Question

If $x$ is real, maximum and minimum values of $\frac{x^2+14x+9}{x^2+2x+3}$ are respectively

• A. 4,-5
• B. 5, -4
• C. 9, 3
• D. 24, 6

Answer 1

Answer: (A)

Let $y=\frac{x^2+14x+9}{x^2+2x+3}$
$\Rightarrow x^{2}y+2xy+3y=x^2+14x+9$
$\Rightarrow x^2(y-1)+2x(y-7)+3y-9=0$
Here, $x\in R$
$\therefore 4(y-7{)}^2-4(y-1)(3y-9)\ge 0$
$(y-7{)}^2-\left(3y^2-12y+9\right)\ge 0$
$y^2-14y+49-3y^2+12y-9\ge 0$
$\Rightarrow 2y^2+2y-40\le 0$
$\Rightarrow y^2+y-20\le 0(y+5)(y-4)\le 0$
$\therefore y\in [-5,4]$
Maximum and minimum value of
$\frac{x^2+14x+9}{x^2+2x+3}$ are $4$ and $-5$ respectively.