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 find the value of \( f'(0) \) given the limit:

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

### Step-by-Step Solution:

1. **Understanding the Limit**:
Since the limit evaluates to a constant (12) and the denominator is \( x^3 \), the numerator must also be a polynomial of degree 3. This means that \( f(x) \) must be a polynomial of degree at least 3.

2. **Assuming a Form for \( f(x) \)**:
Let's assume that \( f(x) \) can be expressed as:
\[ f(x) = n x^3 \]
where \( n \) is a constant.

3. **Substituting into the Limit**:
We substitute \( f(x) \) into the limit:
\[ f(4x) = n(4x)^3 = 64nx^3 \]
\[ f(3x) = n(3x)^3 = 27nx^3 \]
\[ f(2x) = n(2x)^3 = 8nx^3 \]
\[ f(x) = nx^3 \]
Now substituting these into the limit expression:
\[ \lim_{x \to 0} \frac{64nx^3 - 3(27nx^3) + 3(8nx^3) - nx^3}{x^3} \]

4. **Simplifying the Expression**:
Simplifying the numerator:
\[ 64nx^3 - 81nx^3 + 24nx^3 - nx^3 = (64n - 81n + 24n - n)x^3 = (6n)x^3 \]
Therefore, the limit becomes:
\[ \lim_{x \to 0} \frac{6nx^3}{x^3} = 6n \]

5. **Setting the Limit Equal to 12**:
We set the limit equal to 12:
\[ 6n = 12 \]
Solving for \( n \):
\[ n = 2 \]

6. **Finding \( f(x) \)**:
Thus, we have:
\[ f(x) = 2x^3 \]

7. **Finding the Derivative**:
Now we find the derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(2x^3) = 6x^2 \]

8. **Evaluating \( f'(0) \)**:
Finally, we evaluate \( f'(0) \):
\[ f'(0) = 6(0)^2 = 0 \]

### Final Answer:
The value of \( f'(0) \) is \( \boxed{0} \).