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CAT 2025Slot 1QAQues & Sol

AlgebraMedium

Question

Let $3\leq x\leq6$ and $\left[x^{2}\right] =\left[x\right]^{2}$ , where [x] is the greatest integer not exceeding x . If set S represents all feasible values of x, then a possible subset of Sis

Options

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 ${6} \subset S$. ✅

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.