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=0`
C. `3x ^(2) +12x -5`
D. `3x ^(2) -12x -5`

Answer

To solve the problem step by step, we will find the cubic polynomial \( f(x) \) whose roots are 3 units less than the roots of the given cubic polynomial \( x^3 - 3x^2 - 4x + 12 = 0 \) and then differentiate it to find \( f'(x) \).

### Step 1: Identify the roots of the given polynomial
The given polynomial is:
\[ x^3 - 3x^2 - 4x + 12 = 0 \]
Let the roots of this polynomial be \( \alpha, \beta, \gamma \).

### Step 2: Determine the roots of the new polynomial
The roots of the required polynomial \( f(x) \) are 3 units less than the roots of the given polynomial. Thus, the roots of \( f(x) \) will be:
\[ \alpha - 3, \beta - 3, \gamma - 3 \]

### Step 3: Write the polynomial \( f(x) \)
The polynomial \( f(x) \) can be expressed in terms of its roots:
\[ f(x) = (x - (\alpha - 3))(x - (\beta - 3))(x - (\gamma - 3)) \]
This can be rewritten as:
\[ f(x) = (x + 3 - \alpha)(x + 3 - \beta)(x + 3 - \gamma) \]

### Step 4: Substitute \( x + 3 \) with a new variable
Let \( X = x + 3 \). Then we have:
\[ f(x) = (X - \alpha)(X - \beta)(X - \gamma) \]
Since \( \alpha, \beta, \gamma \) are the roots of the original polynomial, we can express this as:
\[ f(X) = (X - \alpha)(X - \beta)(X - \gamma) = X^3 - 3X^2 - 4X + 12 \]

### Step 5: Substitute back for \( x \)
Substituting \( X = x + 3 \) back into the polynomial:
\[ f(x + 3) = (x + 3)^3 - 3(x + 3)^2 - 4(x + 3) + 12 \]

### Step 6: Expand the polynomial
Now we will expand \( f(x + 3) \):
1. Expand \( (x + 3)^3 \):
\[ (x + 3)^3 = x^3 + 9x^2 + 27x + 27 \]
2. Expand \( -3(x + 3)^2 \):
\[ -3(x + 3)^2 = -3(x^2 + 6x + 9) = -3x^2 - 18x - 27 \]
3. Expand \( -4(x + 3) \):
\[ -4(x + 3) = -4x - 12 \]

Combining these:
\[ f(x + 3) = (x^3 + 9x^2 + 27x + 27) + (-3x^2 - 18x - 27) + (-4x - 12) + 12 \]
Simplifying:
\[ f(x + 3) = x^3 + (9x^2 - 3x^2) + (27x - 18x - 4x) + (27 - 27 - 12 + 12) \]
\[ = x^3 + 6x^2 + 5x \]

### Step 7: Write the final form of \( f(x) \)
Thus, we have:
\[ f(x) = x^3 + 6x^2 + 5x \]

### Step 8: Differentiate \( f(x) \)
Now we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}(x^3 + 6x^2 + 5x) = 3x^2 + 12x + 5 \]

### Final Answer
Thus, the derivative \( f'(x) \) is:
\[ f'(x) = 3x^2 + 12x + 5 \]