The cubic polynomial with leading coefficient unity all whose roots are 3 units less than the roots of the equation `x ^(3) -3x ^(2) -4x +12=0` is denoted as `f (x) `, then `f '(x)` is equal to :

Options

A. `3x ^(2) +12x +5`
B. `3x ^(2) -4x+12`
C. `3x ^(2) +12x -5`
D. `3x ^(2) -12x -5`

Answer

To solve the problem, we need to find the derivative of a cubic polynomial whose roots are 3 units less than the roots of the given polynomial \(x^3 - 3x^2 - 4x + 12 = 0\).

### Step-by-Step Solution

1. **Identify the given polynomial**:
The polynomial given is \(x^3 - 3x^2 - 4x + 12 = 0\).

2. **Find the roots of the polynomial**:
Let's denote the roots of this polynomial as \(\alpha\), \(\beta\), and \(\gamma\).

3. **Determine the new roots**:
The roots of the cubic polynomial \(f(x)\) we need to find are 3 units less than the roots of the given polynomial. Therefore, the roots of \(f(x)\) will be:
\[ \alpha - 3, \beta - 3, \gamma - 3 \]

4. **Express the new polynomial in terms of its roots**:
The polynomial can be expressed as:
\[ f(x) = (x - (\alpha - 3))(x - (\beta - 3))(x - (\gamma - 3)) \]
This can be rewritten as:
\[ f(x) = (x + 3 - \alpha)(x + 3 - \beta)(x + 3 - \gamma) \]

5. **Substitute \(X = x + 3\)**:
Let \(X = x + 3\). Then, the polynomial becomes:
\[ f(X - 3) = (X - \alpha)(X - \beta)(X - \gamma) \]

6. **Substitute back into the original polynomial**:
Since \(\alpha\), \(\beta\), and \(\gamma\) are the roots of the original polynomial, we can express \(f(X)\) in terms of the original polynomial:
\[ f(X) = (X - \alpha)(X - \beta)(X - \gamma) = X^3 - 3X^2 - 4X + 12 \]

7. **Replace \(X\) with \(x + 3\)**:
Now, substituting back \(X = x + 3\):
\[ f(x + 3) = (x + 3)^3 - 3(x + 3)^2 - 4(x + 3) + 12 \]

8. **Expand the polynomial**:
Expanding each term:
\[ (x + 3)^3 = x^3 + 9x^2 + 27x + 27 \]
\[ -3(x + 3)^2 = -3(x^2 + 6x + 9) = -3x^2 - 18x - 27 \]
\[ -4(x + 3) = -4x - 12 \]
Combining these, we have:
\[ f(x + 3) = x^3 + 9x^2 + 27x + 27 - 3x^2 - 18x - 27 - 4x - 12 + 12 \]

9. **Combine like terms**:
Simplifying this gives:
\[ f(x + 3) = x^3 + (9 - 3)x^2 + (27 - 18 - 4)x + (27 - 27 - 12 + 12) \]
\[ = x^3 + 6x^2 + 5x + 0 \]
Therefore, the polynomial \(f(x)\) is:
\[ f(x) = x^3 + 6x^2 + 5x \]

10. **Differentiate \(f(x)\)**:
Now we differentiate \(f(x)\) to find \(f'(x)\):
\[ f'(x) = 3x^2 + 12x + 5 \]

### Final Result
Thus, the derivative \(f'(x)\) is:
\[ \boxed{3x^2 + 12x + 5} \]