[5 marks sum]

Question 15 Marks

If $x$ and $y$ both are positive and $(2x^2- 5y^2): xy = 1: 3,$ find $x: y.$

Answer

$\left(2 x^2-5 y^2\right): x y=1: 3 $
$ \Rightarrow \frac{2 x^2-5 y^2}{x y}=\frac{1}{3} $
$\Rightarrow \frac{2 x}{y}-\frac{5 y}{x}=\frac{1}{3}$
Put $\frac{ x }{ v }=$ a we get
$\Rightarrow 2 a-5 \frac{1}{a}=\frac{1}{3} $
$\Rightarrow 3\left(2 a^2-5\right)=a $
$\Rightarrow 6 a 2-a-15=0 $
$ \Rightarrow 6 a^2+9 a-10 a-15=0$
$\Rightarrow 3 a(2 a+3)-5(2 a+3)=0$
$ \Rightarrow(2 a+3)(3 a-5)=0$
$ \Rightarrow(2 a+3)=0 \text { or }(3 a-5)=0$
$\Rightarrow a =-\frac{3}{2}$ or $a =\frac{5}{3}$
$a=-\frac{3}{2}$ is not acceptable, as $x$ and $y$ both are positive.
$a=\frac{5}{3} \Rightarrow \frac{x}{y}=\frac{5}{3} $
$ \Rightarrow x: y=5: 3$

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Question 25 Marks

Given $\frac{x^3+12 x}{6 x^2+8}=\frac{y^3+27 y}{9 y^2+27}$
using componendo and dividendo, find $x : y$

Answer

$\frac{x^3+12 x}{6 x^2+8}=\frac{y^3+27 y}{9 y^2+27}$
Applying componendo and dividendo, we get
$\frac{x^3+12 x+6 x^2+8}{x^3+12 x-6 x^2-8}=\frac{y^3+27+9 y^2+27}{y^3+27 y-9 y^2-27}$
$\Rightarrow\ \frac{ x ^3+3(1)(4) x +3(1)(2) x ^2+2^3}{ x ^3+3(1)(4) x +3(1)(2) x ^2+2^3}=\frac{ y ^3+3(1)(9) y +3(1)(3) y ^2+3^3}{ y ^3+3(1)(9) y +3(1)(3) y ^2+3^3}$
$ \Rightarrow \frac{ x ^3+3(1)(4) x +3(1)(2) x ^2+2^3}{ x ^3+3(1)(2) x +3(1)(4) x ^2+2^3}=\frac{ y ^3+3(1)(9) y +3(1)(3) y ^2+3^3}{ y ^3+3(1)(3) y +3(1)(9) y ^2+3^3} $
$ \Rightarrow \frac{( x +2)^3}{( x -2)^3}=\frac{( y +3)^3}{( y -3)^3}$
Again applying componendo and dividendo, we get
$ \frac{x+2+x-2}{x+2-x+2}=\frac{y+3+y-3}{y+3-y+3} $
$ \Rightarrow \frac{2 x}{4}=\frac{2 y}{6}$
$ \Rightarrow \frac{x}{2}=\frac{y}{3} $
Applying alternendo, we get
$ \frac{x}{y}=\frac{2}{3} $
$ \Rightarrow x: y=2: 3$

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Question 35 Marks

If $x=\frac{2 a b}{a+b}$ find the value of $\frac{x+a}{x-a}+\frac{x+b}{x-b}$

Answer

$ \begin{aligned} & x=\frac{2 a b}{a+b} \\ & \frac{x}{a}=\frac{2 b}{a+b} \end{aligned} $
Applying componendo and dividendo
$ \begin{aligned} & \frac{x+a}{x-a}=\frac{2 b+a+b}{2 b-a-b} \\ & \frac{x=a}{x-a}=\frac{3 b+a}{b-a} \ldots(1) \end{aligned} $
Also $\mathrm{x}=\frac{2 a b}{a+b}$
$ \frac{x}{b}=\frac{2 a}{a+b} $
Apllying componendo and dividendo
$ \begin{aligned} & \frac{x+b}{x-b}=\frac{2 a+a+b}{2 a-a-b} \\ & \frac{x+a}{x-a}=\frac{3 b+a}{b-a} \ldots(1) \end{aligned} $
Also $\mathrm{x}=\frac{2 a b}{a+b}$
$ \frac{x}{b}=\frac{2 a}{a+b} $
Applying componendo and dividendo
$ \begin{aligned} & \frac{x+b}{x-b}=\frac{2 a+a+b}{2 a-a-b} \\ & \frac{x+b}{x-b}=\frac{3 a+b}{a-b} ....... (2)\end{aligned} $
From (1) and (2)
$ \begin{aligned} & \frac{x+a}{x-a}=\frac{2 a+a+b}{2 a-a-b} \\ & \frac{x+b}{x-b}=\frac{3 a+b}{a-b} \ldots(2) \end{aligned} $
From (1) and (2)
$ \begin{aligned} & \frac{x+a}{x-a}+\frac{x+a}{x-b}=\frac{3 b+a}{b-a}+\frac{3 a+b}{a-b} \\ & \frac{x+a}{x-a}+\frac{x+b}{x-b}=\frac{-3 b-a+3 a+b}{a-b} \\ & \frac{x+a}{x-a}+\frac{x+b}{x-b}=\frac{2 a-2 b}{a-b} \\ & \frac{x+a}{x-a}+\frac{x+b}{x-b}=\frac{2 a-2 b}{a-b}=2 \end{aligned} $

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Mathematics [5 marks sum]

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