[4 marks sum]

Question 14 Marks

Draw a line segment of length $\sqrt{5} \mathrm{~cm}$.

Answer

Construct a right$-$angled triangel $\mathrm{OAB}$ with

$ \mathrm{OA}=2 \mathrm{~cm}$
$ \angle \mathrm{OAB}=90^{\circ}$ and
$ \mathrm{AB}=1 \mathrm{~cm}$
Using $ \mathrm{OB}^2=\mathrm{OA}^2+\mathrm{AB}^2$
$ \mathrm{OB}^2=2^2+1^2$
$ \mathrm{OB}^2=4+1$
$ \mathrm{OB}^2=5$
$ \mathrm{OB}=\sqrt{5}$

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Question 24 Marks

Evaluate, correct to one place of decimal, the expression$\frac{5}{\sqrt{20}-\sqrt{10}}$, if $\sqrt{5}=2.2$ and $\sqrt{10}=3.2$

Answer

given:$\frac{5}{\sqrt{20}-\sqrt{10}}$
$=\frac{5}{\sqrt{20}-\sqrt{10}} \times \frac{\sqrt{20}+\sqrt{10}}{\sqrt{20}+\sqrt{10}}$
$=\frac{5(\sqrt{20}+\sqrt{10})}{20-10}$
$=\frac{5^1(\sqrt{20}+\sqrt{10})}{10}$
$=\frac{\sqrt{20}+\sqrt{10}}{2}$
$=\frac{\sqrt{4 \times 5}+\sqrt{10}}{2}$
$=\frac{2 \sqrt{5}+\sqrt{10}}{2}$
$=\frac{2(2.2)+3.2}{2}$
$=\frac{4.4+3.2}{2}$
$=\frac{7.6}{2}$
$=3.8$

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Question 34 Marks

Show that: $x^3+\frac{1}{x^3}=52$, if $\mathrm{x}=2+\sqrt{3}$

Answer

Given: $x=2+\sqrt{ } 3$
$\frac{1}{x}=\frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}$
$=\frac{2-\sqrt{3}}{(2)^2-(\sqrt{3})^2}$
$=\frac{2-\sqrt{3}}{4-3}$
$=2-\sqrt{3}$
Now,
$x+\frac{1}{2}=2+\sqrt{3}+2-\sqrt{3}$
$x+\frac{1}{x}=2+2$
$x+\frac{1}{x}=4$
$\therefore x^3+\frac{1}{x^3}$
$=\left(x+\frac{1}{x}\right)^3-3 \cdot \not x \cdot \frac{1}{\not x}\left(x+\frac{1}{x}\right)$
$=(4)^3-3 \times 4$
$=64-12$
$=52$

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Question 44 Marks

If $x=1-\sqrt{2 } $, find the value of $\left(x-\frac{1}{x}\right)^3$

Answer

Given that $x=1-\sqrt{2 } $
We need to find the value of $\left(x-\frac{1}{x}\right)^3$
Since $x=1-\sqrt{2}$,
we have
$\frac{1}{x}=\frac{1}{1-\sqrt{2}} \times \frac{1+\sqrt{2}}{1+\sqrt{2}}$
$\Rightarrow \frac{1}{x}=\frac{1+\sqrt{2}}{(1)^2-(\sqrt{2})^2} \quad\left[\right.$ Since $\left.(a-b)(a+b)=a^2-b^2\right]$
$\begin{gathered}\Rightarrow \frac{1}{x}=\frac{1+\sqrt{2}}{1-2}\Rightarrow \frac{1}{x}=\frac{1+\sqrt{2}}{-1}\end{gathered}$
$\Rightarrow \frac{1}{x}=-(1+\sqrt{2}) .....(1)$
Thus, $ \left(x-\frac{1}{x}\right)=(1-\sqrt{ 2} )-(-(1+\sqrt{2}))$
$ \Rightarrow\left(x-\frac{1}{x}\right)=1-\sqrt{2 } +1+\sqrt{ 2} $
$ \Rightarrow\left(x-\frac{1}{x}\right)=2$
$ \Rightarrow\left(x-\frac{1}{x}\right)^3=2^3$
$ \Rightarrow\left(x-\frac{1}{x}\right)^3=8$

