If `f (x)` is a thrice differentiable function such that `lim _(xto0)(f (4x) -3 f(3x) +3f (2x) -f (x))/(x ^(3))=12` then the vlaue of `f '''(0)` equais to :

Options

A. 0
B. 1
C. 12
D. None of these

Answer

To solve the problem, we need to evaluate the limit given and find the value of \( f'''(0) \).

Given:
\[ \lim_{x \to 0} \frac{f(4x) - 3f(3x) + 3f(2x) - f(x)}{x^3} = 12 \]

### Step 1: Taylor Expansion
Since \( f(x) \) is a thrice differentiable function, we can use the Taylor series expansion around \( x = 0 \):
\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \frac{f'''(0)}{6}x^3 + o(x^3) \]

### Step 2: Substitute \( 4x, 3x, 2x, x \)
Now, we substitute \( 4x, 3x, 2x, \) and \( x \) into the Taylor expansion:
- For \( f(4x) \):
\[ f(4x) = f(0) + f'(0)(4x) + \frac{f''(0)}{2}(4x)^2 + \frac{f'''(0)}{6}(4x)^3 + o(x^3) \]
\[ = f(0) + 4f'(0)x + 8f''(0)x^2 + \frac{64}{6}f'''(0)x^3 + o(x^3) \]

- For \( f(3x) \):
\[ f(3x) = f(0) + f'(0)(3x) + \frac{f''(0)}{2}(3x)^2 + \frac{f'''(0)}{6}(3x)^3 + o(x^3) \]
\[ = f(0) + 3f'(0)x + \frac{9}{2}f''(0)x^2 + \frac{27}{6}f'''(0)x^3 + o(x^3) \]

- For \( f(2x) \):
\[ f(2x) = f(0) + f'(0)(2x) + \frac{f''(0)}{2}(2x)^2 + \frac{f'''(0)}{6}(2x)^3 + o(x^3) \]
\[ = f(0) + 2f'(0)x + 2f''(0)x^2 + \frac{8}{6}f'''(0)x^3 + o(x^3) \]

- For \( f(x) \):
\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \frac{f'''(0)}{6}x^3 + o(x^3) \]

### Step 3: Combine the Expansions
Now we substitute these expansions into the limit expression:
\[ f(4x) - 3f(3x) + 3f(2x) - f(x) = [f(0) + 4f'(0)x + 8f''(0)x^2 + \frac{64}{6}f'''(0)x^3] - 3[f(0) + 3f'(0)x + \frac{9}{2}f''(0)x^2 + \frac{27}{6}f'''(0)x^3] + 3[f(0) + 2f'(0)x + 2f''(0)x^2 + \frac{8}{6}f'''(0)x^3] - [f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + \frac{f'''(0)}{6}x^3] \]

### Step 4: Simplify
Combining like terms:
- Constant terms: \( f(0) - 3f(0) + 3f(0) - f(0) = 0 \)
- Linear terms: \( 4f'(0)x - 9f'(0)x + 6f'(0)x - f'(0)x = 0 \)
- Quadratic terms: \( 8f''(0)x^2 - \frac{27}{2}f''(0)x^2 + 6f''(0)x^2 - \frac{1}{2}f''(0)x^2 = 0 \)

For cubic terms:
\[ \frac{64}{6}f'''(0)x^3 - \frac{81}{6}f'''(0)x^3 + 4f'''(0)x^3 - \frac{1}{6}f'''(0)x^3 \]
This simplifies to:
\[ \left( \frac{64 - 81 + 24 - 1}{6} \right) f'''(0)x^3 = \frac{6f'''(0)}{6}x^3 = f'''(0)x^3 \]

### Step 5: Set the Limit
Now we have:
\[ \lim_{x \to 0} \frac{f'''(0)x^3}{x^3} = f'''(0) = 12 \]

### Conclusion
Thus, the value of \( f'''(0) \) is:
\[ \boxed{12} \]