Let 'f' be a fifferentiable real valued function satisfying `f (x+2y) =f (x) +f (2y) + 6xy (x+2y) AA x, y in R.` Then `f ' (0), f" (1), f'(2)…..` are in

Options

A. AP
B. GP
C. HP
D. None of these

Answer

To solve the problem, we need to analyze the given functional equation and find the values of \( f'(0) \), \( f(1) \), and \( f'(2) \).

### Step-by-Step Solution:

1. **Start with the given functional equation**:
\[ f(x + 2y) = f(x) + f(2y) + 6xy(x + 2y) \]
for all \( x, y \in \mathbb{R} \).

2. **Substitute \( a = x \) and \( b = 2y \)**:
Rewrite the equation as:
\[ f(a + b) = f(a) + f(b) + 3ab(a + b) \]
This resembles the form of the identity for \( a^3 + b^3 \).

3. **Recognize the form of the equation**:
The equation \( f(a + b) = f(a) + f(b) + 3ab(a + b) \) suggests that \( f(x) \) could be a cubic function. We can assume:
\[ f(x) = x^3 \]
because the cubic polynomial satisfies the form of the equation.

4. **Verify the assumption**:
If \( f(x) = x^3 \), then:
\[ f(x + 2y) = (x + 2y)^3 = x^3 + 3x^2(2y) + 3x(2y)^2 + (2y)^3 \]
Expanding this gives:
\[ = x^3 + 6xy + 12y^2 + 8y^3 \]
Now, calculate \( f(x) + f(2y) + 6xy(x + 2y) \):
\[ f(x) = x^3, \quad f(2y) = (2y)^3 = 8y^3 \]
Thus:
\[ f(x) + f(2y) + 6xy(x + 2y) = x^3 + 8y^3 + 6xy(x + 2y) \]
Simplifying \( 6xy(x + 2y) \):
\[ = 6xyx + 12xy^2 = 6x^2y + 12xy^2 \]
Adding these gives:
\[ = x^3 + 8y^3 + 6x^2y + 12xy^2 \]
This matches the expansion of \( (x + 2y)^3 \).

5. **Find derivatives**:
Now, we calculate:
- \( f'(x) = 3x^2 \)
- \( f'(0) = 3(0)^2 = 0 \)
- \( f(1) = 1^3 = 1 \)
- \( f'(2) = 3(2)^2 = 12 \)

6. **Collect the values**:
We have:
- \( f'(0) = 0 \)
- \( f(1) = 1 \)
- \( f'(2) = 12 \)

7. **Check the type of progression**:
- The values are \( 0, 1, 12 \).
- For Arithmetic Progression (AP): The difference between \( 0 \) and \( 1 \) is \( 1 \), and between \( 1 \) and \( 12 \) is \( 11 \). Since the differences are not equal, they are not in AP.
- For Geometric Progression (GP): Since one of the terms is \( 0 \), they cannot be in GP.
- For Harmonic Progression (HP): We check if \( b = \frac{2ac}{a+c} \). Here, \( a = 0, b = 1, c = 12 \):
\[ 1 \neq \frac{2(0)(12)}{0 + 12} = 0 \]
Thus, they are not in HP either.

8. **Conclusion**:
Since \( 0, 1, 12 \) are not in AP, GP, or HP, we conclude that they do not lie in any standard progression.

### Final Answer:
None of these (Option D).