If `P = 1/2 sin^(2)theta + 1/3 cos^(2)theta`, then

Options

A. `1/3 le P le 1/2`
B. `P ge 1/2`
C. `2 le P le 3`
D. `-sqrt(13)/6 le P le sqrt(13)/6`

Answer

To find the range of \( P = \frac{1}{2} \sin^2 \theta + \frac{1}{3} \cos^2 \theta \), we can follow these steps:

### Step 1: Rewrite \( P \) in terms of \( \cos^2 \theta \)
Using the identity \( \sin^2 \theta = 1 - \cos^2 \theta \), we can express \( P \) as:
\[ P = \frac{1}{2} (1 - \cos^2 \theta) + \frac{1}{3} \cos^2 \theta \]

### Step 2: Simplify the expression
Distributing the terms, we get:
\[ P = \frac{1}{2} - \frac{1}{2} \cos^2 \theta + \frac{1}{3} \cos^2 \theta \]
Combining the \( \cos^2 \theta \) terms:
\[ P = \frac{1}{2} + \left(-\frac{1}{2} + \frac{1}{3}\right) \cos^2 \theta \]

### Step 3: Find a common denominator for the coefficients of \( \cos^2 \theta \)
The common denominator for \( -\frac{1}{2} \) and \( \frac{1}{3} \) is 6. Thus:
\[ -\frac{1}{2} = -\frac{3}{6}, \quad \frac{1}{3} = \frac{2}{6} \]
So,
\[ -\frac{1}{2} + \frac{1}{3} = -\frac{3}{6} + \frac{2}{6} = -\frac{1}{6} \]
Thus, we can rewrite \( P \) as:
\[ P = \frac{1}{2} - \frac{1}{6} \cos^2 \theta \]

### Step 4: Determine the range of \( P \)
Since \( \cos^2 \theta \) ranges from 0 to 1, we can substitute these values into the expression for \( P \):
- When \( \cos^2 \theta = 0 \):
\[ P = \frac{1}{2} - \frac{1}{6} \cdot 0 = \frac{1}{2} \]
- When \( \cos^2 \theta = 1 \):
\[ P = \frac{1}{2} - \frac{1}{6} \cdot 1 = \frac{1}{2} - \frac{1}{6} = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \]

### Step 5: Conclusion on the range of \( P \)
Thus, the range of \( P \) is:
\[ \frac{1}{3} \leq P \leq \frac{1}{2} \]

### Final Answer
The range of \( P \) is \( \left[\frac{1}{3}, \frac{1}{2}\right] \).

---