If the function ` f(x) = x^3 + 3(a-7)x^2 +3(a^2-9)x-1 ` has a positive point Maximum , then

Options

A. `a in (3,oo)cup(-oo,-3)`
B. `a in (-oo,-3)cup(3,29//7)`
C. `(-oo,7)`
D. `(-oo,29)`

Answer

To solve the problem, we need to analyze the function \( f(x) = x^3 + 3(a-7)x^2 + 3(a^2-9)x - 1 \) for conditions that ensure it has a positive point of maximum.

### Step 1: Differentiate the function
To find the critical points, we first differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}(x^3 + 3(a-7)x^2 + 3(a^2-9)x - 1) = 3x^2 + 6(a-7)x + 3(a^2-9) \]

### Step 2: Set the derivative to zero for critical points
We need to find the critical points by setting the derivative equal to zero:
\[ 3x^2 + 6(a-7)x + 3(a^2-9) = 0 \]
Dividing the entire equation by 3 gives:
\[ x^2 + 2(a-7)x + (a^2-9) = 0 \]

### Step 3: Apply the condition for a maximum
For \( f(x) \) to have a maximum at a positive point, the discriminant of this quadratic must be greater than or equal to zero (to ensure real roots):
\[ D = b^2 - 4ac = [2(a-7)]^2 - 4 \cdot 1 \cdot (a^2-9) > 0 \]
Calculating the discriminant:
\[ D = 4(a-7)^2 - 4(a^2-9) > 0 \]
\[ D = 4[(a-7)^2 - (a^2-9)] > 0 \]
Expanding the terms:
\[ (a-7)^2 = a^2 - 14a + 49 \]
Thus,
\[ D = 4[a^2 - 14a + 49 - a^2 + 9] > 0 \]
This simplifies to:
\[ D = 4[-14a + 58] > 0 \]
Dividing by 4:
\[ -14a + 58 > 0 \implies 58 > 14a \implies a < \frac{29}{7} \]

### Step 4: Analyze the first derivative at \( x = 0 \)
Next, we need to ensure that \( f'(0) > 0 \):
\[ f'(0) = 3(a^2 - 9) > 0 \]
This implies:
\[ a^2 - 9 > 0 \implies (a - 3)(a + 3) > 0 \]
The critical points are \( a = -3 \) and \( a = 3 \). Analyzing the sign:
- For \( a < -3 \): \( (a - 3)(a + 3) > 0 \) (positive)
- For \( -3 < a < 3 \): \( (a - 3)(a + 3) < 0 \) (negative)
- For \( a > 3 \): \( (a - 3)(a + 3) > 0 \) (positive)

### Step 5: Combine the conditions
From the two conditions:
1. \( a < \frac{29}{7} \)
2. \( a < -3 \) or \( a > 3 \)

The intersection of these conditions gives:
- For \( a < -3 \): valid since \( -3 < \frac{29}{7} \)
- For \( a > 3 \): we need \( a < \frac{29}{7} \), which is valid.

Thus, the final valid intervals for \( a \) are:
\[ a < -3 \quad \text{or} \quad 3 < a < \frac{29}{7} \]

### Final Answer:
The values of \( a \) for which \( f(x) \) has a positive point maximum are:
\[ a < -3 \quad \text{or} \quad 3 < a < \frac{29}{7} \]