Question

If $x$ is real, then the maximum and minimum values of the expression $\frac{x^2-3x+4}{x^2+3x+4}$ are

• A. $2$, $1$
• B. $7$, $\frac{1}{7}$
• C. $5$, $\frac{1}{5}$
• D. None

Answer 1

The correct option is B $7$, $\frac{1}{7}$

$y=\frac{x^2-3x+4}{x^2+3x+4}$
$\Rightarrow (y-1)x^2+3(y+1)x+4(y-1)=0$
As $x$ is real, discriminant $\ge 0$
$\Rightarrow 9(y+1{)}^2-16(y-1{)}^2\ge 0$
$\Rightarrow -7y^2+50y-7\ge 0$
$\Rightarrow (y-7)(7y-1)\le 0$
$\therefore y\in [\frac{1}{7},7]$
i.e. Maximum value of expression is $7$ and minimum value is $\frac{1}{7}.$
Hence, option B.