[5 marks sum]

Question 15 Marks

If $9 a^2+\frac{1}{9 a^2}=23$; find the value of $27 a^3+\frac{1}{27 a^3}$

Answer

$9 a^2+\frac{1}{9 a^2}=23 $
Using$\left(3 a+\frac{1}{3 a}\right)^2 $
$=(3 a)^2+\left(\frac{1}{3 a}\right)^2+2(3 a)\left(\frac{1}{3 a}\right) $
$\Rightarrow\left(3 a+\frac{1}{3 a}\right)^2 $
$=9 a^2+\frac{1}{9 a^2}+2 $
$=23+2 $
$=25 $
$\Rightarrow 3 a+\frac{1}{3 a}=5$
Cubing both sides, we get:
$(3 a)^3+\left(\frac{1}{3 a}\right)^3+3(3 a)\left(\frac{1}{3 a}\right)\left(3 a+\frac{1}{3 a}\right)=(5)^3 $
$\Rightarrow 27 a^3+\frac{1}{27 a^3}+3(5)=125 $
$\Rightarrow 27 a^3+\frac{1}{27 a^3} $
$=125-15 $
$=110$

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Question 25 Marks

If $m ^2+\frac{1}{ m ^2}=51$; find the value of $m ^3-\frac{1}{ m ^3}$

Answer

$m ^2+\frac{1}{ m ^2}=51$
We know that
$\left(m-\frac{1}{m}\right)^2 $
$=m^2+\frac{1}{m^2}-2 $
$\Rightarrow\left(m-\frac{1}{m}\right)^2=51-2 $
$\Rightarrow\left(m-\frac{1}{m}\right)^2=49=7^2 $
$\Rightarrow m-\frac{1}{m}=7 $
$\Rightarrow\left(m-\frac{1}{m}\right)^3=7^3 $
$\Rightarrow m^3-\frac{1}{m^3}-3\left(m-\frac{1}{m}\right)=343 $
$\Rightarrow m^3-\frac{1}{m^3}-3 \times 7=343 $
$\Rightarrow m^3-\frac{1}{m^3} $
$=343+21 $
$=364 .$

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Question 35 Marks

If $a ^2+\frac{1}{ a ^2}=14$; find the value of $a ^3+\frac{1}{ a ^3}$

Answer

$\text { Using }(a+b)^2=a^2+2 a b+b^2 $
$\left(a+\frac{1}{a}\right)^2$
$ =a^2+2 a\left(\frac{1}{a}\right)+\left(\frac{1}{a}\right)^2$
$ \Rightarrow\left(a+\frac{1}{a}\right)^2=a^2+2+\frac{1}{a^2} $
$ \Rightarrow\left(a+\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}+2 $
$ \Rightarrow\left(a+\frac{1}{a}\right)^2=14+2$
$ \Rightarrow\left(a+\frac{1}{a}\right)^2=16$
$ \Rightarrow a+\frac{1}{a}= \pm 4$
$ a^3+\frac{1}{a^3} $
$ =\left(a+\frac{1}{a}\right)\left(a^2+\frac{1}{a^2}-1\right) \ldots\left[\text { [Using } a^3+b^3=(a+b)\left(a^2+b^2-a b\right)\right] $
$ =( \pm 4)(14-1)$
$=( \pm 4)(13) $
$ = \pm 52$

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Question 45 Marks

If $a -\frac{1}{ a }=7$, find $a ^2+\frac{1}{ a ^2}, a ^2-\frac{1}{ a ^2}$ and $a ^3-\frac{1}{ a ^3}$

