Let `f:R to R` be a function given by
`f(x+y)=f(x)+2y^(2)+"kxy for all "x,y in R`
If` f(1)=2` . Find the value of f(x)

Options

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

Explanation

To solve the problem step by step, we will analyze the functional equation given and derive the function \( f(x) \).

### Step 1: Analyze the functional equation
We are given the functional equation:
\[ f(x+y) = f(x) + 2y^2 + kxy \]
for all \( x, y \in \mathbb{R} \).

### Step 2: Substitute specific values
Let's substitute \( x = 0 \) and \( y = 1 \) into the functional equation:
\[ f(0 + 1) = f(0) + 2(1^2) + k(0)(1) \]
This simplifies to:
\[ f(1) = f(0) + 2 \]
We know from the problem statement that \( f(1) = 2 \). Therefore, we have:
\[ 2 = f(0) + 2 \]
From this, we can deduce:
\[ f(0) = 0 \]

### Step 3: Differentiate the functional equation
Next, we will differentiate the functional equation with respect to \( y \):
\[ \frac{d}{dy}(f(x+y)) = \frac{d}{dy}(f(x) + 2y^2 + kxy) \]
Using the chain rule on the left side, we get:
\[ f'(x+y) = 4y + kx \]
Now, let’s set \( y = 0 \):
\[ f'(x+0) = 4(0) + kx \]
This simplifies to:
\[ f'(x) = kx \]

### Step 4: Integrate to find \( f(x) \)
Now we will integrate \( f'(x) \):
\[ f(x) = \int kx \, dx = \frac{k}{2} x^2 + C \]
where \( C \) is the constant of integration.

### Step 5: Use known values to find constants
We know \( f(0) = 0 \):
\[ f(0) = \frac{k}{2}(0)^2 + C = C \]
Thus, \( C = 0 \).

Now we also know \( f(1) = 2 \):
\[ f(1) = \frac{k}{2}(1)^2 + 0 = \frac{k}{2} \]
Setting this equal to 2 gives:
\[ \frac{k}{2} = 2 \implies k = 4 \]

### Step 6: Write the final form of \( f(x) \)
Substituting \( k = 4 \) back into the expression for \( f(x) \):
\[ f(x) = \frac{4}{2} x^2 + 0 = 2x^2 \]

### Conclusion
Thus, the value of \( f(x) \) is:
\[ \boxed{2x^2} \]