Let `f:NtoN` be defined by
`f(x)=(x+1)^(2)+x-[sqrt((x+1)^(2)+(x+1))]^(2)`
([] denotes the greatest integer function). Suppose a,b,c are three distinct natural numbers. Let
`Delta=|(f(a),a^(2),a),(f(b),b^(2),b),(f(c),c^(2),c)|`
Then `Delta` is equal to

Options

A. `-(a+b+c)`
B. `a+b+c`
C. `-1`
D. `0`

Answer

To solve the problem, we need to compute the determinant \( \Delta = |(f(a), a^2, a), (f(b), b^2, b), (f(c), c^2, c)| \) where \( f(x) = (x+1)^2 + x - \lfloor \sqrt{(x+1)^2 + (x+1)} \rfloor^2 \).

### Step 1: Simplify the function \( f(x) \)

First, we simplify \( f(x) \):
\[ f(x) = (x+1)^2 + x - \lfloor \sqrt{(x+1)^2 + (x+1)} \rfloor^2 \]
Calculating \( (x+1)^2 + (x+1) \):
\[ (x+1)^2 + (x+1) = x^2 + 2x + 1 + x + 1 = x^2 + 3x + 2 \]
Now, we need to find \( \lfloor \sqrt{x^2 + 3x + 2} \rfloor \).

### Step 2: Find \( \sqrt{x^2 + 3x + 2} \)

The expression \( x^2 + 3x + 2 \) can be factored:
\[ x^2 + 3x + 2 = (x+1)(x+2) \]
Thus,
\[ \sqrt{x^2 + 3x + 2} = \sqrt{(x+1)(x+2)} \]

### Step 3: Determine \( \lfloor \sqrt{(x+1)(x+2)} \rfloor \)

For natural numbers \( x \), \( (x+1)(x+2) \) is always a perfect square or close to it. Therefore, we can analyze the values of \( \lfloor \sqrt{(x+1)(x+2)} \rfloor \):
- For \( x = 1 \): \( \sqrt{(1+1)(1+2)} = \sqrt{6} \approx 2.45 \Rightarrow \lfloor \sqrt{6} \rfloor = 2 \)
- For \( x = 2 \): \( \sqrt{(2+1)(2+2)} = \sqrt{12} \approx 3.46 \Rightarrow \lfloor \sqrt{12} \rfloor = 3 \)
- For \( x = 3 \): \( \sqrt{(3+1)(3+2)} = \sqrt{20} \approx 4.47 \Rightarrow \lfloor \sqrt{20} \rfloor = 4 \)

Continuing this process, we find that \( \lfloor \sqrt{(x+1)(x+2)} \rfloor \) will yield \( x + 1 \) for \( x \geq 1 \).

### Step 4: Substitute back into \( f(x) \)

Thus, we can simplify \( f(x) \):
\[ f(x) = (x+1)^2 + x - (x+1)^2 = x \]

### Step 5: Calculate the determinant \( \Delta \)

Now substituting \( f(a) = a \), \( f(b) = b \), \( f(c) = c \):
\[ \Delta = |(a, a^2, a), (b, b^2, b), (c, c^2, c)| \]

### Step 6: Evaluate the determinant

The determinant can be expressed as:
\[ \Delta = \begin{vmatrix} a & a^2 & a \\ b & b^2 & b \\ c & c^2 & c \end{vmatrix} \]
Notice that the first and third columns are identical. Therefore, the determinant evaluates to zero:
\[ \Delta = 0 \]

### Final Answer
\[ \Delta = 0 \]