The ratio of the maximum value to minimum value of `2cos^(2)theta+cos theta+1` is :

Options

A. `32:7`
B. `32:9`
C. `4:1`
D. `2:1`

Answer

To find the ratio of the maximum value to the minimum value of the expression \( f(\theta) = 2\cos^2\theta + \cos\theta + 1 \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the function:
\[ f(\theta) = 2\cos^2\theta + \cos\theta + 1 \]

### Step 2: Substitute \( x = \cos\theta \)
Let \( x = \cos\theta \). The function becomes:
\[ f(x) = 2x^2 + x + 1 \]
where \( x \) varies from -1 to 1.

### Step 3: Find the derivative
To find the critical points, we differentiate \( f(x) \):
\[ f'(x) = 4x + 1 \]

### Step 4: Set the derivative to zero
Setting the derivative equal to zero to find critical points:
\[ 4x + 1 = 0 \implies x = -\frac{1}{4} \]

### Step 5: Evaluate the function at critical points and endpoints
We need to evaluate \( f(x) \) at \( x = -\frac{1}{4} \), \( x = -1 \), and \( x = 1 \).

1. **At \( x = -\frac{1}{4} \)**:
\[ f\left(-\frac{1}{4}\right) = 2\left(-\frac{1}{4}\right)^2 + \left(-\frac{1}{4}\right) + 1 = 2\left(\frac{1}{16}\right) - \frac{1}{4} + 1 = \frac{1}{8} - \frac{2}{8} + \frac{8}{8} = \frac{7}{8} \]

2. **At \( x = -1 \)**:
\[ f(-1) = 2(-1)^2 + (-1) + 1 = 2 - 1 + 1 = 2 \]

3. **At \( x = 1 \)**:
\[ f(1) = 2(1)^2 + (1) + 1 = 2 + 1 + 1 = 4 \]

### Step 6: Determine maximum and minimum values
From the evaluations:
- \( f(-1) = 2 \)
- \( f(-\frac{1}{4}) = \frac{7}{8} \)
- \( f(1) = 4 \)

Thus, the maximum value is \( 4 \) and the minimum value is \( \frac{7}{8} \).

### Step 7: Calculate the ratio of maximum to minimum
Now we find the ratio of the maximum value to the minimum value:
\[ \text{Ratio} = \frac{\text{Maximum}}{\text{Minimum}} = \frac{4}{\frac{7}{8}} = 4 \times \frac{8}{7} = \frac{32}{7} \]

### Final Result
The ratio of the maximum value to the minimum value of \( 2\cos^2\theta + \cos\theta + 1 \) is:
\[ \boxed{32:7} \]