Question

If $x$ is real, then the expression $\frac{x^2-bc}{2x-b-c}$ has no value lying between

• A. $0$ and $b$
• B. $b$ and $c$
• C. $b$ and $-c$
• D. $-b$ and $-c$

Answer 1

The correct option is B $b$ and $c$

Let $y=\frac{x^2-bc}{2x-b-c}$
$\Rightarrow x^2-2xy+(b+c)y-bc=0$
Now, $x$ is real $\Rightarrow D\ge 0$
$\Rightarrow 4y^2-4\left[\right(b+c)y-bc]\ge 0\Rightarrow (y-b)(y-c)\ge 0$
$\Rightarrow$ $y$ has no value lying between $b$ and $c$ .