Question

If $x$ is real number, then $\frac{x}{x^{2}-5 x+9}$ must lie between

• A. $\frac{1}{11}$ and $1$
• B. $-1$ and $\frac{1}{11}$
• C. $-11$ and $1$
• D. $- \frac{1}{11}$ and $1$

Answer 1

Answer: (D)

$- \frac{1}{11}$ and $1$

Answer 2

Let $y=\frac{x}{x^{2}-5 x+9}$
$\Rightarrow x^{2} y-5 x y+9 y=x$
$\Rightarrow x^{2} y-(5 y+1) x+9 y=0$
For real, discriminant $\geq 0$
$\Rightarrow(5 y+1)^{2}-36 y^{2} \geq 0$
$\Rightarrow 25 y ^{2}+10 y +1-36 y ^{2} \geq 0$
$\Rightarrow-11 y ^{2}+10 y +1 \geq 0$
$\Rightarrow 11 y ^{2}-10 y -1 \leq 0$
$\Rightarrow 11 y ^{2}-11 y + y -1 \leq 0$
$\Rightarrow 11 y ( y -1)+1( y -1) \leq 0$
$\Rightarrow( y -1)(11 y +1) \leq 0$
$\Rightarrow y \in\left[\frac{-1}{11}, 1\right]$