Let the maximum and minimum value of the expression `2 cos^(2) theta + cos theta + 1` is M and m respectively, then the value of `[M/m]` is (where [.] is the greatest integer function)

Answer

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

### Step 1: Rewrite the Expression
First, we can rewrite the expression in terms of \( x = \cos \theta \). The range of \( x \) is \([-1, 1]\).

\[ f(x) = 2x^2 + x + 1 \]

### Step 2: Find the Derivative
Next, we need to find the derivative of \( f(x) \) to locate the critical points.

\[ f'(x) = \frac{d}{dx}(2x^2 + x + 1) = 4x + 1 \]

### Step 3: Set the Derivative to Zero
To find the critical points, we set the derivative to zero:

\[ 4x + 1 = 0 \]

Solving for \( x \):

\[ 4x = -1 \implies x = -\frac{1}{4} \]

### Step 4: Evaluate the Function at Critical Points and Endpoints
Now we evaluate \( f(x) \) at the critical point \( x = -\frac{1}{4} \) and at the endpoints \( x = -1 \) 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 \( x = -1 \)**:
\[ f(-1) = 2(-1)^2 + (-1) + 1 = 2(1) - 1 + 1 = 2 \]

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

### Step 5: Determine Maximum and Minimum Values
From the evaluations:
- \( f\left(-\frac{1}{4}\right) = \frac{7}{8} \)
- \( f(-1) = 2 \)
- \( f(1) = 4 \)

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

### Step 6: Calculate \( \left\lfloor \frac{M}{m} \right\rfloor \)
Now we calculate \( \frac{M}{m} \):

\[ \frac{M}{m} = \frac{4}{\frac{7}{8}} = 4 \cdot \frac{8}{7} = \frac{32}{7} \approx 4.5714 \]

Taking the greatest integer function:

\[ \left\lfloor \frac{M}{m} \right\rfloor = 4 \]

### Final Answer
The value of \( \left\lfloor \frac{M}{m} \right\rfloor \) is \( \boxed{4} \).

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