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, $$x \geqslant - 2,{x^2} - x - 2 + x > 0$$
$$\eqalign{
& \Rightarrow {x^2} > 2 \cr
& \Rightarrow x \in \left( { - \infty , - \sqrt 2 } \right) \cup \left( {\sqrt 2 ,\infty } \right) \cr
& \Rightarrow x \in \left[ { - 2, - \sqrt 2 } \right) \cup \left( {\sqrt 2 ,\infty } \right)\,\,{\text{For, }}x < - 2 \cr
& {x^2} + x + 2 + x > 0{\text{ or }}{x^2} + 2x + 2 > 0 \cr} $$
which is true for all $$x.$$
Hence, $$x \in \left( { - \infty , - \sqrt 2 } \right) \cup \left( {\sqrt 2 ,\infty } \right)$$