Let `f: R rarr R` be a function satisfying `f(x+y)= f(x) + 2y^(3) + kxy` for all `x, y in R`. If f(1)=2 and f(2)= 8, then f(x) is equal to

Options

A. `2x^(2)`
B. `6x-4`
C. `x^(2)+3x-2`
D. `-x^(2) + 9x-6`

Answer

To solve the problem, we need to find the function \( f(x) \) that satisfies the equation:

\[ f(x+y) = f(x) + 2y^3 + kxy \]

for all \( x, y \in \mathbb{R} \), given that \( f(1) = 2 \) and \( f(2) = 8 \).

### Step 1: Substitute Values into the Functional Equation

Let's first substitute \( x = 1 \) and \( y = 1 \) into the functional equation:

\[ f(1+1) = f(1) + 2(1)^3 + k(1)(1) \]

This simplifies to:

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

### Step 2: Use Given Values

We know that \( f(1) = 2 \) and \( f(2) = 8 \). Substituting these values gives:

\[ 8 = 2 + 2 + k \]

This simplifies to:

\[ 8 = 4 + k \]

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

From the equation above, we can solve for \( k \):

\[ k = 8 - 4 = 4 \]

### Step 4: Substitute \( k \) Back into the Functional Equation

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

\[ f(x+y) = f(x) + 2y^3 + 4xy \]

### Step 5: Differentiate the Functional Equation

Next, we differentiate both sides with respect to \( y \):

\[ \frac{d}{dy} f(x+y) = \frac{d}{dy} (f(x) + 2y^3 + 4xy) \]

Using the chain rule on the left side, we get:

\[ f'(x+y) = 6y^2 + 4x \]

### Step 6: Set \( y = 0 \)

Now, set \( y = 0 \):

\[ f'(x) = 6(0)^2 + 4x = 4x \]

### Step 7: Integrate to Find \( f(x) \)

Integrating \( f'(x) \):

\[ f(x) = \int 4x \, dx = 2x^2 + C \]

where \( C \) is the constant of integration.

### Step 8: Use Given Values to Find \( C \)

We know \( f(1) = 2 \):

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

This gives:

\[ C = 2 - 2 = 0 \]

### Step 9: Write the Final Form of \( f(x) \)

Thus, the function \( f(x) \) is:

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

### Final Answer

The final answer is:

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