The solution set of the equation `[x]^(2) +[x+1] -3=0,` where `[.]` represents greatest integeral function is :

Options

A. `[-1, o)uu[1,2)`
B. `[-2,-1)uu[1,2)`
C. `[1,2]`
D. `[-3,-2)uu[2,3)`

Answer

To solve the equation \([x]^2 + [x+1] - 3 = 0\), where \([.]\) represents the greatest integer function (also known as the floor function), we will follow these steps:

### Step 1: Rewrite the equation
The equation can be rewritten as:
\[ [x]^2 + [x] + 1 - 3 = 0 \]
This simplifies to:
\[ [x]^2 + [x] - 2 = 0 \]

### Step 2: Let \([x] = t\)
Let \(t = [x]\). Then the equation becomes:
\[ t^2 + t - 2 = 0 \]

### Step 3: Factor the quadratic equation
We can factor the quadratic equation:
\[ t^2 + 2t - t - 2 = 0 \]
This can be grouped and factored as:
\[ (t + 2)(t - 1) = 0 \]

### Step 4: Find the values of \(t\)
Setting each factor to zero gives us:
1. \(t + 2 = 0 \Rightarrow t = -2\)
2. \(t - 1 = 0 \Rightarrow t = 1\)

### Step 5: Translate back to \(x\)
Now we will translate these values of \(t\) back to \(x\) using the definition of the greatest integer function:
1. If \([x] = -2\), then:
\[ -2 \leq x < -1 \quad \text{(i.e., } x \in [-2, -1) \text{)} \]
2. If \([x] = 1\), then:
\[ 1 \leq x < 2 \quad \text{(i.e., } x \in [1, 2) \text{)} \]

### Step 6: Combine the solution sets
The solution set for the equation is:
\[ x \in [-2, -1) \cup [1, 2) \]

### Final Answer
Thus, the solution set of the equation \([x]^2 + [x+1] - 3 = 0\) is:
\[ [-2, -1) \cup [1, 2) \]

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