Question

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

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

Answer 1

Answer: (C)

$-5$ and $4$

$y=\frac{x^2+14x+9}{x^2+2x+3}=1+\frac{12x+6}{x^2+2x+3}{y}^{\prime }=\frac{(x^2+2x+3)\left(12\right)-(12x+6)(2x+2)}{{(x^2+2x+3)}^2}{y}^{\prime }=\frac{-12x^2-12x+24}{{(x^2+2x+3)}^2}$

for max-min, $y^{\prime }=0$

so, $x^2+x-2=0⟹x=1,-2$

at $x=1,y=4$

at $x=-2,y=-5$

and ${lim}_{x\to \infty }y=1$

and $y(x=0)=3$

concluding all these graph of $y$ is shown above.

So, we can say that $y$ lies between $-5$ and $4$.