If f (x + y) = f(x) + `2y^(2) + kxy and ` f(a) = 2, f(b) = 8, then f(x) is of the form

Options

A. `2x^(2)`
B. `2x^(2) + 1 `
C. `2x^(2) - 1`
D. `x^(2)`

Answer

To solve the problem, we start with the functional equation given:

\[ f(x + y) = f(x) + 2y^2 + kxy \]

We also know two specific values of the function:

\[ f(1) = 2 \]
\[ f(2) = 8 \]

### Step 1: Substitute \( x = 1 \) and \( y = 1 \)

We can use the functional equation to find \( k \). Let's substitute \( x = 1 \) and \( y = 1 \):

\[ f(1 + 1) = f(1) + 2(1^2) + k(1)(1) \]

This simplifies to:

\[ f(2) = f(1) + 2 + k \]

### Step 2: Substitute known values

Now, we substitute the known values \( f(1) = 2 \) and \( f(2) = 8 \):

\[ 8 = 2 + 2 + k \]

### Step 3: Solve for \( k \)

Now, we can solve for \( k \):

\[ 8 = 4 + k \implies k = 8 - 4 = 4 \]

### Step 4: Rewrite the functional equation

Now that we have \( k = 4 \), we can rewrite the functional equation as:

\[ f(x + y) = f(x) + 2y^2 + 4xy \]

### Step 5: Substitute \( y = 1 \)

Next, let's substitute \( y = 1 \) into the modified functional equation:

\[ f(x + 1) = f(x) + 2(1^2) + 4x(1) \]

This simplifies to:

\[ f(x + 1) = f(x) + 2 + 4x \]

### Step 6: Rearranging the equation

Rearranging gives us:

\[ f(x + 1) = f(x) + 4x + 2 \]

### Step 7: Find a general form for \( f(x) \)

Let's assume \( f(x) \) is a quadratic function of the form:

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

### Step 8: Use known values to find coefficients

Using \( f(1) = 2 \):

\[ a(1^2) + b(1) + c = 2 \implies a + b + c = 2 \quad (1) \]

Using \( f(2) = 8 \):

\[ a(2^2) + b(2) + c = 8 \implies 4a + 2b + c = 8 \quad (2) \]

### Step 9: Solve the system of equations

Now, we have two equations:

1. \( a + b + c = 2 \)
2. \( 4a + 2b + c = 8 \)

Subtract (1) from (2):

\[ (4a + 2b + c) - (a + b + c) = 8 - 2 \]

This simplifies to:

\[ 3a + b = 6 \quad (3) \]

### Step 10: Express \( b \) in terms of \( a \)

From equation (3):

\[ b = 6 - 3a \quad (4) \]

### Step 11: Substitute \( b \) back into equation (1)

Substituting (4) into (1):

\[ a + (6 - 3a) + c = 2 \]

This simplifies to:

\[ -2a + 6 + c = 2 \implies c = 2a - 4 \quad (5) \]

### Step 12: Choose a value for \( a \)

To find \( a \), we can test values. Let's assume \( a = 2 \):

From (4):

\[ b = 6 - 3(2) = 0 \]

From (5):

\[ c = 2(2) - 4 = 0 \]

### Final Form of \( f(x) \)

Thus, we have:

\[ f(x) = 2x^2 + 0x + 0 = 2x^2 \]

### Conclusion

The function \( f(x) \) is of the form:

\[ \boxed{2x^2} \]