CAT 2025 Slot 2 QA Question 20

Multiple choice (+3 / −1) · Algebra · Modulus · Try it, then check the answer and solution below.

The set of all real values of

x

for which $(x^2 - |x + 9| + x) > 0 , is

⌨️ Shortcuts:[1-4 / A-D] Select[Enter] Check[B] Star[S] Solution

Official Convenor Answer Key: D

Answer and solution

Answer: D) $(-\infty,-3) \cup (3,\infty)$

📌 Core Concept

The problem involves solving the inequality

x^2 - |x + 9| + x > 0

. To handle the absolute value, we consider two cases based on the critical point where the expression inside the absolute value changes sign.

🔢 Step-by-Step Solution

Identify the Critical Point:

The expression inside the absolute value is

x + 9 , which equals zero when

x = -9$.

Case 1:

x \geq -9

Here,

|x + 9| = x + 9

Substitute into the inequality:

x^2 - (x + 9) + x > 0 \implies x^2 - x - 9 + x > 0 \implies x^2 - 9 > 0

Simplify:

x^2 > 9 \implies |x| > 3 \implies x > 3 \text{ or } x < -3

Considering

x \geq -9 , the solution is - 9 \leq x < -3

x > 3

Case 2:

x < -9

Here, \( |x + 9| =

Keep going