Let `[x]` denote the greatest integer function. The number of solutions of the equation `x^(2)-3x+[x]=0` in `[0,3]` is

Options

A. `4`
B. `3`
C. `2`
D. `1`

Answer

To solve the equation \( x^2 - 3x + [x] = 0 \) where \([x]\) denotes the greatest integer function, we will analyze the equation in different intervals based on the value of \([x]\).

### Step 1: Define the intervals based on \([x]\)

The greatest integer function \([x]\) takes integer values. In the interval \([0, 3]\), we have three cases to consider:
1. \(0 \leq x < 1\) where \([x] = 0\)
2. \(1 \leq x < 2\) where \([x] = 1\)
3. \(2 \leq x < 3\) where \([x] = 2\)
4. \(x = 3\) where \([x] = 3\)

### Step 2: Analyze each case

**Case 1: \(0 \leq x < 1\)**

Here, \([x] = 0\). The equation becomes:
\[ x^2 - 3x + 0 = 0 \implies x^2 - 3x = 0 \]
Factoring gives:
\[ x(x - 3) = 0 \]
The solutions are \(x = 0\) and \(x = 3\). However, since \(x\) must be in the interval \([0, 1)\), we only accept \(x = 0\) as a valid solution.

**Case 2: \(1 \leq x < 2\)**

Here, \([x] = 1\). The equation becomes:
\[ x^2 - 3x + 1 = 0 \]
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \]
Calculating the roots:
\[ x_1 = \frac{3 + \sqrt{5}}{2} \quad \text{and} \quad x_2 = \frac{3 - \sqrt{5}}{2} \]
Approximating the values:
- \(x_1 \approx 2.618\) (not in the interval \([1, 2)\))
- \(x_2 \approx 0.382\) (not in the interval \([1, 2)\))

Thus, there are no valid solutions in this interval.

**Case 3: \(2 \leq x < 3\)**

Here, \([x] = 2\). The equation becomes:
\[ x^2 - 3x + 2 = 0 \]
Factoring gives:
\[ (x - 1)(x - 2) = 0 \]
The solutions are \(x = 1\) and \(x = 2\). However, since \(x\) must be in the interval \([2, 3)\), we only accept \(x = 2\) as a valid solution.

**Case 4: \(x = 3\)**

Here, \([x] = 3\). The equation becomes:
\[ x^2 - 3x + 3 = 0 \]
Calculating the discriminant:
\[ D = b^2 - 4ac = 9 - 12 = -3 \]
Since the discriminant is negative, there are no real solutions at \(x = 3\).

### Step 3: Count the valid solutions

From the analysis:
- From Case 1: 1 solution (\(x = 0\))
- From Case 2: 0 solutions
- From Case 3: 1 solution (\(x = 2\))
- From Case 4: 0 solutions

Thus, the total number of solutions in the interval \([0, 3]\) is:
\[ \text{Total Solutions} = 1 + 0 + 1 + 0 = 2 \]

### Final Answer
The number of solutions of the equation \(x^2 - 3x + [x] = 0\) in \([0, 3]\) is **2**.