Question

If $[x]^2 - 5[x] + 6 = 0$, where $[.]$ denotes the greatest integer function, then ______.

• A. $x \in (2, 4)$
• B. $x \in [2, 4]$
• C. $x \in [2, 4)$
• D. $x \in (2, 4]$

Hint:

If $[x] = n$, the valid interval is ALWAYS $[n, n+1)$. If you have consecutive integer solutions like $n$ and $n+1$, they merge seamlessly into $[n, n+2)$!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a quadratic equation where the variable is the Greatest Integer Function (GIF), denoted by $[x]$. We need to solve for $[x]$ first, and then translate those specific integer values into an active interval range for the real variable $x$.

Step 2: Detailed Explanation:
Treat $[x]$ as a single continuous variable (let $t = [x]$) to solve the quadratic equation:
$t^2 - 5t + 6 = 0$
Factor the quadratic equation. We need two numbers that multiply to $+6$ and add to $-5$. These are $-2$ and $-3$:
$(t - 2)(t - 3) = 0$
This yields two solutions for $t$:
$t = 2 \implies [x] = 2$
$t = 3 \implies [x] = 3$
Now we evaluate what physical range of real numbers $x$ produces these specific GIF outputs.
By definition, the Greatest Integer Function $[x]$ equals the largest integer less than or equal to $x$.
1. If $[x] = 2$, then $x$ can be any real number starting precisely at 2, up to (but strictly not including) 3.
Range 1: $2 \le x < 3$, or $x \in [2, 3)$.
2. If $[x] = 3$, then $x$ can be any real number starting precisely at 3, up to (but strictly not including) 4.
Range 2: $3 \le x < 4$, or $x \in [3, 4)$.
The total valid solution set is the complete union of these two adjacent intervals:
$x \in [2, 3) \cup \[3, 4)$
Because the intervals connect perfectly back-to-back at exactly $x = 3$, they merge seamlessly into one single unbroken interval:
$x \in [2, 4)$

Step 3: Final Answer:
The interval is $x \in [2, 4)$, matching option (c).