Sets and Relations — Page 5 | JEE Mathematics

Q41

A survey shows that 63% of the people in a city read newspaper A whereas 76% read newspaper B. If x% of the people read both the newspapers, then a possible value of x can be:

A 37

B 65

C 29

D 55

Correct Answer

Option D

Solution

A $\cup$ B = 63 - x + x + 76 - x = 139 - x As 139 - x $\le$ 100 $\Rightarrow$ x $\ge$ 39 From venn diagram, you can see x should be less than 76 and 63 otherwise only A newspaper or only B newspaper reader will be negative number.

Intersection of x $\le$ 63 and x $\le$ 76 is = x $\le$ 63. $\therefore$ 39 $\le$ x $\le$ 63 From options possible value of x = 55.

Q42

If R is the smallest equivalence relation on the set

\{1,2,3,4\}

such that

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

, then the number of elements in

\mathrm{R}

is __________.

A 15

B 10

C 12

D 8

Correct Answer

Option B

Solution

Given set

\{1,2,3,4\}

Minimum order pairs are

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

Thus no. of elements

=10

Q43

Let

A=\{2,3,6,8,9,11\}

and

B=\{1,4,5,10,15\}

. Let

R

be a relation on

A \times B

defined by

(a, b) R(c, d)

if and only if

3 a d-7 b c

is an even integer. Then the relation

R

is

A reflexive but not symmetric.

B an equivalence relation.

C reflexive and symmetric but not transitive.

D transitive but not symmetric.

Correct Answer

Option C

Solution

(a, b) R(c, d) \Rightarrow 3 a d-7 b c \in

even For reflexive

(a, b) R(a, b) \Rightarrow 3 a b-7 b a=-4 a b \in

even For symmetric

(a, b) R(c, d)

then

(c, d) R(a, b)=3 b c-7 a d

\in

even Now check for transitive

\begin{aligned} & \begin{array}{ll} (a, b) R(c, d) \text{ and }(c, d) R(e, f) \text{ then }(a, b) R(e, f) \\ 3 a d-7 b c=2 m \quad \Rightarrow (2,5) R(6,8) \text{ and } \\ 3 c f-7 e d=2 n \quad (6,8) R(9,4) \\ \text{ then } 3 a f-7 e b \neq \text{ even } \notin(2,5) R(9,4) \\ \Rightarrow \text{ Not transitive option }(3) \end{array} \end{aligned}

Q44

For

\alpha \in \mathbf{N}

, consider a relation

\mathrm{R}

on

\mathbf{N}

given by

\mathrm{R}=\{(x, y): 3 x+\alpha y

is a multiple of 7

\}

. The relation

R

is an equivalence relation if and only if :

A

\alpha=14

B

\alpha

is a multiple of 4

C 4 is the remainder when

\alpha

is divided by 10

D 4 is the remainder when

\alpha

is divided by 7

Correct Answer

Option D

Solution

R = \{ (x,y):3x + \alpha y

is multiple of 7

\}

, now R to be an equivalence relation (1) R should be reflexive :

(a,a) \in R\,\forall \,a \in N

$\therefore$

3a + a\alpha = 7k

$\therefore$

(3 + \alpha )a = 7k

$\therefore$

3 + \alpha = 7{k_1} \Rightarrow \alpha = 7{k_1} - 3

= 7{k_1} + 4

(2) R should be symmetric :

aRb \Leftrightarrow bRa

aRb:3a + (7k - 3)b = 7\,m

\Rightarrow 3(a - b) + 7kb = 7\,m

\Rightarrow 3(b - a) + 7ka = 7\,m

So,

aRb \Rightarrow bRa

$\therefore$ R will be symmetric for

a = 7{k_1} - 3

(3) Transitive : Let

(a,b) \in R,\,(b,c) \in R

\Rightarrow 3a + (7k - 3)b = 7{k_1}

and

3b + (7{k_2} - 3)c = 7{k_3}

Adding

3a + 7kb + (7{k_2} - 3)\,c = 7({k_1} + {k_3})

3a + (7{k_2} - 3)\,c = 7\,m

$\therefore$

(a,c) \in R

$\therefore$ R is transitive $\therefore$

\alpha = 7k - 3 = 7k + 4

Q45

Define a relation R on the interval

\left[0, \dfrac{\pi}{2}\right)

by

x

R

y

if and only if

\sec^2x - \tan^2y = 1

. Then R is :

