CAT 2025 Slot 2 · Quant · Question 14

\left| x+9 \right| changes form at x=-9, so split the problem there.

For x\ge -9 we have \left| x+9 \right|=x+9, and the expression becomes {x}^{2}-\left( x+9 \right)+x={x}^{2}-9.

{x}^{2}-9>0 gives x<-3 or x>3, which combined with x\ge -9 contributes \left[ -9,-3 \right)\cup \left( 3,\infty \right).

For x<-9 we have \left| x+9 \right|=-\left( x+9 \right), and the expression becomes {x}^{2}+2x+9.

Completing the square, {x}^{2}+2x+9={\left( x+1 \right)}^{2}+8, which is always positive, so the whole of \left( -\infty ,-9 \right) qualifies.

The two pieces meet with no gap, giving \left( -\infty ,-3 \right)\cup \left( 3,\infty \right).

Note that -9 is where the cases split, not where the answer splits — it sits comfortably inside the solution set, which is why options 1 and 4 are wrong.

Shortcut. Test one value from each option. x=0 gives 0-9+0=-9, which is negative, so every option containing 0 dies — that leaves only options 1 and 2. Then x=-5 gives 25-4-5=16>0, which option 1 excludes. Option 2 survives both tests.