Question

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

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

Answer 1

The correct option is C – 5 and 4

$\frac{x^2+14x+9}{x^2+2x+3}=y\Rightarrow x^2+14x+9=x^{2}y+2xy+3y$

$\Rightarrow x^2(y-1)+2x(y-7)+(3y-9)=0$

Since x is real,
$\therefore 4(y-7{)}^2-4(3y-9\left)\right(y-1)>0$

$\Rightarrow 4(y^2+49-14y)-4(3y^2+9-12y)>0$

$\Rightarrow 4y^2+196-56y-12y^2-36+48y>0$

$\Rightarrow 8y^2+8y-160<0\Rightarrow y^2+y-20<0$

$\Rightarrow (y+5)(y-4)<0;$
$\therefore$ y lies between -5 and 4