Question

answer this with steps

Answer 1

Answer:

I think the answer is option c

Answer 2

SOLUTION :

GIVEN

Let N be the set of natural numbers

$\sf{A = \{ \: {n}^{2} : n \in N \: \} }$

$\sf{B = \{ \: {n}^{3} : n \in N \: \} }$

TO CHOOSE THE CORRECT OPTION

$\sf{}(a)\: \: A \cup \: B = N$

$\sf{}(b) \: The \: Complement \: of \: A \cup \: B \: \: is \: infinite$

$\sf{} (c) \: \: A \cap \: B \: \: must \: be \: finite \: \: set$

$\sf{} (d)\: A \cap \: B \: is \: a \:proper \: subset \: of \: \{ \: {m}^{6} : m \in \: N \: \}$

EVALUATION

$\sf{A = \{ \: {n}^{2} : n \in N \: \} }$

$= \{1, 4,9,16,25,...... \}$

$\sf{B = \{ \: {n}^{3} : n \in N \: \} }$

$= \{1, 8,27,64,125,...... \}$

CHECKING FOR OPTION : (a)

$\sf{}A \cup \: B = \{1,4,8,9,26,27,........ \}$

$\sf{}Here \: 2 \in N \: \: but \: \: 2 \notin A \cup \: B$

$\sf{} \therefore \: \: A \cup \: B \ne N$

So this option is not correct

CHECKING FOR OPTION : (b)

$\sf{}A \cup \: B = \{1,4,8,9,26,27,........ \}$

So

$\sf{}{(A \cup \: B)}^{c} = \{2,3,5,6,7,10,........ \}$

This is an infinite set

$\therefore\sf{}The \: Complement \: of \: A \cup \: B \:is \: infinite$

So this option is CORRECT

CHECKING FOR OPTION : (c)

$\sf{}A \cap \: B = \{1,64,729,........ \}$

This is an infinite set

$\therefore \: \sf{}A \cap \: B \: can \: not\: be \: finite \: \: set$

So this option is not correct

CHECKING FOR OPTION : (d)

$\sf{}A \cap \: B = \{1,64,729,........ \}$

$\sf{} A \cap \: B \: = \{ \: {m}^{6} : m \in \: N \: \}$

$\sf{} \therefore A \cap \: B \: is \: not\: a \:proper \: subset \: of \: \{ \: {m}^{6} : m \in \: N \: \}$

Infact they are equal sets

So this option is not correct

FINAL RESULT

Let N be the set of natural numbers

$\sf{A = \{ \: {n}^{2} : n \in N \: \} }$

$\sf{B = \{ \: {n}^{3} : n \in N \: \} }$

Then

$\sf{}(b) \: The \: Complement \: of \: A \cup \: B \: \: is \: infinite$

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LEARN MORE FROM BRAINLY

If P = {x:x < 7, x ∈ N} and Q ={x: x²< 49,x ∈ N}

then

P ⊂ Q

Q ⊂ P

P = Q

P∪{7} = Q

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