given, $(a, b) R(c, d) \Rightarrow a d(b-c)=b c(a-d)$ Symmetric : (c, d) $R(a, b) \Rightarrow \operatorname{cb}(\mathrm{d}-\mathrm{a})=\mathrm{da}(\mathrm{c}-\mathrm{b})$ $\Rightarrow$ Symmetric.

Reflexive : (a, b) R (a, b) $\Rightarrow a b(b-a) \neq b a(a-b)$ $\Rightarrow$ Not reflexive.

Transitive : $(2,3) \mathrm{R}(3,2)$ and $(3,2) \mathrm{R}(5,30)$ but $((2,3),(5,30)) \notin \mathrm{R}$ $\Rightarrow$ Not transitive.

Q59

Let

\mathrm{R}=\{(1,2),(2,3),(3,3)\}

be a relation defined on the set

\{1,2,3,4\}

. Then the minimum number of elements, needed to be added in R so that R becomes an equivalence relation, is:

A 9

B 8

C 7

D 10

Correct Answer

Option C

Solution

\mathrm{A}=\{1,2,3,4\}

For relation to be reflexive

\mathrm{R}=\{(1,2),(2,3),(3,3)\}

Minimum elements added will be

\begin{aligned} & (1,1),(2,2),(4,4)(2,1)(3,2)(3,2)(3,1)(1,3) \\ & \therefore \text{ Minimum number of elements }=7 \end{aligned}

Q60

Two newspapers A and B are published in a city. It is known that 25% of the city populations reads A and 20% reads B while 8% reads both A and B. Further, 30% of those who read A but not B look into advertisements and 40% of those who read B but not A also look into advertisements, while 50% of those who read both A and B look into advertisements. Then the percentage of the population who look into advertisement is :-

A 13.5

B 13

C 12.8

D 13.9

Correct Answer

Option D

Solution

The total population to be 100 (for simplicity's sake) and the percentages can be treated as actual numbers of people in this context.

The percentage of people who read newspaper A is given as 25.

However, among these, there are people who read both newspapers A and B, given as 8.

To find the number of people who read only newspaper A, we subtract the number of people who read both from the total number of people who read A.

That is, n(A only) = 25 – 8 = 17 Similarly, the number of people who read only newspaper B is calculated as : n(B only) = 20 – 8 = 12 Now, we are given the percentage of each of these groups that look into the advertisements: 30% of those who read A but not B, 40% of those who read B but not A, 50% of those who read both A and B.

To find the total percentage of the population that looks into advertisements, we add up the contributions from each of these groups.

We calculate each group's contribution by multiplying the size of the group by the percentage of that group that looks at advertisements : =

{{30} \over {100}} \times 17

(from A only) +

{{40} \over {100}} \times 12

(from B only) +

{{50} \over {100}} \times 8

(from both A and B) = 5.1 (from A only) + 4.8 (from B only) + 4 (from both A and B) Adding these up, we get = 13.9 This means that 13.9% of the total population looks into the advertisements.

So, the correct answer is : Option D : 13.9.