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Question 54 Marks

Show that:$\frac{1}{3-2 \sqrt{ } 2}-\frac{1}{2 \sqrt{ } 2-\sqrt{ } 7}+\frac{1}{\sqrt{ } 7-\sqrt{ } 6}-\frac{1}{\sqrt{ } 6-\sqrt{ } 5}+\frac{1}{\sqrt{ } 5-2}=5$

Answer

$ \text { L.H.S }$
$= \frac{1}{3-2 \sqrt{2 } }-\frac{1}{2 \sqrt{2 } -\sqrt{7 } }+\frac{1}{\sqrt{7 } -\sqrt{ 6} }-\frac{1}{\sqrt{6 } -\sqrt{ 5} }+\frac{1}{\sqrt{ 5} -2}$
$ =\frac{1}{3-\sqrt{8 } }-\frac{1}{\sqrt{ 8} -\sqrt{ 7} }+\frac{1}{\sqrt{ 7 }-\sqrt{6 } }-\frac{1}{\sqrt{6 } -\sqrt{ 5} }+\frac{1}{\sqrt{ 5} -2}$
$=\frac{1}{3-\sqrt{8 } } \times \frac{3+\sqrt{ 8} }{3+\sqrt{ 8} }-\frac{1}{\sqrt{ 8} -\sqrt{ 7} } \times \frac{\sqrt{8 } +\sqrt7{ } }{\sqrt{8 } +\sqrt{ 7} }+\frac{1}{\sqrt{ 7} -\sqrt{6 } } \times \frac{\sqrt{ 7} +\sqrt{6 } }{\sqrt{7 } +\sqrt{ 6} }$$-\frac{1}{\sqrt{ 6} -\sqrt{5 } } \times \frac{\sqrt{ 6} +\sqrt{5 } }{\sqrt{6 } +\sqrt{ 5} }+\frac{1}{\sqrt{5 } -2} \times \frac{\sqrt{ 5} +2}{\sqrt{5 } +2}$
$=\frac{3+\sqrt{8 } }{(3)^2-(\sqrt{ 8} )^2}-\frac{\sqrt{8 } +\sqrt{7 } }{(\sqrt{8 } )^2-(\sqrt{7 } )^2}+\frac{\sqrt{7 } +\sqrt{ 6} }{(\sqrt{ 7} )^2-(\sqrt{ 6} )^2}-\frac{\sqrt{ 6} -\sqrt{5 } }{(\sqrt{6 } )^2-(\sqrt{5 } )^2}$$+\frac{\sqrt{5 } -2}{(\sqrt{5 } )^2-(2)^2}$
$ =\frac{3+\sqrt{ 8} }{9-8}-\frac{\sqrt{ 8} +\sqrt{ 7} }{8-7}+\frac{\sqrt{7 } +\sqrt{6 } }{7-6}-\frac{\sqrt{6 } -\sqrt{ 5} }{6-5}+\frac{\sqrt{5 } -2}{5-4}$
$ =3+\sqrt{ 8} -\sqrt{8 } -\sqrt{7 } +\sqrt{ 7} +\sqrt{ 6} -\sqrt{6 } -\sqrt{5 } +\sqrt{5 } +2$
$ =3+2$
$ =5$
$ =\text { R.H.S. }$

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Question 64 Marks

Given universal set $=$$\left\{-6,-5 \frac{3}{4},-\sqrt{4},-\frac{3}{5},-\frac{3}{8}, 0, \frac{4}{5}, 1,1 \frac{2}{3}, \sqrt{8}, 3.01, \pi, 8.47\right\}$From the given set, find : set of rational numbers