Answer

$a -\frac{1}{ a }=7$
Squaring both sides of $(1),$
$\left(a-\frac{1}{a}\right)^2=(7)^2$
$\Rightarrow a^2+\frac{1}{a^2}-2=49 $
$\Rightarrow a^2+\frac{1}{a^2}$
$=49+2 $
$=51$
Now, $\left(a+\frac{1}{a}\right)^2$
$= a ^2+\frac{1}{ a ^2}+2$
$=51+2$
$=53$
$\Rightarrow a+\frac{1}{a}$
$= \pm \sqrt{53}$
Now $a ^2-\frac{1}{ a ^2}$
$=\left(a +\frac{1}{ a }\right)^{ a ^2}\left( a -\frac{1}{ a }\right) $
$=(\pm \sqrt{53})(7) $
$= \pm 7 \sqrt{53}$
Cubing both sides of $(1),$
$\left(a-\frac{1}{a}\right)^3=(7)^3 $
$\Rightarrow a^3-\frac{1}{a^3}-3\left(a-\frac{1}{a}\right)=343 $
$\Rightarrow a^3-\frac{1}{a^3}-3(7)=343 $
$\Rightarrow a^3-\frac{1}{a^3} $
$=343+21 $
$=364$

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Question 55 Marks

If $x+\frac{1}{x}=5$, find the value of $x^2+\frac{1}{x^2}, x^3+\frac{1}{x^3}$ and $x^4+\frac{1}{x^4}$.

Answer

$x+\frac{1}{x}=5$
Squaring both sides of $(1)$
$\left(x+\frac{1}{x}\right)^2=(5)^2 $
$\Rightarrow x^2+\frac{1}{x^2}+2=25 $
$\Rightarrow x^2+\frac{1}{x^2} $
$=25-2 $
$=23$
Cubing both sides of $(1),$
$\left(x+\frac{1}{x}\right)^3=95^3 $
$x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)=125 $
$\Rightarrow x^3+\frac{1}{x^3}+3(5)=125 $
$\Rightarrow x^3+\frac{1}{x^3} $
$=125-15 $
$=110$
Squaring both sides of $(2),$
$\left(x^2+\frac{1}{x^2}\right)^2=(23)^2 $
$\Rightarrow x^4=\frac{1}{x^4}+=529 $
$=x^4+\frac{1}{x^4} $
$=529-2 $
$=527 .$

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Question 65 Marks

If $x+\frac{1}{x}= p , x-\frac{1}{x}= q$ find the relation between $p$ and $q$.

Answer

$x+\frac{1}{x}= p , x-\frac{1}{x}= q $
$\left(x+\frac{1}{x}\right)^2 $
$=x^2+\frac{1}{x^2}+2 $
$\Rightarrow p ^2=x^2+\frac{1}{x^2}+2 $
$\Rightarrow x^2+\frac{1}{x^2}= p ^2-2$
Also, $\left(x-\frac{1}{x}\right)^2$
$=x^2+\frac{1}{x^2}-2 $
$\Rightarrow q ^2=x^2+\frac{1}{x^2}-2 $
$\Rightarrow x^2+\frac{1}{x^2}= q ^2+2$
Equating the value $x^2+\frac{1}{x^2}$ from and $(2),$ we get :
$p^2-2=q^2+2 $
$\Rightarrow p^2-q^2=4$

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Question 75 Marks

If $a+\frac{1}{a}=2$, then show that $a^2+\frac{1}{a^2}=a^3+\frac{1}{a^3}=a^4+\frac{1}{a^4}$

Answer

$a+\frac{1}{a}=2 $
$\left(a+\frac{1}{a}\right)^2 $
$=a^2+\frac{1}{a^2}+2 $
$\Rightarrow(2)^2=a^2+\frac{1}{a^2}+2 $
$\Rightarrow a^2+\frac{1}{a^2} $
$=4-2 $
$=2 $
$\left(a+\frac{1}{a}\right)^3 $
$=a^3+\frac{1}{a^3}+3\left(a+\frac{1}{a}\right) $
$\Rightarrow(2)^3=a^3+\frac{1}{a^3}+3(2) $
$\Rightarrow a^3+\frac{1}{a^3} $
$=8-6 $
$=2 $
$\left(a^2+\frac{1}{a^2}\right)^2 $
$=a^4+\frac{1}{a^4}+2 $
$\Rightarrow(2 a)^2=a^4+\frac{1}{a^4}+2 $
$\Rightarrow a^4+\frac{1}{a^4} $
$=4-2 $
$=2$
Thus, $a^2+\frac{1}{a^2}=a^3+\frac{1}{a^3}=a^4+\frac{1}{a^4}$

