3 Marks Question

Question 13 Marks

Let $\text{T}=\Big\{\text{x}|\frac{\text{x}+5}{\text{x}-7}-5=\frac{4\text{x}-40}{13-\text{x}}\Big\}.$ Is T an empty set? Justify your answer.

Answer

Given that: $\text{T}=\Big\{\text{x}\Big|\frac{\text{x}+5}{\text{x}-7}-5=\frac{4\text{x}-40}{13-\text{x}}\Big\}$
$\Rightarrow\frac{\text{x}+5}{\text{x}-7}-5=\frac{4\text{x}-40}{13-\text{x}}$
$\Rightarrow \frac{(\text{x}+5)-5(\text{x}-7)}{\text{x}-7}=\frac{4\text{x}-40}{13-\text{x}}$
$\Rightarrow \frac{\text{x}5-5\text{x}+35}{(\text{x}-7)}=\frac{4\text{x}-40}{13-\text{x}}$
$\Rightarrow \frac{-4\text{x}+40}{(\text{x}-7)}=\frac{4\text{x}-40}{13-\text{x}}$
$\Rightarrow -4(\text{x}-10)(12-\text{x})=4(\text{x}-10)(\text{x}-7)$
$\Rightarrow -4(\text{x}-10)(13-\text{x}+\text{x}-7)=0$
$\Rightarrow -4(\text{x}-10)6=0$
$\Rightarrow -24(\text{x}-10)=0$
$\Rightarrow \text{x}-10=0$
$\Rightarrow \text{x}=10$
$\therefore \text{T}={10}$
Hence, T is not an empty set.

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Question 23 Marks

If A and B are subsets of the universal set U, then show that.
$\text{A} \subset \text{A} \Leftrightarrow \text{A}\cap\text{B}=\text{B}$

Answer

If $\text{A}\subset\text{B}$
Then let $\text{x}\in\text{A}\cup\text{B}$
$\Rightarrow \text{x}\in\text{A}$ or $\text{x}\in\text{B}$
$\Rightarrow \text{x}\in\text{B}\ \big[\because \text{A}\subset\text{B}\big]$
$\Rightarrow \text{A}\cup\text{B}\subset\text{B}\ .....(\text{i})$
But $ \text{B}\subset\text{A}\cup\text{B}\ .....(\text{ii})$
From eqn. (i) and (ii), we get
$\text{A}\cup\text{B}=\text{B}.$
Now if $\text{A}\cup\text{B}=\text{B}$
Let $\text{y}\in\text{A}\Rightarrow\text{y}\in(\text{A}\cup\text{B})$
$ \Rightarrow \text{y}\in\text{B} \big[\because \text{A}\cup\text{B}=\text{B}\big]$
Hence $\text{A}\subset\text{B}$
So we proved $\text{A}\subset\text{B} \Leftrightarrow \text{A}\cup\text{B}=\text{B}$

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