Answer

Given Universal set is
$\left\{-6,-5 \frac{3}{4},-\sqrt{4},-\frac{3}{5},-\frac{3}{8}, 0, \frac{4}{5}, 1,1 \frac{2}{3}, \sqrt{8}, 3.01, \pi, 8.47\right\}$
We need to find the set of rational numbers.
Rational numbers are numbers of the form $\frac{P}{q}$, where $\mathrm{q} \neq 0$.
$U=\left\{-6,-5 \frac{3}{4},-\sqrt{4},-\frac{3}{5},-\frac{3}{8}, 0, \frac{4}{5}, 1,1 \frac{2}{3}, \sqrt{8}, 3.01, \pi, 8.47\right\}$
Clearly, $-5 \frac{3}{4},-\frac{3}{5},-\frac{3}{8}, \frac{4}{5}$ and $1 \frac{2}{3}$ are of the form $\frac{P}{q}$.
Hence, they are rational numbers.
Since the set of integers is a subset of rational numbers, $-6,0$ and $1$ are also rational numbers.
Thus, decimal numbers $3.01$ and $8.47$ are also rational numbers.
because they are terminating decimals.
Hence, from the above set, the set of rational numbers is $Q$,
and
$Q=\left\{-6,-5 \frac{3}{4},-\sqrt{4},-\frac{3}{5},-\frac{3}{8}, 0, \frac{4}{5}, 1,1 \frac{2}{3}, 3.01,8.47\right\}$

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Question 74 Marks

Show that: $\frac{4-\sqrt{5}}{4+\sqrt{5}}+\frac{2}{5+\sqrt{3}}+\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{2}{5-\sqrt{3}}=\frac{52}{11}$

Answer

$\frac{4-\sqrt{5}}{4+\sqrt{5}}+\frac{2}{5+\sqrt{3}}+\frac{4+\sqrt{5}}{4-\sqrt{5}}+\frac{2}{5-\sqrt{3}}$
$=\frac{4-\sqrt{5}}{4+\sqrt{5}} \times \frac{4-\sqrt{5}}{4+\sqrt{5}}+\frac{2}{5+\sqrt{3}} \times \frac{5-\sqrt{3}}{5-\sqrt{3}}+\frac{4+\sqrt{5}}{4-\sqrt{5}} \times \frac{4+\sqrt{5}}{4+\sqrt{5}}+\frac{2}{5-\sqrt{3}} \times \frac{5+\sqrt{3}}{5+\sqrt{3}}$
$=\frac{(4-\sqrt{5})^2}{(4)^2-(\sqrt{5})^2}+\frac{2(5-\sqrt{3})}{(5)^2-(\sqrt{3})^2}+\frac{(4+\sqrt{5})^2}{(4)^2-(\sqrt{5})}+\frac{2(5+\sqrt{3})}{(5)^2-(\sqrt{3})^2}$
$=\frac{16+5-8 \sqrt{5}}{16-5}+\frac{10-2 \sqrt{3}}{25-3}+\frac{16+5+8 \sqrt{5}}{16-5}+\frac{2(5+\sqrt{3})}{25-3}$
$=\frac{21-8 \sqrt{5}}{11}+\frac{10-2 \sqrt{3}}{22}+\frac{21+8 \sqrt{5}}{11}+\frac{\not 2^1(5+\sqrt{3})}{\not 22_{11}}$
$=\frac{21- \not8 \not \sqrt{5}}{11}+\frac{\not 2^1(5-\sqrt{3})}{\not 22_{11}}+\frac{21+8 \sqrt{5}}{11}+\frac{5+\sqrt{\not3}}{11}$
$=\frac{21-\not8 \sqrt{\not5}+5- \sqrt{\not3}+21+\not8 \sqrt{\not5}+5+ \sqrt{\not3}}{11}$
$=\frac{21+5+21+5}{11}$
$=\frac{52}{11}$

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Question 84 Marks

Using the following figure, show that $BD =\sqrt{x}$.

