Question

if x be real and 0 < b< c then $\frac{x^2-bc}{2x-b-c}$

• A. cannot lie between b and c.
• B. cannot lie between b+1 and c.
• C. cannot lie between b and c+1.
• D. None of these.

Answer 1

Answer: (A)

cannot lie between b and c.

Let $y=\frac{x^2-bc}{2x-b-c}$
$\Rightarrow x^2-bc=2xy-(b+c)y\Rightarrow x^2-(2x-b-c)y-bc=0$
The discriminant of this expression must be non-negative.
$\Rightarrow {(2x-b-c)}^2+4bc\ge 0\Rightarrow 4x^2+b^2+c^2-4xb-4xc+2bc+4bc\ge 0\Rightarrow 4x^2-4(b+c)x+b^2+c^2+6bc\ge 0\Rightarrow {(2x-(b+c))}^2+4bc\ge 0$
This is possible only when $bc\ge 0$