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Question 85 Marks

If $a +\frac{1}{ a }=6$;find $a -\frac{1}{ a }$

Answer

$\left(a+\frac{1}{a}\right)^2 $
$ =\left(a^2\right)+2(a)\left(\frac{1}{a}\right)+\left(\frac{1}{a}\right)^2 $
$ =a^2+\frac{1}{a^2}+2 $
$=36=a^2+\frac{1}{a^2}+2$
$=a^2+\frac{1}{a^2} $
$=34 $
$ \left(a-\frac{1}{a}\right)^2 $
$ =(a)^2-2(a)\left(\frac{1}{a}\right)+\left(\frac{1}{a}\right)^2 $
$ =a^2+\frac{1}{a^2}-2 $
$ =34-2$
$=32$
$ \Rightarrow a-\frac{1}{a} $
$ = \pm \sqrt{32} $
$ = \pm 4 \sqrt{2} .$

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Question 95 Marks

If $x + y = 9, xy = 20$ ; find: $x - y$

Answer

$x + y = 9, xy = 20$
We know $(a + b)$
$= a^2 + 2ab + b^2$
$\therefore (x + y)^2$
$= 81 x^2 + y^2 + 2xy$
$\Rightarrow x^2 + y^2$
$= 81 - 2(120)$
$= 41$
We also know $(a - b)^2$
$= a^2 - 2ab + b^2$
$\Rightarrow (x - y)^2$
$= x^2- 2xy + y^2$
$\Rightarrow (x - y)^2$
$= 41 - 2(20)$
$= 1$
$\Rightarrow x - y$
$= ±1.$

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Question 105 Marks

If $a - b = 10$ and $ab = 11;$ find $a + b.$

Answer

$a - b =10, ab =11$
We know that:
$(a-b)^2=a^2-2 a b+b^2 $
$\Rightarrow(10)^2=a^2+b^2-2 \times 11 $
$\Rightarrow 100=a^2+b^2-22 $
$\Rightarrow a^2+b^2 $
$=100+22 $
$=122
$
Using $(a+b)^2=a^2+b^2+2 a b$, we get
$(a+b)^2 $
$=122+2(11) $
$=122+22 $
$=144 $
$\Rightarrow(a+b) $
$=\sqrt{144} $
$= \pm 12$

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Question 115 Marks

If $a ^2-3 a -1=0$ and $a \neq 0$, find $: a +\frac{1}{ a }$

Answer

$a-\frac{1}{a}=3$
Squaring both sides, we get
$\left(a-\frac{1}{a}\right)^2 $
$=a^2+\frac{1}{a^2}-2 $
$=9 $
$=a^2+\frac{1}{a^2} $
$=11 .$
Now,
$\left(a+\frac{1}{a}\right)^2 $
$=a^2+\frac{1}{a^2} $
$=11+2 $
$=13 $
$\Rightarrow a+\frac{1}{a^2} $
$= \pm \sqrt{13} .$

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Question 125 Marks

If $a -\frac{1}{ a }=10$; find $a +\frac{1}{ a }$

Answer

$a-\frac{1}{a}=10 $
$\left(a-\frac{1}{a}\right)^2 $
$=a^2+\frac{1}{a^2}-2(a)\left(\frac{1}{a}\right) $
$\Rightarrow(10)^2 $
$=a^2+\frac{1}{a^2}-2 $
$=a^2+\frac{1}{a^2} $
$=102$
Now $_t\left( a +\frac{1}{ a ^2}\right)$
$=a^2+\frac{1}{a^2}+2(a)\left(\frac{1}{a}\right) $
$=102+2 $
$=104 $
$=a^2-\frac{1}{a^2} $
$=\sqrt{104} $
$= \pm 2 \sqrt{26} .$

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MATHEMATICS [5 marks sum]

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