Q53 / 160 · AP EAMCET 2019 (21 Apr Shift 1) · Mathematics
The set of all real numbers satisfying the inequation $x^2-|x+2|+x>0$ is
- A$[-2,-\sqrt{2}) \cup(\sqrt{2}, \infty)$
- B$(-\infty,-2) \cup(2, \infty)$
- Correct answerC$(-\infty,-\sqrt{2}) \cup(\sqrt{2}, \infty)$
- D$(-\infty,-2) \cup(\sqrt{2}, \infty)$
Solution
$x^2-|x+2|+x>0$
Case I If $x+2 \geq 0$
$$
\begin{aligned}
& x^2-(x+2)+x>0 \quad\left\{\because|x|=\left\{\begin{array}{cc}
x, & x \geq 0 \\
-x, & x < 0
\end{array}\right\}\right. \\
& \Rightarrow \quad x^2-2>0 \\
& \Rightarrow \quad(x-\sqrt{2})(x+\sqrt{2})=0 \\
& \frac{+}{-2-\sqrt{2} \quad \sqrt{2}} \\
& x \in[-2,-\sqrt{2}] \cup[\sqrt{2}, \infty] \\
&
\end{aligned}
$$
Case II $x+2 < 0$
$$
\begin{aligned}
& x^2+(x+2)+x>0 \\
\Rightarrow & x^2+2 x+2>0 \\
\Rightarrow & (x+1)^2+1>0 \Rightarrow x < -2
\end{aligned}
$$
Hence, the solution set is
$$
x \in(-\infty,-\sqrt{2}) \cup(\sqrt{2}, \infty)
$$
Hence, the correct answer is option: 'C'.