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

Step 1: Write \(x = n + f\), where \(n = [x]\) is an integer and \(0 \le f < 1\). Then \(x^2 = n^2 + 2nf + f^2\), so \([x^2] = n^2\) exactly when \(0 \le 2nf + f^2 < 1\).
Step 2: Solve \(f^2 + 2nf - 1 < 0\) for \(f \ge 0\): the positive root of \(f^2+2nf-1=0\) is \(f = -n + \sqrt{n^2+1}\), so the condition becomes \(0 \le f < -n + \sqrt{n^2+1}\), i.e. \(n \le x < \sqrt{n^2+1}\).
Step 3: Apply this for \(n = 3, 4, 5\): \(x \in [3,\sqrt{10}) \cup [4,\sqrt{17}) \cup [5,\sqrt{26})\), plus the isolated point \(x = 6\) (checked directly, \(6^2=36=[6]^2\)). Matching against the options, only \((3,\sqrt{10}) \cup [5,\sqrt{26}) \cup \{6\}\) sits entirely inside this set.
\[ \boxed{\text{Option (1)}} \]