Question

If $x$ is real, then the value of the expression $\frac{x^2+14x+9}{x^2+2x+3}$ lies between

• A. $4$ and $5$
• B. $-4$ and $5$
• C. $-5$ and $4$
• D. None of these

Answer 1

The correct option is C $-5$ and $4$

Let $y=\frac{x^2+14x+9}{x^2+2x+3}$
$\Rightarrow x^2+14x+9=x^{2}y+2xy+3y$
$\Rightarrow x^2(y-1)+2x(y-7)+(3y-9)=0$
Since, $x$ is real, we have
$4(y-7{)}^2-4(3y-9\left)\right(y-1)>0$
$\Rightarrow (y^2+49-14y)-(3y^2+9-12y)>0$
$\Rightarrow (y+5)(y-4)<0$
Therefore, $y$ lies between $-5$ and $4$.