IPMAT Rohtak 2019 (QA) - The set of all real numbers x for which x^2 - |x + 2| + x > 0, is | PYQs + Solutions

IPMAT Rohtak 2019

Algebra

Quadratic Equations

Medium

The set of all real numbers x for which $x^2 - |x + 2| + x > 0$ , is

(-\infty, -2) \cup (2, \infty)

(-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)

(-\infty, -1) \cup (1, \infty)

(\sqrt{2}, \infty)

Correct Option: 2

From the given data,

x^2 - |x + 2| + x > 0

We can discuss this problem by observing two different cases.

Case 1: When

(x + 2) \geq 0

Therefore,

x^2 - (x + 2) + x > 0

Hence,

x^2 - 2 > 0

x^2 > 2

\Rightarrow x < -\sqrt{2}

x > \sqrt{2}

Hence,

x \in (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)

......

(1)

Case 2: When

(x + 2) < 0

Then

x^2 + x + 2 + x > 0

So,

x^2 + 2x + 2 > 0

This gives

(x + 1)^2 + 1 > 0

and this is true for every

x

(Because

(x + 1)^2 \geq 0 \Rightarrow (x + 1)^2 + 1 \geq 1

)

Hence,

x \in (-\infty, \infty)

......

(2)

From equations

(1)

and

(2)

, we take the intersection, hence we get

x \in (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)

Therefore,

x \in (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty)

is the required answer.