Question

If $x$ be 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 the above.

Answer 1

The correct option is B $1$ and $-\frac{1}{11}$

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

or $y{x}^2-5xy+9y=x⟹y{x}^2-(5y+1)x+9y=0$

Now, $x$ is real, so

$\Delta \ge 0{⟹(-(5y+1\left)\right)}^2-4y\left(9y\right)\ge 0$

or $-11y^2+10y+1\ge 0⟹11y^2-10y-1\le 0⟹(11y+1)(y-1)\le 0⟹\frac{-1}{11}\le y\le 1\therefore y\in [\frac{-1}{11},1]$