Question

If $x$ is real and $y=\frac{x^2-3x+4}{x^2+3x+4}$, then

• A. Minimum value of $y$ is $1$
• B. Maximum value of $y$ is $7$
• C. $y\in \left[\frac{1}{7},7\right]$
• D. $y$ can assume negative values

Answer 1

Answer: (B), (C)

The correct options are
B $y\in \left[\frac{1}{7},7\right]$
C Maximum value of $y$ is $7$

Let $y=\frac{(x^2-3x+4)}{(x^2+3x+4)}$
$x^2(y-1)+3x(y+1)+4(y-1)=0$
Since, $x$ is real
$\Delta \ge 0$
$9(y+1{)}^2-16(y-1{)}^2\ge 0$
$7y^2-50y+7\le 0$
$\Rightarrow y\in [\frac{1}{7},7]$
The minimum value is $\frac{1}{7}$ and maximum is $7$.