Question

Find the range of the rational expression $y=\frac{x^2+2x-4}{x^2-2x+3}\forall x\in R$

• A. $[1,3]$
• B. $[-3,3]$
• C. $[-1,3]$
• D. $[-3,1]$

Answer 1

The correct option is D $[-3,1]$

Given: $y=\frac{x^2+2x-4}{x^2-2x+3}\forall x\in R$

$\Rightarrow y(x^2-2x+3)=x^2+2x-4$
$\Rightarrow (y-1)x^2-(2y+2)x+3y+4=0$

On comparing with standard quadratic equation $y=a{x}^2+bx+c$ we get, $a=(y-1),b=-(2y+2),c=3y+4.$

Here, we have a quadratic expression in $x,$ where $x\in R$ this means the quadratic in $x$ will have two roots and the $D$ of the quadratic will be greater than or equal to $0.$

$\therefore D\ge 0$
$\Rightarrow b^2-4ac\ge 0$
$\Rightarrow (-(2y+2){)}^2-4(y-1\left)\right(3y+4)\ge 0$
$\Rightarrow 4y^2+4y+8-12y^2-20y+16\ge 0$
$\Rightarrow 8y^2+16y-24\le 0$
$\Rightarrow y^2+2y-3\le 0$
$\Rightarrow y^2+3y-y-3\le 0$
$\Rightarrow y(y+3)-(y+3)\le 0$
$\Rightarrow (y+3)(y-1)\le 0$
$\Rightarrow y\in [-3,1]$