The set of values of p for which the points of extremum of the function, `f(x) = x^3 -3px^2 + 3 (p^2 -1)x + 1` lin in the interval `(-2,4)` is

Options

A. `(-3,5)`
B. `(-3,3)`
C. `(-1,3)`
D. `(-1,4)`

Answer

To solve the problem, we need to find the set of values of \( p \) for which the points of extremum of the function

\[ f(x) = x^3 - 3px^2 + 3(p^2 - 1)x + 1 \]

lie in the interval \((-2, 4)\).

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

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

Calculating the derivative:

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

### Step 2: Set the derivative equal to zero
Next, we set the derivative equal to zero to find the critical points:

\[ 3x^2 - 6px + 3(p^2 - 1) = 0 \]

Dividing the entire equation by 3 simplifies it to:

\[ x^2 - 2px + (p^2 - 1) = 0 \]

### Step 3: Use the quadratic formula
The critical points (extremum points) can be found using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1 \), \( b = -2p \), and \( c = p^2 - 1 \):

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

Calculating the discriminant:

\[ (-2p)^2 - 4(p^2 - 1) = 4p^2 - 4(p^2 - 1) = 4p^2 - 4p^2 + 4 = 4 \]

Thus, the critical points are:

\[ x = \frac{2p \pm 2}{2} = p \pm 1 \]

### Step 4: Determine the intervals for \( p \)
We need both critical points \( p - 1 \) and \( p + 1 \) to lie within the interval \((-2, 4)\).

1. For \( p - 1 \):
\[ -2 < p - 1 < 4 \]
Adding 1 to all parts:
\[ -1 < p < 5 \]

2. For \( p + 1 \):
\[ -2 < p + 1 < 4 \]
Subtracting 1 from all parts:
\[ -3 < p < 3 \]

### Step 5: Find the intersection of the intervals
Now, we need to find the intersection of the two intervals:

- From \( p - 1 \): \( -1 < p < 5 \)
- From \( p + 1 \): \( -3 < p < 3 \)

The intersection of these intervals is:

\[ -1 < p < 3 \]

### Final Answer
Thus, the set of values of \( p \) for which the points of extremum of the function lie in the interval \((-2, 4)\) is:

\[ p \in (-1, 3) \]