Answer

$ \mathrm{AB}=\mathrm{x}$ and $\mathrm{BC}=1$
$ \mathrm{AC}=\mathrm{AB}+\mathrm{BC}$
$ =\mathrm{x}+1$
diameter$=\mathrm{x}+1$
radius $\mathrm{OA}=\mathrm{OD}=\mathrm{OC}=\mathrm{OB}=\mathrm{OC}-\mathrm{BC}$
$ =\frac{x+1}{2}-1=\frac{x+1-2}{2}=\frac{x-1}{2}$
Using pythagoras in $\triangle \mathrm{BOD}$
$\mathrm{P}^2+\mathrm{B}^2=\mathrm{H}^2$
$ \mathrm{P}^2+\left(\frac{x-1}{2}\right)^2=\left(\frac{x+1}{2}\right)^2$
$ \mathrm{P}^2=\left(\frac{x+1}{2}\right)^2-\left(\frac{x-1}{2}\right)^2$
$ =\frac{(x+1)^2-(x-1)^2}{4}$
$ =\frac{\left(x^2+1+2 x\right)-\left(x^2+1-2 x\right)}{4}$
$\mathrm{P}^2=\frac{4 x}{4}$
$P^2=x$
$\mathrm{P}=\sqrt{x}$

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Question 94 Marks

If $\frac{2+\sqrt{ } 5}{2-\sqrt{ } 5}=x$ and $\frac{2-\sqrt{ } 5}{2+\sqrt{ } 5}=y ;$ find the value of $x^2-y^2$

Answer

$ x =\frac{2+\sqrt{ 5}}{2-\sqrt{5 }}$
$ =\frac{2+\sqrt{5 }}{2-\sqrt{ 5}} \times \frac{2+\sqrt{ 5} }{2+\sqrt{5 } }$
$ =\frac{(2+\sqrt{ 5} )^2}{2^2-(\sqrt{ 5})^2}$
$ =\frac{4+4 \sqrt{ 5} +5}{4}-5$
$ =\frac{9+4 \sqrt{5 }}{-1}$
$ =-(9-4 \sqrt{ 5})$
$y =\frac{2-\sqrt{5 }}{2+\sqrt{5 }}$
$ =\frac{2-\sqrt{5 }}{2+\sqrt{ 5}} \times \frac{2-\sqrt{ 5}}{2-\sqrt{ 5}}$
$ =\frac{(2-\sqrt{ 5})^2}{2^2-(\sqrt{5 })^2}$
$ =\frac{4-4 \sqrt{ 5}+5}{4}-5$
$ =\frac{9-4 \sqrt{ 5}}{-1}$
$=-(9 +4 \sqrt{5})$
$\therefore x^2-y^2 =(-9-4 \sqrt{5})^2-(-9+4 \sqrt{5})^2$
$ =81+72 \sqrt{5}+80-(81-72 \sqrt{ 5} +80)$
$ =81+72 \sqrt{5}+80-81+72 \sqrt{5}-80$
$ =144 \sqrt{5 }$

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Question 104 Marks

Rationalise the denominator of $: \frac{1}{\sqrt{ 3} -\sqrt{2 } +1}$

Answer

$ \frac{1}{\sqrt{3}-\sqrt{2}+1}$
$ =\frac{1}{(\sqrt{3}-\sqrt{2})+1} \times \frac{(\sqrt{3}-\sqrt{2})-1}{(\sqrt{3}-\sqrt{2})-1}$
$ =\frac{\sqrt{3}-\sqrt{2}-1}{(\sqrt{3}-\sqrt{2})^2-(1)^2}$
$ =\frac{\sqrt{3}-\sqrt{2}-1}{(\sqrt{3})^2-2 \sqrt{6}+(\sqrt{2})^2-1}$
$ =\frac{\sqrt{3}-\sqrt{2}-1}{3-2 \sqrt{6}+2-1}$
$ =\frac{\sqrt{3}-\sqrt{2}-1}{4-2 \sqrt{6}}$
$ =\frac{(\sqrt{3}-\sqrt{2})-1}{2(2-\sqrt{6})}$
$ =\frac{\sqrt{3}-\sqrt{2}-1}{2(2-\sqrt{6})} \times \frac{2+\sqrt{6}}{2+\sqrt{6}}$
$ =\frac{2 \sqrt{3}-2 \sqrt{2}-2+\sqrt{18}-\sqrt{12}-\sqrt{6}}{2\left[(2)^2-(\sqrt{6})^2\right]}$
$ =\frac{2 \sqrt{3}-2 \sqrt{2}-2+3 \sqrt{2}-2 \sqrt{3}-\sqrt{6}}{2[4-6]}$
$ =\frac{\sqrt{2}-2-\sqrt{6}}{2(-2)}$
$ =\frac{\sqrt{2}-2-\sqrt{6}}{-4}$
$ =\frac{1}{4}(2+\sqrt{6}-\sqrt{2})$