A both reflexive and symmetric but not transitive

B both reflexive and transitive but not symmetric

C reflexive but neither symmetric not transitive

D an equivalence relation

Correct Answer

Option D

Solution

\begin{aligned} & \sec ^2 x-\tan ^2 x=1 \quad(\text{ on replacing } y \text{ with } x) \\ & \Rightarrow \text{ Reflexive } \\ & \sec ^2 x-\tan ^2 y=1 \\ & \Rightarrow 1+\tan ^2 x+1-\sec ^2 y=1 \\ & \Rightarrow \sec ^2 y-\tan ^2 x=1 \\ & \Rightarrow \operatorname{symmetric} \\ & \sec ^2 x-\tan ^2 y=1 \\ & \sec ^2 y-\tan ^2 z=1 \end{aligned}

Adding both

\begin{aligned} & \Rightarrow \sec ^2 x-\tan ^2 y+\sec ^2 y-\tan ^2 z=1+1 \\ & \sec ^2 x+1-\tan ^2 z=2 \\ & \sec ^2 x-\tan ^2 z=1 \\ & \Rightarrow \text{ Transitive } \end{aligned}

hence equivalence relation

Q46

Let R1 = {(a, b)

\in

N

\times

N : |a

-

b|

\le

13} and R2 = {(a, b)

\in

N

\times

N : |a

-

b|

\ne

13}. Then on N :

A Both R1 and R2 are equivalence relations

B Neither R1 nor R2 is an equivalence relation

C R1 is an equivalence relation but R2 is not

D R2 is an equivalence relation but R1 is not

Correct Answer

Option B

Solution

$R_{1}=\{(a, b) \in N \times N:|a-b| \leq 13\}$ and $R_{2}=\{(a, b) \in N \times N:|a-b| \neq 13\}$ In $R_{1}: \because|2-11|=9 \leq 13$ $\therefore \quad(2,11) \in R_{1}$ and $(11,19) \in R_{1}$ but $(2,19) \notin R_{1}$ $\therefore \quad R_{1}$ is not transitive Hence $R_{1}$ is not equivalence In $R_{2}:(13,3) \in R_{2}$ and $(3,26) \in R_{2}$ but $(13,26) \notin R_{2}$ $(\because|13-26|=13)$ $\therefore R_{2}$ is not transitive Hence $R_{2}$ is not equivalence.

Q47

Let

\mathrm{A}=\{1,3,4,6,9\}

and

\mathrm{B}=\{2,4,5,8,10\}

. Let

\mathrm{R}

be a relation defined on

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

such that

\mathrm{R}=\left\{\left(\left(a_{1}, b_{1}\right),\left(a_{2}, b_{2}\right)\right): a_{1} \leq b_{2}\right.

and

\left.b_{1} \leq a_{2}\right\}

. Then the number of elements in the set R is :

A 180

B 26

C 52

D 160

Correct Answer

Option D

Solution

Given that the sets are $A = \{1, 3, 4, 6, 9\}$ and $B = \{2, 4, 5, 8, 10\}$ , for the relation $\mathrm{R}$ on the set $A \times B$ , we need to find the combinations of pairs that satisfy the conditions $a_1 \leq b_2$ and $b_1 \leq a_2$ .

We find the number of combinations by considering the possible values for $b_2$ for each $a_1$ and the possible values for $a_2$ for each $b_1$ : For each $a_1$ in $A = \{1, 3, 4, 6, 9\}$ , the number of valid $b_2$ values in $B = \{2, 4, 5, 8, 10\}$ are : - For $a_1 = 1$ , there are 5 choices for $b_2$ . - For $a_1 = 3$ , there are 4 choices for $b_2$ . - For $a_1 = 4$ , there are 4 choices for $b_2$ . - For $a_1 = 6$ , there are 2 choices for $b_2$ . - For $a_1 = 9$ , there is 1 choice for $b_2$ .

This results in a total of $5+4+4+2+1 = 16$ possible pairs $(a_1, b_2)$ .

