Let `f : RtoR` be a function such that `f(x)=x^3+x^2f'(1)+xf''(2)+f'''(3),x in R`. Then f(2) equals

Options

A. 8
B. -2
C. -4
D. 30

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the function
Given the function:
\[ f(x) = x^3 + x^2 f'(1) + x f''(2) + f'''(3) \]
We note that \( f'(1) \), \( f''(2) \), and \( f'''(3) \) are constants because they are evaluated at specific points.

### Step 2: Differentiate the function
We will differentiate \( f(x) \) three times to find \( f'(x) \), \( f''(x) \), and \( f'''(x) \).

1. **First Derivative**:
\[ f'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(x^2 f'(1)) + \frac{d}{dx}(x f''(2)) + \frac{d}{dx}(f'''(3)) \]
\[ f'(x) = 3x^2 + 2x f'(1) + f''(2) \]

2. **Second Derivative**:
\[ f''(x) = \frac{d}{dx}(3x^2) + \frac{d}{dx}(2x f'(1)) + \frac{d}{dx}(f''(2)) \]
\[ f''(x) = 6x + 2 f'(1) \]

3. **Third Derivative**:
\[ f'''(x) = \frac{d}{dx}(6x) + \frac{d}{dx}(2 f'(1)) \]
\[ f'''(x) = 6 \]

### Step 3: Evaluate the derivatives at specific points
Now we can evaluate these derivatives at the specified points:

- \( f'''(3) = 6 \)
- For \( f''(2) \):
\[ f''(2) = 6(2) + 2 f'(1) = 12 + 2 f'(1) \]
- For \( f'(1) \):
\[ f'(1) = 3(1^2) + 2(1) f'(1) + f''(2) \]
Substituting \( f''(2) \):
\[ f'(1) = 3 + 2 f'(1) + (12 + 2 f'(1)) \]
\[ f'(1) = 3 + 12 + 4 f'(1) \]
\[ -3 f'(1) = 15 \implies f'(1) = -5 \]

### Step 4: Substitute back to find \( f''(2) \)
Now substituting \( f'(1) = -5 \) into \( f''(2) \):
\[ f''(2) = 12 + 2(-5) = 12 - 10 = 2 \]

### Step 5: Substitute everything back into \( f(2) \)
Now we can substitute \( f'(1) \), \( f''(2) \), and \( f'''(3) \) back into the original function to find \( f(2) \):
\[ f(2) = 2^3 + 2^2 f'(1) + 2 f''(2) + f'''(3) \]
\[ f(2) = 8 + 4(-5) + 2(2) + 6 \]
\[ f(2) = 8 - 20 + 4 + 6 \]
\[ f(2) = 8 - 20 + 10 = -2 \]

### Final Answer
Thus, the value of \( f(2) \) is:
\[ \boxed{-2} \]