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Question 114 Marks

If $x=5-2 \sqrt{ 6} $, find $x^2+\frac{1}{x^2}$

Answer

Given $x=5-2 \sqrt{6 } $
We need to find $x^2+\frac{1}{x^2}$
Since $x=5-2 \sqrt{6}$, we have
$\frac{1}{x}=\frac{1}{5-2 \sqrt{6}}$
$ \Rightarrow \frac{1}{x}=\frac{1}{5-2 \sqrt{6}} \times \frac{5+2 \sqrt{6}}{5+2 \sqrt{6}}$
$ \Rightarrow \frac{1}{x}=\frac{5-2 \sqrt{6}}{(5-2 \sqrt{6})(5+2 \sqrt{6})}$
$ \Rightarrow \frac{1}{x}=\frac{5+2 \sqrt{6}}{5^2-(2 \sqrt{6})^2}$
$ \Rightarrow \frac{1}{x}=\frac{5+2 \sqrt{6}}{25-24}$
$ \Rightarrow \frac{1}{x}=\frac{5+2 \sqrt{6}}{1}$
$\Rightarrow \frac{1}{x}=(5+2 \sqrt{6}) \dots....(1)$
Thus, $\left(x-\frac{1}{x}\right)=(5- \not 2 \sqrt{\not6})-(5+\not 2 \sqrt{\not 6})=10$
$\left(x^2\right)+\left(\frac{1}{x^2}\right)=(-4 \sqrt{6})^2+2$
$ \Rightarrow\left(x^2\right)+\left(\frac{1}{x^2}\right)=96+2$
$ \Rightarrow\left(x^2\right)+\left(\frac{1}{x^2}\right)=98$

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Question 124 Marks

Prove that the following number is irrational:$ \sqrt{5} - 2$

Answer

$\sqrt{5}-2$
Let $\sqrt{5}-2$ be a rational number.
$\Rightarrow \sqrt{5}-2=x$
Squaring on both the sides, we get
$(\sqrt{5}-2)^2=x^2$
$ \Rightarrow 5+4-2 \times 2 \times \sqrt{5}=x^2$
$ \Rightarrow 9-x^2=4 \sqrt{5}$
$ \Rightarrow \sqrt{5}=\frac{9-x^2}{4}$
Here, $\mathrm{x}$ is a rational number.
$\Rightarrow x^2$ is a rational number.
$\Rightarrow 9-x^2$ is a rational number.
$\Rightarrow \frac{9-x^2}{4}$ is also a rational number.
$\Rightarrow \sqrt{2}=\frac{9-x^2}{4}$ is a rational number
But $\sqrt{2}$ is an irrational number.
$\Rightarrow \sqrt{5}=\frac{9-x^2}{4}$ is an irrational number.
$\Rightarrow 9-x^2$ is an irrational number.
$\Rightarrow x^2$ is an irrational number.
$\Rightarrow x$ is an irrational number.
But we have assume that $\mathrm{x}$ is a rational number.
$\therefore$ we arrive at a contradiction.
So, our assumption that $\sqrt{5}-2$ is a rational number is wrong.
$\therefore \sqrt{5}-2$ is an irrational number.

