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 \( f(x) \):

\[ f'(x) = \frac{d}{dx}(x^3 + 3ax^2 + 3(a^2 - 1)x + 1) \]

Calculating the derivative, we get:

\[ f'(x) = 3x^2 + 6ax + 3(a^2 - 1) \]

### 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 to find the roots
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 these values into the formula gives:

\[ x = \frac{-2a \pm \sqrt{(2a)^2 - 4 \cdot 1 \cdot (a^2 - 1)}}{2 \cdot 1} \]

Simplifying the discriminant:

\[ x = \frac{-2a \pm \sqrt{4a^2 - 4(a^2 - 1)}}{2} \]
\[ = \frac{-2a \pm \sqrt{4a^2 - 4a^2 + 4}}{2} \]
\[ = \frac{-2a \pm \sqrt{4}}{2} \]
\[ = \frac{-2a \pm 2}{2} \]
\[ = -a \pm 1 \]

Thus, the critical points are:

\[ x_1 = -a + 1 \quad \text{and} \quad x_2 = -a - 1 \]

### Step 4: Determine the conditions for the critical points to lie in (2, 4)
We need both critical points to lie in the interval \( (2, 4) \):

1. For \( x_1 = -a + 1 \):
- \( -a + 1 > 2 \) implies \( -a > 1 \) or \( a < -1 \)
- \( -a + 1 < 4 \) implies \( -a < 4 \) or \( a > -4 \)

2. For \( x_2 = -a - 1 \):
- \( -a - 1 > 2 \) implies \( -a > 3 \) or \( a < -3 \)
- \( -a - 1 < 4 \) implies \( -a < 5 \) or \( a > -5 \)

### Step 5: Combine the inequalities
From \( x_1 \):
- \( -4 < a < -1 \)

From \( x_2 \):
- \( -5 < a < -3 \)

The intersection of these intervals is:

\[ -4 < a < -3 \]

### Conclusion
Thus, the value of \( a \) for which all extremum of the function \( f(x) \) lie in the interval \( (2, 4) \) is:

\[ \boxed{(-4, -3)} \]