CAT 2025

Slot 1

QA

Question & Solution

1. Concept Used

• Topic: Floor Function (Greatest Integer Function) and Inequalities
• Formula: For any integer $$n$$ and $$x \in [n, n+1)$$: $$\lfloor x \rfloor = n$$, and the condition $$\lfloor x^2 \rfloor = \lfloor x \rfloor^2 = n^2$$ requires $$x^2 \in [n^2, n^2 + 1)$$, i.e., $$x \in [n, \sqrt{n^2 + 1})$$

2. Calculation

We need $$\lfloor x^2 \rfloor = \lfloor x \rfloor^2$$ for $$3 \leq x \leq 6$$. We analyze each integer interval separately.

Case 1: $$x \in [3, 4)$$, so $$\lfloor x \rfloor = 3$$ and $$\lfloor x \rfloor^2 = 9$$.

We need $$\lfloor x^2 \rfloor = 9$$, which means $$x^2 \in [9, 10)$$, i.e., $$x \in [3, \sqrt{10})$$.

Since $$\sqrt{10} \approx 3.162$$, this interval is valid within $$[3, 4)$$. So the feasible values here are $$x \in [3, \sqrt{10})$$.

Case 2: $$x \in [4, 5)$$, so $$\lfloor x \rfloor = 4$$ and $$\lfloor x \rfloor^2 = 16$$.

We need $$\lfloor x^2 \rfloor = 16$$, which means $$x^2 \in [16, 17)$$, i.e., $$x \in [4, \sqrt{17})$$.

Since $$\sqrt{17} \approx 4.123$$, the feasible values here are $$x \in [4, \sqrt{17})$$.

Case 3: $$x \in [5, 6)$$, so $$\lfloor x \rfloor = 5$$ and $$\lfloor x \rfloor^2 = 25$$.

We need $$\lfloor x^2 \rfloor = 25$$, which means $$x^2 \in [25, 26)$$, i.e., $$x \in [5, \sqrt{26})$$.

Since $$\sqrt{26} \approx 5.099$$, the feasible values here are $$x \in [5, \sqrt{26})$$.

Case 4: $$x = 6$$, so $$\lfloor x \rfloor = 6$$ and $$\lfloor x \rfloor^2 = 36$$.

$$\lfloor x^2 \rfloor = \lfloor 36 \rfloor = 36 = 6^2 = 36$$. ✓ So $$x = 6$$ is included.

Therefore, the complete set is: $$S = [3, \sqrt{10}) \cup [4, \sqrt{17}) \cup [5, \sqrt{26}) \cup {6}$$

Now we check each option against $$S$$:

• Option B includes $$\sqrt{10}$$ (closed bracket $$[3, \sqrt{10}]$$), but $$\sqrt{10} otin S$$ since $$\lfloor (\sqrt{10})^2 \rfloor = \lfloor 10 \rfloor = 10 eq 9 = \lfloor \sqrt{10} \rfloor^2$$. ❌
• Option C also includes $$\sqrt{10}$$ and $$\sqrt{17}$$ (closed brackets), both of which are NOT in $$S$$. ❌
• Option D includes $$\sqrt{18}$$ and $$\sqrt{27}$$, neither of which are in $$S$$. ❌
• Option A: $$(3, \sqrt{10}) \cup [5, \sqrt{26}) \cup {6}$$ — every element here is a subset of $$S$$. The open interval $$(3, \sqrt{10}) \subset [3, \sqrt{10})$$, and $$[5, \sqrt{26}) \subset S$$, and $${6} \subset S$$. ✅

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3. Solution

Answer = Option A ✅

The complete feasible set is $$S = [3,\sqrt{10}) \cup [4,\sqrt{17}) \cup [5,\sqrt{26}) \cup {6}$$. Option A, which is $$(3,\sqrt{10}) \cup [5,\sqrt{26}) \cup {6}$$, is the only choice that is fully contained within $$S$$ (a valid subset), as Options B, C, and D all include boundary points like $$\sqrt{10}$$, $$\sqrt{17}$$, $$\sqrt{18}$$, or $$\sqrt{27}$$ that do not satisfy the given condition.