If `f(x)=ax^(2)+bx+c and f(x+1)=f(x)+x+1`, then the value of (a+b) is __

Options

A. -2
B. -1
C. 0
D. 1

Answer

To solve the problem, we start with the given function:

\[ f(x) = ax^2 + bx + c \]

We also know that:

\[ f(x+1) = f(x) + x + 1 \]

### Step 1: Calculate \( f(x+1) \)

To find \( f(x+1) \), we substitute \( x+1 \) into the function \( f(x) \):

\[ f(x+1) = a(x+1)^2 + b(x+1) + c \]

Expanding this:

\[ = a(x^2 + 2x + 1) + b(x + 1) + c \]
\[ = ax^2 + 2ax + a + bx + b + c \]
\[ = ax^2 + (2a + b)x + (a + b + c) \]

### Step 2: Calculate \( f(x) + x + 1 \)

Now, we calculate \( f(x) + x + 1 \):

\[ f(x) + x + 1 = (ax^2 + bx + c) + x + 1 \]
\[ = ax^2 + (b + 1)x + (c + 1) \]

### Step 3: Set the two expressions equal

Now we set the two expressions equal to each other:

\[ ax^2 + (2a + b)x + (a + b + c) = ax^2 + (b + 1)x + (c + 1) \]

### Step 4: Compare coefficients

Since the coefficients of \( x^2 \) are equal, we can ignore them. Now we compare the coefficients of \( x \) and the constant terms:

1. For the coefficients of \( x \):
\[ 2a + b = b + 1 \]
Subtract \( b \) from both sides:
\[ 2a = 1 \quad \Rightarrow \quad a = \frac{1}{2} \]

2. For the constant terms:
\[ a + b + c = c + 1 \]
Subtract \( c \) from both sides:
\[ a + b = 1 \]

### Step 5: Substitute the value of \( a \)

Now we substitute \( a = \frac{1}{2} \) into the equation \( a + b = 1 \):

\[ \frac{1}{2} + b = 1 \]
Subtract \( \frac{1}{2} \) from both sides:
\[ b = 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 6: Find \( a + b \)

Now we can find \( a + b \):

\[ a + b = \frac{1}{2} + \frac{1}{2} = 1 \]

Thus, the value of \( a + b \) is:

\[ \boxed{1} \]