Let $x = \left\lbrack \frac{a + 2b}{a + b} \right\rbrack$ and $y = \frac{a}{b}$, where $a$ and $b$ are positive integers. If $y^{2} 2$, then

Question

Let $x = \left\lbrack \frac{a + 2b}{a + b} \right\rbrack$ and $y = \frac{a}{b}$, where $a$ and $b$ are positive integers. If $y^{2} > 2$, then

Dr. Abubacker · Accepted Answer

(b)
Given that, $x = \left\lbrack \frac{a + 2b}{a + b} \right\rbrack$ and $y = \frac{a}{b}$
$\therefore\ \ \ x = \frac{a + 2b}{a + b} = \frac{\frac{a}{b} + 2}{1 + \frac{a}{b}} = 1 + \frac{1}{\frac{a}{b} + 1}$
$\Rightarrow \ \ x = 1 + \frac{1}{y + 1}\ \ \ \ \ \ \ \lbrack\because y = \frac{a}{b}\ and\ y^{2} > 2\ (given)\rbrack$
Which shows $x^{2} < 2$ $\left\lbrack \because\frac{1}{y + 1} < as\ y > 1 \right\rbrack$

Q. Let $x = \left\lbrack \frac{a + 2b}{a + b} \right\rbrack$ and $y = \frac{a}{b}$, where $a$ and $b$ are positive integers. If $y^{2} > 2$, then

Explanation

Reviewed by Dr. Abubacker

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