Question
If $$x$$ be real and $$b < c,$$ then $$\frac{{{x^2} - bc}}{{2x - b - c}}$$ lies in
A.
$$\left( {b,c} \right)$$
B.
$$\left[ {b,c} \right]$$
C.
$$\left( { - \infty ,b} \right] \cup \left[ {c,\infty } \right)$$
D.
$$\left( { - \infty ,b} \right) \cup \left( {c,\infty } \right)$$
Answer :
$$\left( { - \infty ,b} \right] \cup \left[ {c,\infty } \right)$$
Solution :
Let, $$y = \frac{{{x^2} - bc}}{{2x - b - c}}$$
$$\eqalign{
& \Rightarrow {x^2} - 2yx + \left( {b + c} \right)y - bc = 0 \cr
& \because x \in R,\,\,{\text{so }}4{y^2} - 4\left( {b + c} \right)y + 4bc \geqslant 0 \cr
& \Rightarrow x \leqslant b{\text{ or }}x \geqslant c\,\,\,\,\,\left( {\because b < c} \right) \cr} $$