Question

If $x$ is real, then the value of $\frac{x^2-3x+4}{x^2+3x+4}$ lies in the interval

• A. $[\frac{1}{3},3]$
• B. $[\frac{1}{5},5]$
• C. $[\frac{1}{6},6]$
• D. $[\frac{1}{7},7]$

Answer 1

Answer: (D)

$[\frac{1}{7},7]$

Let $y$ $=\frac{x^2-3x+4}{x^2+3x+4}$

$(y-1)x^2+3(y+1)x+4(y-1)=0$

Since $x$ is real , $D\ge 0$

$9(y^2+2y+1)-16(y^2-2y+1)\ge 0$

$-7(y^2+1)+50y\ge 0$

$7y^2-50y+7\le 0$

$(7y-1)(y-7)\le 0$

$y$ lies in $[\frac{1}{7},7]$