Complex Numbers And Quadratic Equations question 667
Question: The set of all real numbers x for which $ x^{2}-[x+2]+x>0, $ is
Options:
A) $ ( -\infty ,-2 )\cup ( 2,\infty ) $
B) $ ( -\infty ,-\sqrt{2} )\cup ( \sqrt{2},\infty ) $
C) $ ( -\infty ,-1 )\cup ( 1,\infty ) $
D) $ ( \sqrt{2},\infty ) $
Show Answer
Answer:
Correct Answer: B
Solution:
For $ x\ge -2,x^{2}-x-2+x>0 $
$ \Rightarrow x^{2}>2\Rightarrow x\in (-\infty ,-\sqrt{2})\cup (\sqrt{2},\infty ) $
$ \Rightarrow x\in [ -2,-\sqrt{2} )\cup ( \sqrt{2},\infty ) Forx<-2 $ $ {x^{2}}+x+2+x>0 orx^{2}+2x+2>0 $ which is true for all x. Hence $ x\in (-\infty ,-\sqrt{2})\cup (\sqrt{2},\infty ) $
BETA