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Question 134 Marks

Prove that the following number is irrational$: 3 - \sqrt{2}$

Answer

$3-\sqrt{2}$
Let $3-\sqrt{2}$ be a rational number.
$\Rightarrow 3-\sqrt{2}=x$
Squaring on both the sides, we get
$(3-\sqrt{2})^2=x^2$
$ \Rightarrow 9+2-2 \times 3 \times \sqrt{2}=x^2$
$ \Rightarrow 11-x^2=6 \sqrt{2}$
$ \Rightarrow \sqrt{2}=\frac{11-x^2}{6}$
Here, $x$ is a rational number.
$\Rightarrow x^2$ is a rational number.
$\Rightarrow 11-x^2$ is a rational number.
$\Rightarrow \frac{11-x^2}{6}$ is also a rational number.
$\Rightarrow \sqrt{2}=\frac{11-x^2}{6}$ is a rational number.
But $\sqrt{2}$ is an irrational number.
$\Rightarrow \frac{11-x^2}{6}=\sqrt{2}$ is an irrational number.
$\Rightarrow 11-x^2$ is an irrational number.
$\Rightarrow x^2$ is an irrational number.
$\Rightarrow \mathrm{x}$ is an irrational number.
But we have assume that $\mathrm{x}$ is a rational number.
$\therefore$ we arrive at a contradiction.
So, our assumption that $3-\sqrt{2}$ is a rational number is wrong.
$\therefore 3-\sqrt{2}$ is an irrational number.

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Question 144 Marks

Prove that the following number is irrational:$ \sqrt{3} + \sqrt{2}$

Answer

$\sqrt{3}+\sqrt{2}$
Let $\sqrt{3}+\sqrt{2}$ be a rational number.
$\Rightarrow \sqrt{3}+\sqrt{2}=x$
Squaring on both the sides, we get
$(\sqrt{3}+\sqrt{2})^2=x^2$
$ \Rightarrow 3+2+2 \times \sqrt{3} \times \sqrt{2}=x^2$
$ \Rightarrow \mathrm{x}^2-5=2 \sqrt{6 } $
$ \Rightarrow \sqrt{6}=\frac{x^2-5}{2}$
Here, $x$ is a rational number.
$\Rightarrow x^2$ is a rational number.
$\Rightarrow x^2-5$ is a rational number.
$\Rightarrow \frac{x^2-5}{2}$ is also a rational number.
But $\sqrt{6 } $ is an irrational number.
$\Rightarrow \frac{x^2-5}{2}$ is also a irrational number.
$\Rightarrow x^2-5$ is an irrational number.
$\Rightarrow x^2$ is an irrational number.
$\Rightarrow x$ is an irrational number.
But we have assume that $x$ is a rational number.
$\therefore$ we arrive at a contradiction.
So, our assumption that $\sqrt{3}+\sqrt{2}$ is a rational number is wrong.
$\therefore \sqrt{3}+\sqrt{2}$ is an irrational number.

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Question 154 Marks

Use method of contradiction to show that $\sqrt{3}$ and $\sqrt{5}$ are irrational numbers.

Answer

Let us suppose that $\sqrt{3}$ and $\sqrt{5}$ are rational numbers.
$\therefore \sqrt{3}=\frac{a}{b}$ and $\sqrt{5}=\frac{x}{y}$ (Where $\mathrm{a}, \mathrm{b} \in 7$ and $\left.\mathrm{b}, \mathrm{y} \neq 0 \mathrm{x}, \mathrm{y}\right)$
Squaring both sides,
$3=a^{\prime} b^2, 5=\frac{x^2}{y^2}$
$3 b^2=a^2, 5 y^2=x^2$
$\Rightarrow \mathrm{a}^2$ and $\mathrm{x}^2$ are odd as $3 \mathrm{~b}^2$ and $5 \mathrm{y}^2$ are odd.
$\Rightarrow \mathrm{a}$ and $\mathrm{x}$ are odd $\ldots .(1)$
Let $a=3 c, x=5 z$
$a^2=9 c^2, x^2=25 z^2$
$3 b^2=9 c^2, 5 y^2=25 z^2 \quad($From equation$)$
$\Rightarrow \mathrm{b}^2=3 \mathrm{c}^2, \mathrm{y}^2=5 \mathrm{z}^2$
$\Rightarrow \mathrm{b}^2$ and $\mathrm{y}^2$ are odd as $3 \mathrm{c}^2$ and $5 \mathrm{z}^2$ are odd.
$\Rightarrow b$ and $y$ are odd $\ldots(2)$
From equation $(1)$ and $(2)$ we get $a, b, x, y$ are odd integers.
i.e., $a, b$, and $x, y$ have common factors $3$ and $5$ this contradicts our assumption that $\frac{a}{b}$ and $\frac{x}{y}$ are rational
i.e, $\mathrm{a}, \mathrm{b}$ and $\mathrm{x}, \mathrm{y}$ do not have any common factors other than.
$\Rightarrow \frac{a}{b}$ and $\frac{x}{y}$ is not rational.
$\Rightarrow \sqrt{3}$ and $\sqrt{5}$ and are irrational.

