Complex Numbers And Quadratic Equations question 805
Question: The set of all real numbers x for which $ x^{2}-|x+2|+x>0, $ is [IIT Screening 2002]
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 ) $
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Answer:
Correct Answer: B
Solution:
Case I: When $ x+2\ge 0 $ i.e. $ x\ge -2, $ Then given inequality becomes $ x^{2}-(x+2)+x>0 $
Þ $ x^{2}-2>0\Rightarrow |x|>\sqrt{2} $
Þ $ x<-\sqrt{2} $ or $ x>\sqrt{2} $ As $ x\ge -2, $ therefore, in this case the part of the solution set is $ [-2,-\sqrt{2})\cup (\sqrt{2},\infty ). $ Case II: When $ x+2\le 0 $ i.e. $ x\le -2, $ Then given inequality becomes $ x^{2}+(x+2)+x>0 $
$ \Rightarrow x^{2}+2x+2>0 $
$ \Rightarrow {{(x+1)}^{2}}+1>0, $ which is true for all real x Hence, the part of the solution set in this case is $ (-\infty ,-2] $ . Combining the two cases, the solution set is $ (-\infty ,-2)\cup ([-2,-\sqrt{2}]\cup (\sqrt{2},\infty ) $ $ =(-\infty ,-\sqrt{2})\cup (\sqrt{2},\infty ). $
BETA
