Question

Solve the question in the attachment!!​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that,

$\rm :\longmapsto\:A \: = \:\{x\} \:$

and

$\rm :\longmapsto\:B \: = \:\{y\} \:$

$\rm\implies \:A \: \cap \: B = \phi$

and

$\rm :\longmapsto\:A\cup B = \{x, \: y\}$

We know,

Power set of a set A is defined as set of subsets of A.

So,

$\rm :\longmapsto\:P(A) = \{ \phi ,\: x\}$

and

$\rm :\longmapsto\:P(B) = \{ \phi ,\: y\}$

and

$\rm\implies \:P(A \: \cap \: B) =\{ \phi \}$

and

$\rm :\longmapsto\:P(A\cup B) = \{\phi ,x,y,xy\}$

Now, Consider

$\rm :\longmapsto\:P(A)\cup P(B) = \{\phi ,x,y\}$

and

$\rm :\longmapsto\:P(A\cup B) = \{\phi ,x,y,xy\}$

$\rm\implies \:\boxed{\tt{ \: P(A)\cup P(B) \: \ne \: P(A\cup B) \: }}$

Now, Consider

$\rm :\longmapsto\:P(A\cap B) = \{\phi \}$

and

$\rm :\longmapsto\:P(A) \: \cap \: P( B) = \{\phi \}$

$\rm\implies \:\boxed{\tt{ \: P(A)\cap P(B) \: = \: P(A\cap B) \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

1. COMMUTATIVE LAW

$\rm :\longmapsto\:\boxed{\tt{ A\cup B = B\cup A}} \\ \\ \rm :\longmapsto\:\boxed{\tt{ A\cap B = B\cap A}}$

2. ASSOCIATIVE LAW

$\rm :\longmapsto\:\boxed{\tt{ (A\cup B)\cup C = A\cup (B\cup C)}} \\ \\ \rm :\longmapsto\:\boxed{\tt{ (A\cap B)\cap C = A\cap (B\cap C)}}$

3. DISTRIBUTIVE LAW

$\boxed{\tt{ A\cup (B\cap C) = (A\cup B)\cap (A\cup C)}} \\ \\ \boxed{\tt{ A\cap (B\cup C) = (A\cap B)\cup (A\cap C)}}$

Answer 2

$\large\underline{\sf{Solution-}}$

Let assume that,

$\rm :\longmapsto\:A \: = \:\{x\} \:$

and

$\rm :\longmapsto\:B \: = \:\{y\} \:$

$\rm\implies \:A \: \cap \: B = \phi$

and

$\rm :\longmapsto\:A\cup B = \{x, \: y\}$

We know,

Power set of a set A is defined as set of subsets of A.

So,

$\rm :\longmapsto\:P(A) = \{ \phi ,\: x\}$

and

$\rm :\longmapsto\:P(B) = \{ \phi ,\: y\}$

and

$\rm\implies \:P(A \: \cap \: B) =\{ \phi \}$

and

$\rm :\longmapsto\:P(A\cup B) = \{\phi ,x,y,xy\}$

Now, Consider

$\rm :\longmapsto\:P(A)\cup P(B) = \{\phi ,x,y\}$

and

$\rm :\longmapsto\:P(A\cup B) = \{\phi ,x,y,xy\}$

$\rm\implies \:\boxed{\tt{ \: P(A)\cup P(B) \: \ne \: P(A\cup B) \: }}$

Now, Consider

$\rm :\longmapsto\:P(A\cap B) = \{\phi \}$

and

$\rm :\longmapsto\:P(A) \: \cap \: P( B) = \{\phi \}$

$\rm\implies \:\boxed{\tt{ \: P(A)\cap P(B) \: = \: P(A\cap B) \: }}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO KNOW

1. COMMUTATIVE LAW

$\rm :\longmapsto\:\boxed{\tt{ A\cup B = B\cup A}} \\ \\ \rm :\longmapsto\:\boxed{\tt{ A\cap B = B\cap A}}$

2. ASSOCIATIVE LAW

$\rm :\longmapsto\:\boxed{\tt{ (A\cup B)\cup C = A\cup (B\cup C)}} \\ \\ \rm :\longmapsto\:\boxed{\tt{ (A\cap B)\cap C = A\cap (B\cap C)}}$

3. DISTRIBUTIVE LAW

$\boxed{\tt{ A\cup (B\cap C) = (A\cup B)\cap (A\cup C)}} \\ \\ \boxed{\tt{ A\cap (B\cup C) = (A\cap B)\cup (A\cap C)}}$