Similarly, for each $b_1$ in $B$ , the number of valid $a_2$ values in $A$ are : - For $b_1 = 2$ , there are 4 choices for $a_2$ . - For $b_1 = 4$ , there are 3 choices for $a_2$ . - For $b_1 = 5$ , there are 2 choices for $a_2$ . - For $b_1 = 8$ , there is 1 choice for $a_2$ . - For $b_1 = 10$ , there are no choices for $a_2$ .

This results in a total of $4+3+2+1+0 = 10$ possible pairs $(b_1, a_2)$ .

Therefore, the total number of elements in the relation $\mathrm{R}$ , which satisfies the given conditions, is $16 \times 10 = 160$ .

So, the correct answer is 160.

Q48

Let

R

be a relation on

Z \times Z

defined by

(a, b) R(c, d)

if and only if

a d-b c

is divisible by 5. Then

R

is

A Reflexive and transitive but not symmetric

B Reflexive and symmetric but not transitive

C Reflexive but neither symmetric nor transitive

D Reflexive, symmetric and transitive

Correct Answer

Option B

Solution

(a, b) R(a, b)

as

a b-a b=0

Therefore reflexive Let

(a, b) R(c, d) \Rightarrow a d-b c

is divisible by 5

\Rightarrow \mathrm{bc}-\mathrm{ad}

is divisible by

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

Therefore symmetric Relation not transitive as

(3,1) \mathrm{R}(10,5)

and

(10,5) \mathrm{R}(1,1)

but

(3,1)

is not related to

(1,1)

Q49

Let

S=\{1,2,3, \ldots, 10\}

. Suppose

M

is the set of all the subsets of

S

, then the relation

\mathrm{R}=\{(\mathrm{A}, \mathrm{B}): \mathrm{A} \cap \mathrm{B} \neq \phi ; \mathrm{A}, \mathrm{B} \in \mathrm{M}\}

is :

A symmetric only

B reflexive only

C symmetric and reflexive only

D symmetric and transitive only

Correct Answer

Option A

Solution

Let

S=\{1,2,3, \ldots, 10\}

R=\{(A, B): A \cap B \neq \phi ; A, B \in M\}

For Reflexive,

M

is subset of '

S

' So

\phi \in \mathrm{M}

for

\phi \cap \phi=\phi

$\Rightarrow$ but relation is

\mathrm{A} \cap \mathrm{B} \neq \phi

So it is not reflexive. For symmetric,

\begin{array}{ll} \text{ ARB } & \mathrm{A} \cap \mathrm{B} \neq \phi, \\ \Rightarrow \mathrm{BRA} & \Rightarrow \mathrm{B} \cap \mathrm{A} \neq \phi, \end{array}

So it is symmetric. For transitive,

\begin{aligned} \text{ If } A & =\{(1,2),(2,3)\} \\ B & =\{(2,3),(3,4)\} \\ C & =\{(3,4),(5,6)\} \end{aligned}

\mathrm{ARB}

&

\mathrm{BRC}

but

\mathrm{A}

does not relate to

\mathrm{C}

So it not transitive

Q50

Let

A

and

B

be two finite sets with

m

and

n

elements respectively. The total number of subsets of the set

A

is 56 more than the total number of subsets of

B

. Then the distance of the point

P(m, n)

from the point

Q(-2,-3)

is :

A 8

B 10

C 4

D 6

Correct Answer

Option B

Solution

\begin{aligned} & 2^{\mathrm{m}}-2^{\mathrm{n}}=56 \\ & 2^{\mathrm{n}}\left(2^{\mathrm{m}-\mathrm{n}}-1\right)=2^3 \times 7 \\ & 2^{\mathrm{n}}=2^3 \text{ and } 2^{\mathrm{m}-\mathrm{n}}-1=7 \\ & \Rightarrow \mathrm{n}=3 \text{ and } 2^{\mathrm{m}-\mathrm{n}}=8 \\ & \Rightarrow \mathrm{n}=3 \text{ and } \mathrm{m}-\mathrm{n}=3 \\ & \Rightarrow \mathrm{n}=3 \text{ and } \mathrm{m}=6 \\ & \mathrm{P}(6,3) \text{ and } \mathrm{Q}(-2,-3) \\ & \mathrm{PQ}=\sqrt{8^2+6^2}=\sqrt{100}=10 \end{aligned}

Hence option (1) is correct