If ` f(x) = x^(3) + x^(2) * f' (1)+ xf''(2) + f''(2) + f'''(3) , AA x in R `
Where , f(x) is a polynimial of degree 3 , then

Options

A. `f(0) + f(2) = f(1)`
B. `f(0) + f(3) = 0 `
C. `f(1) + f(3) = f(2)`
D. All of these

Explanation

To solve the problem, we start with the given polynomial function:

\[ f(x) = x^3 + x^2 f'(1) + x f''(2) + f''(2) + f'''(3) \]

Since \( f(x) \) is a polynomial of degree 3, we can express it in the general form:

\[ f(x) = ax^3 + bx^2 + cx + d \]

### Step 1: Identify the derivatives

1. **First derivative** \( f'(x) \):
\[ f'(x) = 3ax^2 + 2bx + c \]

2. **Second derivative** \( f''(x) \):
\[ f''(x) = 6ax + 2b \]

3. **Third derivative** \( f'''(x) \):
\[ f'''(x) = 6a \]

### Step 2: Evaluate the derivatives at specific points

- **Evaluate \( f'(1) \)**:
\[ f'(1) = 3a(1)^2 + 2b(1) + c = 3a + 2b + c \]

- **Evaluate \( f''(2) \)**:
\[ f''(2) = 6a(2) + 2b = 12a + 2b \]

- **Evaluate \( f'''(3) \)**:
\[ f'''(3) = 6a \]

### Step 3: Substitute the derivatives back into the function

Substituting these values back into the original equation for \( f(x) \):

\[ f(x) = x^3 + x^2(3a + 2b + c) + x(12a + 2b) + (12a + 2b) + 6a \]

### Step 4: Combine like terms

Now we combine like terms:

\[ f(x) = x^3 + (3a + 2b)x^2 + (12a + 2b)x + (12a + 2b + 6a) \]

This simplifies to:

\[ f(x) = x^3 + (3a + 2b)x^2 + (12a + 2b)x + (18a + 2b) \]

### Step 5: Equate coefficients

Since \( f(x) \) is a polynomial of degree 3, we can equate the coefficients of \( f(x) \) with the general form \( ax^3 + bx^2 + cx + d \):

1. Coefficient of \( x^3 \): \( a = 1 \)
2. Coefficient of \( x^2 \): \( 3a + 2b = 0 \)
3. Coefficient of \( x \): \( 12a + 2b = 0 \)
4. Constant term: \( 18a + 2b = 0 \)

### Step 6: Solve the equations

1. From \( a = 1 \):
\[ 3(1) + 2b = 0 \implies 2b = -3 \implies b = -\frac{3}{2} \]

2. Substitute \( a \) into \( 12a + 2b = 0 \):
\[ 12(1) + 2(-\frac{3}{2}) = 0 \implies 12 - 3 = 0 \text{ (True)} \]

3. Substitute \( a \) into \( 18a + 2b = 0 \):
\[ 18(1) + 2(-\frac{3}{2}) = 0 \implies 18 - 3 = 0 \text{ (True)} \]

### Step 7: Find \( c \) and \( d \)

Now we can find \( c \) and \( d \):

Using \( f'(1) = 3a + 2b + c \):
\[ 3(1) + 2(-\frac{3}{2}) + c = 0 \implies 3 - 3 + c = 0 \implies c = 0 \]

Using \( f(0) = d \):
\[ f(0) = 0^3 + 0^2(-\frac{3}{2}) + 0 + d \implies d = 0 \]

### Final Polynomial

Thus, the polynomial \( f(x) \) is:

\[ f(x) = x^3 - \frac{3}{2}x^2 \]