The value of a for which all extremum of function `f(x)=x^(3)+3ax^(2)+3(a^(2)-1)x+1`, lie in the interval (2, 4) is

Options

A. `(3,4)`
B. `(-1,3)`
C. `(-3, -1)`
D. none of these

Explanation

To find the value of \( a \) for which all extremum of the function \( f(x) = x^3 + 3ax^2 + 3(a^2 - 1)x + 1 \) lie in the interval (2, 4), we will follow these steps:

### Step 1: Find the derivative of the function
To find the extremum points, we first need to calculate the derivative of the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(x^3 + 3ax^2 + 3(a^2 - 1)x + 1) \]
Using the power rule:
\[ f'(x) = 3x^2 + 6ax + 3(a^2 - 1) \]

### Step 2: Set the derivative to zero
To find the critical points (extremum), we set the derivative equal to zero:

\[ 3x^2 + 6ax + 3(a^2 - 1) = 0 \]
Dividing the entire equation by 3 simplifies it to:
\[ x^2 + 2ax + (a^2 - 1) = 0 \]

### Step 3: Use the quadratic formula
The roots of the quadratic equation can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = 2a \), and \( c = a^2 - 1 \). Plugging in these values gives:
\[ x = \frac{-2a \pm \sqrt{(2a)^2 - 4 \cdot 1 \cdot (a^2 - 1)}}{2 \cdot 1} \]
\[ x = \frac{-2a \pm \sqrt{4a^2 - 4(a^2 - 1)}}{2} \]
\[ x = \frac{-2a \pm \sqrt{4a^2 - 4a^2 + 4}}{2} \]
\[ x = \frac{-2a \pm 2}{2} \]
\[ x = -a \pm 1 \]

### Step 4: Identify the critical points
Thus, the critical points are:
\[ x_1 = -a + 1 \quad \text{and} \quad x_2 = -a - 1 \]

### Step 5: Set the interval conditions
We want both critical points to lie within the interval (2, 4):
1. For \( x_1 = -a + 1 \):
\[ 2 < -a + 1 < 4 \]
This gives us two inequalities:
- \( -a + 1 > 2 \) → \( -a > 1 \) → \( a < -1 \)
- \( -a + 1 < 4 \) → \( -a < 3 \) → \( a > -3 \)

2. For \( x_2 = -a - 1 \):
\[ 2 < -a - 1 < 4 \]
This gives us two inequalities:
- \( -a - 1 > 2 \) → \( -a > 3 \) → \( a < -3 \)
- \( -a - 1 < 4 \) → \( -a < 5 \) → \( a > -5 \)

### Step 6: Combine the conditions
From the inequalities derived:
- From \( x_1 \): \( -3 < a < -1 \)
- From \( x_2 \): \( -5 < a < -3 \)

The combined conditions give us:
\[ -3 < a < -1 \]

### Conclusion
The value of \( a \) for which all extremum of the function \( f(x) \) lie in the interval (2, 4) is:
\[ \boxed{(-3, -1)} \]