If the sum of an infinite G.P. is equal to the maximum value of `f(x)=x^(3)+2x-8` in the interval [-1,4] and the sum of first two terms is 8. Then, the common ratio of the G.P. is

Options

A. `(1)/(8)`
B. `(sqrt(3))/(8)`
C. `sqrt((7)/(8))`
D. none of these

Explanation

To solve the given problem step by step, we will follow the outlined approach:

### Step 1: Find the maximum value of the function \( f(x) = x^3 + 2x - 8 \) in the interval \([-1, 4]\).

To find the maximum value, we first need to determine the critical points by taking the derivative of the function.

**Hint:** Differentiate the function to find critical points.

### Step 2: Differentiate the function.

The derivative of the function is:
\[ f'(x) = 3x^2 + 2 \]
Since \( 3x^2 + 2 \) is always positive for all \( x \) (as both terms are positive), the function \( f(x) \) is increasing on the entire real line.

**Hint:** Check if the derivative is positive or negative to determine if the function is increasing or decreasing.

### Step 3: Evaluate the function at the endpoints of the interval.

Since \( f(x) \) is increasing, the maximum value in the interval \([-1, 4]\) will occur at the right endpoint \( x = 4 \).

Calculating \( f(4) \):
\[ f(4) = 4^3 + 2 \cdot 4 - 8 = 64 + 8 - 8 = 64 \]

**Hint:** Always evaluate the function at the endpoints when determining maximum or minimum values in a closed interval.

### Step 4: Set up the equation for the sum of the infinite G.P.

The sum of an infinite geometric progression (G.P.) is given by the formula:
\[ S_{\infty} = \frac{a}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio. We know from the problem statement that:
\[ S_{\infty} = 64 \]

**Hint:** Remember the formula for the sum of an infinite G.P. and how to relate it to the problem.

### Step 5: Relate the first term \( a \) to the common ratio \( r \).

We also know that the sum of the first two terms of the G.P. is 8:
\[ a + ar = 8 \]
Factoring out \( a \):
\[ a(1 + r) = 8 \quad \text{(1)} \]

**Hint:** Use the relationship between the first two terms to express \( a \) in terms of \( r \).

### Step 6: Substitute \( a \) from equation (1) into the sum equation.

From \( a(1 + r) = 8 \), we can express \( a \):
\[ a = \frac{8}{1 + r} \]
Substituting this into the sum equation:
\[ \frac{8}{1 + r} \cdot \frac{1}{1 - r} = 64 \]

**Hint:** Substitute carefully and simplify to find relationships between \( r \) and constants.

### Step 7: Solve for \( r \).

Cross-multiplying gives:
\[ 8 = 64(1 + r)(1 - r) \]
This simplifies to:
\[ 8 = 64(1 - r^2) \]
Dividing both sides by 64:
\[ \frac{1}{8} = 1 - r^2 \]
Rearranging gives:
\[ r^2 = 1 - \frac{1}{8} = \frac{7}{8} \]
Taking the square root:
\[ r = \sqrt{\frac{7}{8}} = \frac{\sqrt{7}}{2\sqrt{2}} = \frac{\sqrt{14}}{4} \]

**Hint:** Remember to consider both positive and negative roots, but since \( r \) must be less than 1 for convergence, we take the positive root.

### Final Answer:
The common ratio \( r \) of the G.P. is:
\[ r = \sqrt{\frac{7}{8}} \]