If `[x]^(2)-3[x]+2=0` where [*] denotes the greatest integer function, then

Options

A. `x in [2,3]`
B. `x in(1,2]`
C. `x in [1,2]`
D. `x in [1,3)`

Answer

To solve the equation \([x]^2 - 3[x] + 2 = 0\), where \([x]\) denotes the greatest integer function (also known as the floor function), we will follow these steps:

### Step 1: Substitute the Greatest Integer Function
Let \(y = [x]\). The equation then becomes:
\[ y^2 - 3y + 2 = 0 \]

### Step 2: Factor the Quadratic Equation
We can factor the quadratic equation:
\[ y^2 - 3y + 2 = (y - 1)(y - 2) = 0 \]
This gives us the solutions:
\[ y - 1 = 0 \quad \Rightarrow \quad y = 1 \]
\[ y - 2 = 0 \quad \Rightarrow \quad y = 2 \]

### Step 3: Interpret the Solutions
Since \(y = [x]\), we have two cases:
1. If \([x] = 1\), then \(x\) lies in the interval:
\[ 1 \leq x < 2 \quad \text{(closed interval at 1, open interval at 2)} \]
2. If \([x] = 2\), then \(x\) lies in the interval:
\[ 2 \leq x < 3 \quad \text{(closed interval at 2, open interval at 3)} \]

### Step 4: Combine the Intervals
Now, we combine the intervals from both cases:
- From case 1: \( [1, 2) \)
- From case 2: \( [2, 3) \)

Combining these intervals gives us:
\[ [1, 2) \cup [2, 3) = [1, 3) \]

### Final Result
Thus, the solution is:
\[ x \in [1, 3) \]