Solve the Following Question.(5 Marks)

Question 15 Marks

For any two sets A and B, prove the following:
$\text{A}\cap\text{(A}'\cup\text{B})=\text{A}\cap\text{B}$

Answer

$\text{LHS}=\text{A}\cap\text{(A}'\cup\text{B)}$
$=\text{(A}\cap\text{A}')\cup\text{(A}\cap\text{B)}$ $[\because\cap$ distributes over (i)$]$
$= \oint\cup\text{ (A}\cap\text{B)}$ $[\because\text{A}\cap\text{A}' = \oint]$
$=\text{A}\cap\text{B}$ $[\because\oint\cup$ x = x for any set x$]$
$= \text{RHS}$
$\therefore$ LHS = RHS Proved.

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Question 25 Marks

For any two sets A and B, prove that
$\text{A}\cup(\text{B}-\text{A})=\text{A}\cup\text{B}$

Answer

Let $\text{x}\in \text{A} \cup \text{(B} - \text{A)}\Rightarrow\text{x}\in\text{A or x}\in\text{(B} - \text{A)}$
$\Rightarrow \text{x} \in \text{A or x} \in \text{B and x}\not\in \text{A}$
$\Rightarrow \text{x}\in \text{B}$
$\Rightarrow \text{x}\in \text{A}\cup\text{B}$ $[\because \text{B}\subset\text{(A}\cup\text{B})]$
This is true for all $\text{x}\in\text{A}\cup\text{(B} - \text{A)}$
$\therefore \text{A}\cup\text{(B} - \text{A)}\subset\text{(A}\cup \text{B})....\text{(i)}$
Conversely,
Let, $\text{x}\in \text{(A}\cup\text{B)}$
$\Rightarrow\text{x} \in \text{A or x} \in\text{B}$
$\Rightarrow \text{x}\in \text{A or x} \in \text{(B – A)}$ $[\because\text{B}\subset \text{(B – A)]}$
$\therefore\text{(A}\cup\text{B}) \subset \text{A}\cup \text{(B}- \text{A)}.....\text{(ii)}$
From (i) and (ii), we get
$\text{A}\cup\text{(B} - \text{A) = (A}\cup\text{B}).$

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Question 35 Marks

For any two sets A and B, prove the following:
$\text{A}\cap\text{(A}\cup\text{B})=\phi$

Answer

$\text{LHS}=\text{A}\cap\text{(A}\cup\text{B}')$
$=\text{A}\cap\text{(A}'\cap\text{B}')$ [By De-morgan's law]
$=\text{(A}\cap\text{A}')\cap\text{B}'$ [By associative law]
$= \oint \cap \text{ B}'$ $[\because \text{A}\cap{A}' =\oint]$
$= \oint$
$= \text{RHS}$
$\therefore$ LHS = RHS Proved.

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Question 45 Marks

For any two sets A and B, prove that
$(\text{A}\cap\text{B})-\text{B = A}- \text{B}$

Answer

$\text{(A} \cup\text{B) – B = (A} - \text{B)}\cup\text{(A} - \text{B)}$
$= \oint\cap \text{ (A} - \text{B)}$
$= \text{A} - \text{B}.$

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Question 55 Marks

For any two sets A and B, prove the following:
$\text{A}-\text{(A}-\text{B})=\text{A}\cap\text{B}$

Answer

For any sets A and B we have by De-morgan's laws
$\text{(A}\cup\text{B)}' = \text{A}'\cap\text{B}',\text{(A}\cap\text{B)}' = \text{A}'\cup\text{B}'$
Also
$\text{LHS}= \text{A} - \text{(A} - \text{B)}$
$=\text{A}\cap\text{(A}- \text{B)}'$
$=\text{A}\cap\text{(A}\cap\text{B}')'$
$=\text{A}\cap\text{(A}\cap\text{(B})')'$ [By De-morgan's law]
$=\text{A}\cap\text{(A}'\cup\text{B})$ $[\because\text{(B}')'=\text{B}]$
$=\text{(A}\cap\text{A}')\cup\text{(A}\cap\text{B})$
$=\oint\cup\text{(A}\cap\text{B})$ $[\because\text{A}\cap\text{A}'=\oint]$
$=\text{A}\cap\text{B}$ $[\because\oint\cup$ x = x, for and set x$]$
$= \text{RHS}$
$\therefore$ LHS = RHS Proved.

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Question 65 Marks

For any two sets A and B, prove that
$\text{(A}-\text{B)}\cup(\text{B}\cap\text{A})=\text{A}$

Answer

Let $\text{x}\in\text{A.}$
Then either $\text{x} \in \text{(A}- \text{B) or x} \in \text{(A}\cap \text{B})$
$\Rightarrow \text{x}\in\text{(A – B)}\cup \text{(A} \cap\text{B})$
$\therefore\text{A} \subset \text{(A – B)}\cup \text{(A}\cap\text{B}).....\text{(i)}$
Conversely,
Let $\text{x}\in\text{(A} - \text{B)} \cup \text{(A}\cap\text{B)}$
$\Rightarrow\text{x} \in \text{(A – B) or x (A} \cap \text{B})$
$\Rightarrow\text{x} \in \text{A and x}\not\in\text{B or x}\in \text{A and x} \in\text{B}$
$\Rightarrow \text{x} \in \text{A}$
$\therefore\text{(A} - \text{B)}\cup\text{(A} \cap \text{B}) \subset \text{A}.....\text{(ii)}$
From (i) and (ii), we get
$\text{(A} - \text{B)}\cup\text{(A} \cap \text{B) = A}.$

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Question 75 Marks

For any two sets A and B, prove the following:
$\text{A} - \text{B}=\text{A }\Delta\text{ (A}\cap\text{B})$

Answer

$\text{RHS}=\text{A }\Delta\text{ (A} \cap8)$ $=\text{(A} - \text{(A}\cap\text{B}))\cup\text{(A}\cap\text{B} - \text{A})$ $[\because\text{E }\Delta\text{ F} =\text{(E} - \text{F)}\cup\text{(F} - \text{E)]}$ $=\text{(A}\cap\text{(A}\cap\text{B})')\cup\text{(A}\cap\text{B}\cap\text{A}')$ $[\because \text{E - F = E }\cap \text{F}']$ $=\text{(A}\cap \text{(A}'\cup\text{B}'))\cup\text{(A}\cap\text{A}'\cap\text{B})$ [By de-morgan's law & associative law] $=\text{(A}\cap\text{A}')\cup(\text{A}\cap\text{B}')\cup(\oint\cap\text{B})$ $[\because\cap\text{ distributes over}\cup\text{and A }\cap \text{A}' = \oint]$ $= \oint \cup \text{(A}\cap\text{B}')\cup \oint$ $[\because \oint \cap \text{ B} = \oint]$ $= \text{A}\cap\text{B}'$ $[\because\oint\cup\text{ x = x for any set x}]$ $= \text{A} - \text{B }$ $[\because \text{A} \cap\text{B}' = \text{A - B}]$ $=\text{LHS}$ $\therefore$ LHS = RHS Proved.

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Maths Solve the Following Question.(5 Marks)

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