Let `f''(x)` be continuous at x = 0 and `f''(0)=4`, Then value of `lim_(x to 0)(2f(x) -3f(2x) + f(4x))/x^(2)` is:

Options

A. 11
B. 2
C. 12
D. none of these

Answer

To solve the problem, we need to evaluate the limit:

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

Given that \(f''(x)\) is continuous at \(x = 0\) and \(f''(0) = 4\), we can use Taylor's expansion for \(f(x)\) around \(x = 0\).

### Step 1: Taylor Expansion
The Taylor series expansion of \(f(x)\) around \(x = 0\) is given by:

\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2}x^2 + o(x^2) \]

Using this expansion, we can express \(f(2x)\) and \(f(4x)\):

\[ f(2x) = f(0) + f'(0)(2x) + \frac{f''(0)}{2}(2x)^2 + o((2x)^2) = f(0) + 2f'(0)x + 2f''(0)x^2 + o(x^2) \]

\[ f(4x) = f(0) + f'(0)(4x) + \frac{f''(0)}{2}(4x)^2 + o((4x)^2) = f(0) + 4f'(0)x + 8f''(0)x^2 + o(x^2) \]

### Step 2: Substitute into the Limit
Now substitute \(f(x)\), \(f(2x)\), and \(f(4x)\) into the limit expression:

\[ 2f(x) = 2\left(f(0) + f'(0)x + 2x^2 + o(x^2)\right) = 2f(0) + 2f'(0)x + 4x^2 + o(x^2) \]

\[ -3f(2x) = -3\left(f(0) + 2f'(0)x + 2x^2 + o(x^2)\right) = -3f(0) - 6f'(0)x - 6x^2 - 3o(x^2) \]

\[ f(4x) = f(0) + 4f'(0)x + 8x^2 + o(x^2) \]

Now combine these:

\[ 2f(x) - 3f(2x) + f(4x) = (2f(0) - 3f(0) + f(0)) + (2f'(0)x - 6f'(0)x + 4f'(0)x) + (4x^2 - 6x^2 + 8x^2) + o(x^2) \]

This simplifies to:

\[ 0 + 0x + 6x^2 + o(x^2) = 6x^2 + o(x^2) \]

### Step 3: Evaluate the Limit
Now substitute this back into the limit:

\[ \lim_{x \to 0} \frac{6x^2 + o(x^2)}{x^2} = \lim_{x \to 0} \left(6 + \frac{o(x^2)}{x^2}\right) \]

As \(x \to 0\), \(\frac{o(x^2)}{x^2} \to 0\). Therefore, the limit simplifies to:

\[ 6 + 0 = 6 \]

### Final Answer
Thus, the value of the limit is:

\[ \boxed{6} \]