Question

Column I contain rational algebraic expression and column II contains possible integers, which lie in their range.

Answer 1

A) $y=\frac{x^2-2x+4}{x^2+2x+4}$
$\Rightarrow (y-1)x^2+2x(y+1)+4y-4=0$
Since, $x$ is real
$\Rightarrow D\ge 0$
$\Rightarrow 4(y+1{)}^2-16(y-1)\ge 0$
$\Rightarrow 3y^2-10y+3\le 0$
$\Rightarrow (3y-1)(y-3)\le 0$
$\Rightarrow y\in [\frac{1}{3},3]$
So, option A i.e. 1 lies in the range.
B) $y=\frac{x^2-3x+2}{2x-3}$
$\Rightarrow x^2+x(-3-2y)+2+3y=0$
Since, $x$ is real
$\Rightarrow (-3-2y{)}^2-4(2+3y)\ge 0$
$\Rightarrow 4y^2+1\ge 0$
$\Rightarrow y\in R$
So, all options lie in the range
$y=\frac{2x^2-2x+4}{x^2-4x+3}$
$\Rightarrow x^{2}y-4xy+3y=2x^2-2x+4$
$\Rightarrow x^2(y-2)+x(-4y+2)+3y-4=0$
Since, $x$ is real
$D\ge 0$
$\Rightarrow (-4y+2{)}^2-4(y-2\left)\right(3y-4)\ge 0$
$\Rightarrow y^2+6y-7\ge 0$
$\Rightarrow (y+7)(y-1)\ge 0$
$\Rightarrow y\in (-\infty ,-7]\cup [1,\infty )$
All option values except option c values lies in the range
$x^2-(a-3)x+2<0$
$\Rightarrow (a-3{)}^2-8>0$