Theory of Equations 1 Question 25
26. The set of all real numbers $x$ for which $x^{2}-|x+2|+x>0$ is
(2002, 1M)
(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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Solution:
- Given, $x^{2}-|x+2|+x>0$
Case I When $\quad x+2 \geq 0$
$$ \begin{array}{lc} \therefore & x^{2}-x-2+x>0 \Rightarrow x^{2}-2>0 \\ \Rightarrow & x<-\sqrt{2} \text { or } x>\sqrt{2} \\ \Rightarrow & x \in[-2,-\sqrt{2}) \cup(\sqrt{2}, \infty) \end{array} $$
Case II When $x+2<0$
$$ \begin{array}{ll} \therefore & x^{2}+x+2+x>0 \\ \Rightarrow & x^{2}+2 x+2>0 \\ \Rightarrow & (x+1)^{2}+1>0 \end{array} $$
which is true for all $x$.
$$ \therefore \quad x \leq-2 \text { or } x \in(-\infty,-2) $$
From Eqs. (ii) and (iii), we get
$$ x \in(-\infty,-\sqrt{2}) \cup(\sqrt{2}, \infty) $$
