Question

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

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

Answer 1

Answer: (A)

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