Question:
The set of all real numbers x for which $x^2-|x+2|+x>0$ is
The set of all real numbers x for which $x^2-|x+2|+x>0$ is
Updated On: Aug 19, 2023
- $(-8,-2)\cup(2,8)$
- $(-8,-\sqrt2)\cup(\sqrt2,8)$
- $(-8,-1)\cup(1,8)$
- $(\sqrt2,8)$
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The Correct Option is B
Solution and Explanation
Given, $x^2-|x+2|+x>0\, \, \, \, \, \, \, \, \, \, \, ...(i)$
Case I When $\hspace20mmx+2\ge0$
$\therefore\hspace20mmx^2-x-2+x>0 \Rightarrow x^2-2>0$
$\Rightarrow\hspace25mmx\sqrt2$
$\Rightarrow\hspace25mmx\in(-2,-\sqrt2)\cup(\sqrt2,8)\, \, \, \, ...(ii)$
Case II When $x+2<0$
$\therefore\, \, \, \, \, \, \, x^2+x+2+x>0$
$\Rightarrow\, \, \, \, \, \, \, \, \, x^2+2x+2>0$
$\Rightarrow\, \, \, \, \, \, \, \, \, (x+1)^2+1>0$
which is true for all x.
$\therefore\, \, \, \, \, \, \, x\le-2 \, \, or\, x\in(-8,-2) \, \, \, \, \, \, \, \, \, ...(iii)$
From Eqs. (ii) and (iii), we get
$\hspace25mmx\in(-8,-\sqrt2)\cup(\sqrt2,8)$
Case I When $\hspace20mmx+2\ge0$
$\therefore\hspace20mmx^2-x-2+x>0 \Rightarrow x^2-2>0$
$\Rightarrow\hspace25mmx\sqrt2$
$\Rightarrow\hspace25mmx\in(-2,-\sqrt2)\cup(\sqrt2,8)\, \, \, \, ...(ii)$
Case II When $x+2<0$
$\therefore\, \, \, \, \, \, \, x^2+x+2+x>0$
$\Rightarrow\, \, \, \, \, \, \, \, \, x^2+2x+2>0$
$\Rightarrow\, \, \, \, \, \, \, \, \, (x+1)^2+1>0$
which is true for all x.
$\therefore\, \, \, \, \, \, \, x\le-2 \, \, or\, x\in(-8,-2) \, \, \, \, \, \, \, \, \, ...(iii)$
From Eqs. (ii) and (iii), we get
$\hspace25mmx\in(-8,-\sqrt2)\cup(\sqrt2,8)$
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Some important operations on sets include union, intersection, difference, and the complement of a set, a brief explanation of operations on sets is as follows:
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- It is denoted as “A U B”
2. Intersection of Sets:
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