CAT 2025 Slot 1 · Quant · Question 22

Take n=⌊x⌋, so n\le x<n+1 and the condition reads ⌊{x}^{2}⌋={n}^{2}.

That means {n}^{2}\le{x}^{2}<{n}^{2}+1, and since x is positive, n\le x<\sqrt{{n}^{2}+1}.

On 3\le x\le 6 the value of n is 3, 4 or 5, plus the endpoint x=6 (where ⌊36⌋=36={6}^{2}).

So, S=\left[ 3,\sqrt{10}\right)\cup \left[ 4,\sqrt{17}\right)\cup \left[ 5,\sqrt{26}\right)\cup \{6\}.

Now read the question carefully: it asks for a subset of S, not for S itself.

Each upper endpoint is excluded, because at x=\sqrt{{n}^{2}+1} we get ⌊{x}^{2}⌋={n}^{2}+1.

Option 1 includes values up to \sqrt{18}, beyond \sqrt{17}. Options 2 and 3 are closed at \sqrt{10}, which is not in S.

Option 4 is \left( 3,\sqrt{10}\right)\cup \left[ 5,\sqrt{26}\right)\cup \{6\}, and every piece sits inside S, so it is the answer.