The solution of `(x)^(2)+(x+1)^(2)=25` (where (.) denotes the least integer function) is

Options

A. (2, 4)
B. `(-5,-4]uu(2,3]`
C. `[-4,-3)uu[3,4)`
D. none of these

Answer

To solve the equation \(\lfloor x \rfloor^2 + \lfloor x + 1 \rfloor^2 = 25\), where \(\lfloor . \rfloor\) denotes the least integer function (floor function), we can follow these steps:

### Step 1: Rewrite the equation
We know that \(\lfloor x + 1 \rfloor = \lfloor x \rfloor + 1\). Therefore, we can rewrite the equation as:
\[ \lfloor x \rfloor^2 + (\lfloor x \rfloor + 1)^2 = 25 \]

### Step 2: Expand the equation
Now, expand the second term:
\[ \lfloor x \rfloor^2 + (\lfloor x \rfloor^2 + 2\lfloor x \rfloor + 1) = 25 \]
This simplifies to:
\[ 2\lfloor x \rfloor^2 + 2\lfloor x \rfloor + 1 = 25 \]

### Step 3: Rearrange the equation
Next, we rearrange the equation:
\[ 2\lfloor x \rfloor^2 + 2\lfloor x \rfloor + 1 - 25 = 0 \]
This simplifies to:
\[ 2\lfloor x \rfloor^2 + 2\lfloor x \rfloor - 24 = 0 \]

### Step 4: Divide by 2
To simplify further, divide the entire equation by 2:
\[ \lfloor x \rfloor^2 + \lfloor x \rfloor - 12 = 0 \]

### Step 5: Factor the quadratic equation
Now, we need to factor this quadratic equation. We look for two numbers that multiply to -12 and add to 1. The numbers 4 and -3 work:
\[ (\lfloor x \rfloor + 4)(\lfloor x \rfloor - 3) = 0 \]

### Step 6: Solve for \(\lfloor x \rfloor\)
Setting each factor to zero gives us:
1. \(\lfloor x \rfloor + 4 = 0 \Rightarrow \lfloor x \rfloor = -4\)
2. \(\lfloor x \rfloor - 3 = 0 \Rightarrow \lfloor x \rfloor = 3\)

### Step 7: Determine the corresponding \(x\) values
Now we find the ranges for \(x\):
1. If \(\lfloor x \rfloor = -4\), then:
\(-5 < x \leq -4\)
2. If \(\lfloor x \rfloor = 3\), then:
\(3 \leq x < 4\)

### Final Solution
Combining both ranges, we have:
\[ x \in (-5, -4] \cup [3, 4) \]