Let `f(x)=[x^(2)]+[x+2]-8`, where [x] denotes the greater integer than or equal to x , then

Options

A. `f(x) ne 0 ` for all ` x in R `
B. `f(x)=0` only for two real values of x
C. `f(x)=0` for infinity many values of x
D. none of these

Explanation

To solve the problem, we need to analyze the function \( f(x) = [x^2] + [x + 2] - 8 \), where \( [x] \) denotes the greatest integer function (also known as the floor function).

### Step 1: Rewrite the Function
We can rewrite the function as:
\[ f(x) = [x^2] + [x] + 2 - 8 \]
This simplifies to:
\[ f(x) = [x^2] + [x] - 6 \]

### Step 2: Let \( [x] = t \)
Let \( [x] = t \). Since \( t \) is the greatest integer less than or equal to \( x \), we have:
\[ t \leq x < t + 1 \]
Now, we can express \( f(x) \) in terms of \( t \):
\[ f(x) = [x^2] + t - 6 \]

### Step 3: Determine the Range of \( x \)
Since \( [x] = t \), we can express \( x \) as:
\[ t \leq x < t + 1 \]
Now, we need to find \( [x^2] \). The square of \( x \) will fall between:
\[ t^2 \leq x^2 < (t + 1)^2 \]
This gives us:
\[ t^2 \leq [x^2] < t^2 + 2t + 1 \]
Thus, \( [x^2] \) can take values from \( t^2 \) to \( t^2 + 2t \).

### Step 4: Set Up the Equation
We need to find when \( f(x) = 0 \):
\[ [x^2] + t - 6 = 0 \implies [x^2] = 6 - t \]
This means:
\[ t^2 \leq 6 - t < t^2 + 2t \]

### Step 5: Solve the Inequalities
1. **First Inequality**:
\[ t^2 \leq 6 - t \implies t^2 + t - 6 \leq 0 \]
Factoring gives:
\[ (t - 2)(t + 3) \leq 0 \]
The solution to this inequality is:
\[ -3 \leq t \leq 2 \]

2. **Second Inequality**:
\[ 6 - t < t^2 + 2t \implies t^2 + 3t - 6 > 0 \]
Factoring gives:
\[ (t - 2)(t + 3) > 0 \]
The solution to this inequality is:
\[ t < -3 \quad \text{or} \quad t > 2 \]

### Step 6: Combine the Solutions
From the first inequality, we have \( -3 \leq t \leq 2 \), and from the second inequality, we have \( t < -3 \) or \( t > 2 \). The only overlapping solution is:
\[ t = -3 \quad \text{and} \quad t = 2 \]

### Step 7: Find Corresponding \( x \) Values
1. For \( t = -3 \):
\[ -3 \leq x < -2 \quad \text{(i.e., } x \in [-3, -2)) \]
2. For \( t = 2 \):
\[ 2 \leq x < 3 \quad \text{(i.e., } x \in [2, 3)) \]

### Final Result
Thus, the values of \( x \) for which \( f(x) = 0 \) are:
\[ x \in [-3, -2) \cup [2, 3) \]