If `f(x) = x^3 + 3 (a -7) x^2 + 3(a^2 - 9) x - 2` where a > 0, has + ive point of maximum then a varies over an interval of length

Options

A. `8/7`
B. `6/7`
C. `4/7`
D. `3/7`

Answer

To solve the problem, we need to analyze the function \( f(x) = x^3 + 3(a - 7)x^2 + 3(a^2 - 9)x - 2 \) under the conditions provided. We are looking for the values of \( a \) such that the function has a positive point of maximum.

### Step 1: Find the first derivative of \( f(x) \)

The first derivative \( f'(x) \) is calculated as follows:

\[ f'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}(3(a - 7)x^2) + \frac{d}{dx}(3(a^2 - 9)x) - \frac{d}{dx}(2) \]

Calculating each term:

\[ f'(x) = 3x^2 + 6(a - 7)x + 3(a^2 - 9) \]

Thus, we have:

\[ f'(x) = 3x^2 + 6(a - 7)x + 3(a^2 - 9) \]

### Step 2: Set the first derivative equal to zero

To find critical points, we set the first derivative equal to zero:

\[ 3x^2 + 6(a - 7)x + 3(a^2 - 9) = 0 \]

Dividing the entire equation by 3:

\[ x^2 + 2(a - 7)x + (a^2 - 9) = 0 \]

### Step 3: Calculate the discriminant

For the quadratic equation \( Ax^2 + Bx + C = 0 \), the discriminant \( D \) is given by:

\[ D = B^2 - 4AC \]

Here, \( A = 1 \), \( B = 2(a - 7) \), and \( C = a^2 - 9 \). Thus, the discriminant is:

\[ D = [2(a - 7)]^2 - 4 \cdot 1 \cdot (a^2 - 9) \]

Calculating the discriminant:

\[ D = 4(a - 7)^2 - 4(a^2 - 9) \]

Expanding this:

\[ D = 4(a^2 - 14a + 49) - 4a^2 + 36 \]

Simplifying:

\[ D = 4a^2 - 56a + 196 - 4a^2 + 36 = -56a + 232 \]

For the quadratic to have real roots, we require:

\[ -56a + 232 > 0 \]

Solving for \( a \):

\[ 56a < 232 \implies a < \frac{232}{56} = \frac{29}{7} \]

### Step 4: Analyze the critical point

Next, we need to ensure that the critical point is positive. The critical point \( x \) can be found using the formula \( x = -\frac{B}{2A} \):

\[ x = -\frac{2(a - 7)}{2 \cdot 1} = 7 - a \]

We require \( 7 - a > 0 \):

\[ a < 7 \]

### Step 5: Combine conditions

Now we have two conditions:

1. \( a < \frac{29}{7} \)
2. \( a < 7 \)

The more restrictive condition is \( a < 7 \).

### Step 6: Determine the lower bound

Next, we check the condition for \( f'(0) > 0 \):

\[ f'(0) = 3(a^2 - 9) > 0 \]

This implies:

\[ a^2 - 9 > 0 \implies a^2 > 9 \implies a > 3 \quad (\text{since } a > 0) \]

### Final Interval for \( a \)

Combining all conditions, we have:

\[ 3 < a < 7 \]

### Step 7: Length of the interval

The length of the interval \( (3, 7) \) is:

\[ 7 - 3 = 4 \]

Thus, the length of the interval over which \( a \) varies is \( 4 \).