Question

If $x$ is real, then $\frac{x}{(x^2-5x+9)}$ lies between

• A. $-1$ and $\frac{-1}{11}$
• B. $1$ and $\frac{-1}{11}$
• C. $1$ and $\frac{1}{11}$
• D. none of these

Answer 1

Answer: (B)

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

Let,

$\frac{x}{x^2-5x+9}=y$

$\Rightarrow y{x}^2-5yx+9y=x$

$\Rightarrow y{x}^2-(5y+1)x+9y=0$

Now, $x$ is real, so

$D\ge 0$

$\Rightarrow (-(5y+1){)}^2-4\cdot y\cdot (9y)\ge 0$

$\Rightarrow -11y^2+10y+1\ge 0$

$\Rightarrow 11y^2-10y-1\le 0$

$\Rightarrow (11y+1)(y-1)\le 0$

$\Rightarrow -\frac{1}{11}\le y\le 1$