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 \([.]\) denotes 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:
\[ [x]^2 + [x + 1] - 3 = 0 \]
Let \(t = [x]\). Then, we know that \([x + 1] = t + 1\) because the greatest integer function increases by 1 when we add 1 to \(x\).

### Step 2: Substitute into the equation
Substituting \(t\) into the equation gives:
\[ t^2 + (t + 1) - 3 = 0 \]
This simplifies to:
\[ t^2 + t + 1 - 3 = 0 \implies t^2 + t - 2 = 0 \]

### Step 3: Factor the quadratic equation
Next, we factor the quadratic equation:
\[ t^2 + t - 2 = (t + 2)(t - 1) = 0 \]
Setting each factor to zero gives us the roots:
\[ t + 2 = 0 \implies t = -2 \]
\[ t - 1 = 0 \implies t = 1 \]

### Step 4: Determine the corresponding \(x\) values
Now, we need to find the values of \(x\) corresponding to \(t = -2\) and \(t = 1\).

1. For \(t = -2\):
\[ [x] = -2 \implies -2 \leq x < -1 \]
This means \(x\) can take any value in the interval \([-2, -1)\).

2. For \(t = 1\):
\[ [x] = 1 \implies 1 \leq x < 2 \]
This means \(x\) can take any value in the interval \([1, 2)\).

### Step 5: Combine the intervals
The solution set for \(x\) is the union of the two intervals:
\[ x \in [-2, -1) \cup [1, 2) \]

### Final Answer
Thus, the solution set of the equation is:
\[ [-2, -1) \cup [1, 2) \]