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

We are given: \[ 3 \le x \le 6,\quad [x^2] = [x]^2. \] Since $[x]$ is constant on each interval between consecutive integers, we split $[3,6]$ into: \[ [3,4),\ [4,5),\ [5,6),\ \{6\}. \] Case 1: \(3 \le x<4\) Here, \([x] = 3\), so \([x]^2 = 9\). Condition: \[ [x^2] = 9 \;\Longrightarrow\; 9 \le x^2<10. \] Taking square roots (and noting \(x \ge 3\)): \[ 3 \le x<\sqrt{10}. \] This lies inside \([3,4)\) since \(\sqrt{10} \approx 3.16\). So solutions in this case: \[ [3, \sqrt{10}). \] Case 2: \(4 \le x<5\) Here, \([x] = 4\), so \([x]^2 = 16\). Condition: \[ [x^2] = 16 \;\Longrightarrow\; 16 \le x^2<17. \] Thus: \[ 4 \le x<\sqrt{17},\quad\text{where }\sqrt{17} \approx 4.12. \] So solutions: \[ [4, \sqrt{17}). \] Case 3: \(5 \le x<6\) Here, \([x] = 5\), so \([x]^2 = 25\). Condition: \[ [x^2] = 25 \;\Longrightarrow\; 25 \le x^2<26, \] hence: \[ 5 \le x<\sqrt{26},\quad\text{where }\sqrt{26} \approx 5.10. \] So solutions: \[ [5, \sqrt{26}). \] Case 4: \(x = 6\) \[ [x] = 6,\quad [x]^2 = 36,\quad x^2 = 36 \Rightarrow [x^2] = 36, \] so the equality holds and \(x=6\) is a solution. Thus, the full solution set: \[ S = [3, \sqrt{10}) \cup [4, \sqrt{17}) \cup [5, \sqrt{26}) \cup \{6\}. \] Now check each option as a \emph{subset} of $S$:
- Option (A): \((3, \sqrt{10})\) is contained in \([3, \sqrt{10})\); \([5, \sqrt{26})\) matches exactly a part of $S$; \(\{6\}\) is in $S$. Hence (A) \(\subseteq S\); valid.
 - Option (B): \([5, \sqrt{27})\) goes beyond \(\sqrt{26}\); for \(x \in (\sqrt{26}, \sqrt{27})\), we get \([x^2]=26\) but \([x]^2=25\), so those $x$ are \emph{not} in $S$. Hence (B) is not a subset. 
- Option (C): Includes $x=\sqrt{10}$ and $x=\sqrt{26}$ (closed intervals). At \(x=\sqrt{10}\), we have \([x^2]=10\) but \([x]^2 = 3^2 = 9\), so not in $S$. Similarly, \(\sqrt{26}\) is not included in $S$ (only $x<\sqrt{26}$). Hence (C) is not a subset. 
- Option (D): Includes the endpoints \(\sqrt{10}\) and \(\sqrt{17}\), which are not in $S$ for the same reason as above. So (D) is not a subset. Therefore, the only valid subset among the options is: \[ \boxed{(3, \sqrt{10}) \cup [5, \sqrt{26}) \cup \{6\}}. \]