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

$\Rightarrow x^{2}y-5xy+9y-x=0$

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

$x$ is real, therefore discriminant $⩾$ 0

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

$\Rightarrow 25y^2+1+10y-36y^2⩾0$

$\Rightarrow -11y^2+10y+1⩾0$

$\Rightarrow 11y^2-10y-1⩽0$

$\Rightarrow 11y^2-11y+y-1⩽0$

$\Rightarrow 11y(y-1)+1(y-1)⩽0$

$\Rightarrow (11y+1)(y-1)⩽0$

$\Rightarrow -\frac{1}{11}⩽y⩽1$

$\therefore$ $\frac{x}{x^2-5x+9}$ lies between $1$ and $-\frac{1}{11}$

Hence option 'B' is correct