Complex Numbers And Quadratic Equations question 676
Question: If x be real and $ b<c, $ then $ \frac{x^{2}-bc}{2x-b-c} $ lies in
Options:
A) $ (b,c) $
B) $ [b,c] $
C) $ (-\infty ,b]\cup [c,\infty ) $
D) $ (-\infty ,b)\cup (c,\infty ) $
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ y=\frac{x^{2}-bc}{2x-b-c} $
$ \Rightarrow x^{2}-2yx+(b+c)y-bc=0 $ $ \because x\in R,so 4y^{2}-4( b+c )y+4bc\ge 0 $
$ \Rightarrow x\le borx\ge c~~~( \because b<c ) $
BETA
