Question
The set of all real numbers $$x$$ for which $${x^2} - \left| {x + 2} \right| + x > 0,\,{\text{is}}$$
A.
$$\left( { - \infty , - 2} \right) \cup \left( {2,\infty } \right)$$
B.
$$\left( { - \infty , - \sqrt 2 } \right) \cup \left( {\sqrt 2 ,\infty } \right)$$
C.
$$\left( { - \infty , - 1} \right) \cup \left( {1,\infty } \right)$$
D.
$$\left( {\sqrt 2 ,\infty } \right)$$
Answer :
$$\left( { - \infty , - \sqrt 2 } \right) \cup \left( {\sqrt 2 ,\infty } \right)$$
Solution :
For $$x < - 2,$$
$$\left| {x + 2} \right| = - \left( {x + 2} \right)$$ and the inequality becomes
$$\eqalign{
& {x^2} + x + 2 + x > 0 \cr
& \Rightarrow \,{\left( {x + 1} \right)^2} + 1 > 0 \cr
& {\text{which is valid }}\forall \,\,x \in R\,\,{\text{but }}x < - 2 \cr
& \therefore \,\,x \in \left( { - \infty , - 2} \right)\,\,\,\,\,\,\,\,\,.....\left( 1 \right) \cr
& {\text{For }}x \geqslant 2,\left| {x + 2} \right| = x + 2\,\,{\text{and the inequality becomes}} \cr
& {x^2} - x - 2 + x > 0 \cr
& \Rightarrow \,\,{x^2} > 2 \cr
& \Rightarrow \,\,x > \sqrt 2 \,\,{\text{or }}x < - \sqrt 2 \cr
& {\text{i}}{\text{.e}}{\text{., }}x \in \left( { - \infty , - \sqrt 2 } \right) \cup \left( {\sqrt 2 ,\infty } \right) \cr
& {\text{but }}x \geqslant - 2 \cr
& \Rightarrow \,\,x \in \left[ { - 2, - \sqrt 2 } \right) \cup \left( {\sqrt 2 ,\infty } \right)\,\,\,\,\,\,\,.....\left( 2 \right) \cr
& {\text{From}}\left( 1 \right){\text{and}}\left( 2 \right) \cr
& x \in \left( { - \infty , - 2} \right) \cup \left[ { - 2, - \sqrt 2 } \right) \cup \left( {\sqrt 2 ,\infty } \right) \cr
& \Rightarrow \,\,x \in \left( { - \infty , - \sqrt 2 } \right) \cup \left( {\sqrt 2 ,\infty } \right) \cr} $$