Sets and Relations

Given,

Total number of subsets of set $(A \times B)=2^8$ Number of subsets of set $A \times B$ having no element (i.e.

$\phi)=1$ Number of subsets of set $A \times B$ having one element $={ }^8 C_1$ Number of subsets of set $A \times B$ having two elements $={ }^8 C_2$ $\therefore$ Number of subsets having atleast three elements

An equivalence relation on a set must satisfy three properties: reflexivity (every element is related to itself), symmetry (if an element is related to a second, the second is related to the first), and transitivity (if a first element is related to a second, and the second is related to a third, then the first is related to the third).

Statement I : $A={(x, y) \in R \times R: }$ y-x is an integer .

Reflexivity : For all $x$ in $R$, $x - x = 0$ which is an integer.

So, every element is related to itself.

Symmetry : For all $x, y$ in $R$, if $y - x$ is an integer, then $x - y = - (y - x)$ is also an integer.

So, if $x$ is related to $y$, then $y$ is related to $x$.

Transitivity : For all $x, y, z$ in $R$, if $y - x$ and $z - y$ are integers, then $(z - y) + (y - x) = z - x$ is also an integer.

So, if $x$ is related to $y$ and $y$ is related to $z$, then $x$ is related to $z$.

Therefore, $A$ is an equivalence relation on $R$.

Statement II : $B={(x, y) \in R \times R: x=\alpha y}$ for some rational number $\alpha$.

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 $x = \alpha y$ for some rational $\alpha$, then $y = \dfrac{1}{\alpha}x$.

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

Transitivity : If $x = \alpha y$ and $y = \beta z$ for some rational numbers $\alpha$ and $\beta$, then $x = (\alpha \beta)z$.

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

In conclusion, the correct answer is Option B : Statement I is true, Statement II is false.

For any element xi present in X, 4 cases arises while making subsets Y and Z. Case- 1 : xi

Y, xi

Z $\Rightarrow$ Y $\cap$ Z $\ne$ $\phi$ Case- 2 : xi

Y, xi $\notin$ Z $\Rightarrow$ Y $\cap$ Z = $\phi$ Case- 3 : xi $\notin$ Y, xi

Z $\Rightarrow$ Y $\cap$ Z = $\phi$ Case- 4 : xi $\notin$ Y, xi $\notin$ Z $\Rightarrow$ Y $\cap$ Z = $\phi$ $\therefore$ For every element, number of ways = 3 for which Y $\cap$ Z = $\phi$ $\Rightarrow$ Total ways = 3 × 3 × 3 × 3 × 3 [$\because$ no. of elements in set X = 5] = 35

Here both R1 and R2 are symmetric as for any (x, y)

R1, we have (y, x)

R1 and similarly for any (x, y)

R2, we have (y, x)

R2 In R1, (b, c)

R1, (c, a)

R1 but (b,a) $\notin$ R1 Similarly in R2, (b, a)

R2, (a, c)

R2 but (b, c) $\notin$ R2 $\therefore$ R1 and R2 are not transitive.

S = {1,2,3, . . . .100} = Total non empty subsets-subsets with product of element is odd = 2100 $-$ 1 $-$ 1[(250 $-$ 1)] = 2100 $-$ 250 = 250 (250 $-$ 1)

C $\to$ person like coffee T $\to$ person like Tea n(C) = 73 n(T) = 65 n(C $\cup$ T) $\le$ 100 n(C) + n(T) – n (C $\cap$ T) $\le$ 100 73 + 65 – x $\le$ 100 x $\ge$ 38 73 – x $\ge$ 0 $\Rightarrow$ x $\le$ 73 65 – x $\ge$ 0 $\Rightarrow$ x $\le$ 65 $\therefore$ 38 $\le$ x $\le$ 65

A : x

(–2, 2); B : x

(–$\infty$, –1] $\cup$ [5, $\infty$) $\Rightarrow$ B – A = R – (–2, 5)

X1, X2,....., X50 = 50 sets. Given each sets having 10 elements. So total elements = 50 $\times$ 10

\times

{{50 \times 10} \over {20}}

{{50 \times 10} \over {20}}

\therefore

{{50 \times 10} \over {20}}$=${{5 \times n} \over {20}}

\Rightarrow $$ n = 30

ad = bc (a, b) R (4, 3) $\Rightarrow$ 3a = 4b a =

b b must be multiple of 3 b = {3, 6, 9 ..... 30} (a, b) = {(4, 3), (8, 16), (12, 9), (16, 12), (20, 15), (24, 18), (28, 21)} $\Rightarrow$ 7 ordered pair