Question

Let $3 \leq x \leq 6$ and $[x^2] = [x]^2$, where $[x]$ is the greatest integer not exceeding $x$. If set $S$ represents all feasible values of $x$, then which of the following is a possible subset of $S$?

• A. \((3, \sqrt{10}) \cup [5, \sqrt{26}) \cup \{6\}\)
• B. \((4, \sqrt{10}) \cup [5, \sqrt{27}) \cup \{6\}\)
• C. \([3, \sqrt{10}] \cup [5, \sqrt{26}]\)
• D. \([3, \sqrt{10}] \cup [4, \sqrt{17}] \cup \{6\}\)

Hint:

For equations involving the floor function, always: \begin{itemize} \item Break the domain into intervals where $[x]$ is constant. \item Translate conditions like $[x^2]=k$ into inequalities: $k \le x^2<k+1$. \item Carefully handle open vs closed endpoints when forming solution intervals. \end{itemize}

Answer 1

The Correct Option is A

Approach: The greatest-integer function \([x]\) is constant on each unit interval, so \([x]^2\) is a fixed number there. Just demand that \([x^2]\) equals that fixed number on each strip \([3,4),[4,5),[5,6)\) and on \(\{6\}\); this caps how large \(x\) may grow inside each strip.

Step 1: Strip \(3 \le x < 4\). Here \([x] = 3\), so \([x]^2 = 9\). Need \([x^2] = 9\), i.e. \(9 \le x^2 < 10\), giving \(3 \le x < \sqrt{10}\ (\approx 3.162)\). Solutions: \([3, \sqrt{10})\).

Step 2: Strip \(4 \le x < 5\). Here \([x]^2 = 16\). Need \(16 \le x^2 < 17\), i.e. \(4 \le x < \sqrt{17}\ (\approx 4.123)\). Solutions: \([4, \sqrt{17})\).

Step 3: Strip \(5 \le x < 6\). Here \([x]^2 = 25\). Need \(25 \le x^2 < 26\), i.e. \(5 \le x < \sqrt{26}\ (\approx 5.099)\). Solutions: \([5, \sqrt{26})\).

Step 4: The point \(x = 6\). \([x] = 6,\ [x]^2 = 36 = x^2 = [x^2]\). Works.

So the full set is
\[ S = [3, \sqrt{10}) \cup [4, \sqrt{17}) \cup [5, \sqrt{26}) \cup \{6\}. \]

Step 5: Which option is a subset of \(S\)?
Option (1): \((3,\sqrt{10})\) lies inside \([3,\sqrt{10})\); \([5,\sqrt{26})\) is exactly a piece of \(S\); \(\{6\}\in S\). So (1) \(\subseteq S\).
Option (2) reaches \(\sqrt{27} > \sqrt{26}\) — outside \(S\). Options (3) and (4) include closed endpoints like \(\sqrt{10}, \sqrt{26}, \sqrt{17}\) where \([x^2]\) jumps and the equality fails.

Answer: \((3, \sqrt{10}) \cup [5, \sqrt{26}) \cup \{6\}\) — option (1).

Answer 2

Approach: Instead of building the full set \(S\) symbolically, test each option with concrete sample points and boundary checks. A single point that breaks the equality \([x^2] = [x]^2\) is enough to reject an option; checking a few interior points confirms a surviving option.

Step 1: Test option (1)'s interior points. Take \(x = 3.1\): \([x] = 3\), \([x]^2 = 9\); \(x^2 = 9.61\), \([x^2] = 9\). Equal, fine. Take \(x = 5.05\): \([x]=5, [x]^2=25\); \(x^2 = 25.5025\), \([x^2]=25\). Equal, fine. And \(x = 6\) clearly works since \([6]^2 = 36 = [36]\).

Step 2: Test option (2)'s extra range. Option (2) reaches up to \(\sqrt{27} \approx 5.196\), beyond \(\sqrt{26} \approx 5.099\). Take \(x = 5.15\), which lies in \((\sqrt{26}, \sqrt{27})\): \([x] = 5\), \([x]^2 = 25\); \(x^2 = 26.5225\), \([x^2] = 26 \neq 25\). Fails, option (2) is out. (Its piece \((4,\sqrt{10})\) is also nonsensical since \(\sqrt{10} \approx 3.162 < 4\).)

Step 3: Test the closed endpoints in options (3) and (4). At \(x = \sqrt{10}\) exactly: \(x^2 = 10\) so \([x^2] = 10\), but \([x] = 3\) so \([x]^2 = 9 \neq 10\). Since options (3) and (4) both include \(\sqrt{10}\) as a closed endpoint, they fail at that very point.

Answer: only option (1) survives every test, option (1).