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Question 164 Marks

Arrange $\frac{5}{8},-\frac{3}{16},-\frac{1}{4}$ and $\frac{17}{32}$ in descending order of their magnitudes.Also, find the sum of the lowest and largest of these fractions. Express the result obtained as a decimal fraction correct to two decimal places.

Answer

Consider the given numbers : $\frac{5}{8},-\frac{3}{16},-\frac{1}{4}$ and $\frac{17}{32}$
The $\text{LCM}$ of $8,16,4$ and $32$ is $32 .$
Thus, the given numbers are given below:
$ \frac{5}{8},-\frac{3}{16},-\frac{1}{4}$ and $\frac{17}{32}$
$ =\frac{5 \times 4}{8 \times 4},-\frac{3 \times 2}{16 \times 2},-\frac{1 \times 8}{4 \times 8}$ and $\frac{17 \times 1}{32 \times 1}$
$ =\frac{20}{32},-\frac{6}{32},-\frac{8}{32},$ and $\frac{17}{32}$
Thus, the numbers in descending order are shown below:
$\frac{20}{32}, \frac{17}{32},-\frac{6}{32}$ and $-\frac{8}{32}$
Thus, the given numbers in descending order are listed below:
$\frac{5}{8}, \frac{17}{32},-\frac{3}{16}$ and $-\frac{1}{4} \text {. }$
We need to find the sum of the largest and smallest of the above numbers.
Thus, sum $=\frac{5}{8}+\left(-\frac{1}{4}\right)$
$=\frac{5}{8}-\frac{1}{4}$
$ =\frac{5}{8}-\frac{1 \times 2}{4 \times 2}$
$=\frac{5}{8}-\frac{2}{8}$
$=\frac{3}{8}$
We need to express this fraction as a decimal, correct to two decimal places.
Thus, we have $\frac{3}{8}=0.375 \approx 0.38$.

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Question 174 Marks

Arrange $-\frac{5}{7}, \frac{7}{12},-\frac{2}{3}$ and $\frac{11}{18}$ in ascending order of their magnitudes.Also, find the difference between the largest and smallest of these rational numbers. Express this difference as a decimal fraction correct to one decimal place.

Answer

Consider the given number : $-\frac{5}{9}, \frac{7}{12},-\frac{2}{3}$ and $\frac{11}{18}$
The $\text{L.C.M.}$ of $9,12 ,$ and $18$ is $36 .$
Thus, the given numbers are:
$-\frac{5}{9}, \frac{7}{12},-\frac{2}{3}$ and $\frac{11}{18}$
$ =-\frac{5 \times 4}{9 \times 4}, \frac{7 \times 3}{12 \times 3},-\frac{2 \times 12}{3 \times 12}$ and $\frac{11 \times 2}{18 \times 2}$
$ =-\frac{20}{36}, \frac{21}{36},-\frac{24}{36}$ and $\frac{22}{36}$
Thus, the numbers in ascending order are shown below:
$-\frac{24}{36},-\frac{20}{36}, \frac{21}{36}$ and $\frac{22}{36}$
Thus, the given numbers in ascending order are shown below:
$-\frac{2}{3},-\frac{5}{9}, \frac{7}{12}$ and $\frac{11}{18}$
We must find the difference between the largest and smallest of the above numbers.
Thus, difference $=\frac{11}{18}-\left(-\frac{2}{3}\right)$
$=\frac{11}{18}+\frac{2}{3}$
$ =\frac{11}{18}+\frac{2 \times 6}{3 \times 6}$
$ =\frac{11}{18}+\frac{12}{18}$
$ =\frac{11+12}{18}$
$ =\frac{23}{18}$
We need to express this fraction as a decimal, Correct to one decimal place.
Thus, we have $\frac{23}{18}=1.2 \overline{7} \approx 1.3$.

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MATHEMATICS [4 marks sum]

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