Question

The set of all real numbers $x$ for which ${x^2} - \left| {x + 2} \right| + x > 0$ 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 1

Here, in the given question, we are given ${x^2} - \left| {x + 2} \right| + x > 0$ and we need to find the set of all real numbers $x$. As here, we are given modulus in function, so at first we will find the values of $x$ using modulus and solve the two cases separately. From both the cases we will find the set of real numbers. The common interval of both the sets will be our required answer.

Complete step by step answer:
We have, ${x^2} - \left| {x + 2} \right| + x > 0$
Case 1: If $x + 2 \geqslant 0$, that is $x \geqslant - 2$, we get
${x^2} - \left( {x + 2} \right) + x > 0$
On opening the bracket, we get
${x^2} - x - 2 + x > 0$
On subtraction of like terms, we get
${x^2} - 2 > 0$
$\Rightarrow {x^2} - {\left( {\sqrt 2 } \right)^2} > 0$

As we know ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$. Therefore, we get
$\left( {x - \sqrt 2 } \right)\left( {x + \sqrt 2 } \right) > 0$
From here, we get
$x \in \left( { - \infty , - \sqrt 2 } \right) \cup \left( {\sqrt 2 ,\infty } \right)$
But $x \geqslant - 2$. Hence,
$x \in \left( { - 2, - \sqrt 2 } \right) \cup \left( {\sqrt 2 ,\infty } \right).......\left( i \right)$

Case 2: If $x + 2 < 0$, that is $x < \- 2$, we get
$\Rightarrow {x^2} + x + 2 + x > 0$
On addition of like terms, we get
$\Rightarrow {x^2} + 2x + 2 > 0$
We can also write it as,
$\Rightarrow \left( {{x^2} + 2x + 1} \right) + 1 > 0$

On completing the square, we get
$\Rightarrow {\left( {x + 1} \right)^2} + 1 > 0$
Which is true for all $x$. Therefore,
$\Rightarrow x \in \left( { - \infty , - 2} \right).......\left( {ii} \right)$
Thus, from equation $\left( i \right)$ and $\left( {ii} \right)$, we get
$\therefore x \in \left( { - \infty , - \sqrt 2 } \right) \cup \left( {\sqrt 2 ,\infty } \right)$

Therefore, the correct option is B.

Note: To solve this type of questions, one must know the common algebraic identities. One must also know that the sum (or difference) of two like terms is a like term with coefficient equal to the sum (or difference) of coefficients of the two like terms. When we add (or subtract) two algebraic expressions, the like terms are added (or subtracted) and the unlike terms are written as they are.