If `P_(n) = cos^(n) theta + sin^(n) theta` where
`theta [0,pi/2], n `(-infinity, 2)` `then minimum of `P_(n)` will be
1. 1, 2.`1/2`, 3. `sqrt(2)`, 4.`1/sqrt(2)`

Options

A. 1
B. `1/2`
C. `sqrt(2)`
D. `1/sqrt(2)`

Answer

To find the minimum value of \( P_n = \cos^n \theta + \sin^n \theta \) where \( \theta \in [0, \frac{\pi}{2}] \) and \( n \in (-\infty, 2) \), we can follow these steps:

### Step 1: Understand the expression
We start with the expression:
\[ P_n = \cos^n \theta + \sin^n \theta \]
This expression combines the powers of cosine and sine.

**Hint:** Remember that both \( \cos \theta \) and \( \sin \theta \) are non-negative in the interval \( [0, \frac{\pi}{2}] \).

### Step 2: Analyze the behavior of \( P_n \)
Since \( n < 2 \), the terms \( \cos^n \theta \) and \( \sin^n \theta \) will behave differently compared to when \( n \geq 2 \). Specifically, as \( n \) approaches negative infinity, both terms approach 1 when either \( \cos \theta \) or \( \sin \theta \) is 1.

**Hint:** Consider the limits of \( \cos^n \theta \) and \( \sin^n \theta \) as \( n \) varies.

### Step 3: Find critical points
To find the minimum value, we can evaluate \( P_n \) at the endpoints of the interval for \( \theta \):
- When \( \theta = 0 \):
\[ P_n(0) = \cos^n(0) + \sin^n(0) = 1^n + 0^n = 1 \]
- When \( \theta = \frac{\pi}{2} \):
\[ P_n\left(\frac{\pi}{2}\right) = \cos^n\left(\frac{\pi}{2}\right) + \sin^n\left(\frac{\pi}{2}\right) = 0^n + 1^n = 1 \]

**Hint:** Evaluate \( P_n \) at \( \theta = 0 \) and \( \theta = \frac{\pi}{2} \) to check for minimum values.

### Step 4: Check the midpoint
Now, consider \( \theta = \frac{\pi}{4} \):
\[ P_n\left(\frac{\pi}{4}\right) = \left(\frac{1}{\sqrt{2}}\right)^n + \left(\frac{1}{\sqrt{2}}\right)^n = 2 \left(\frac{1}{\sqrt{2}}\right)^n = 2^{1 - \frac{n}{2}} \]
As \( n \) approaches negative infinity, this value approaches infinity.

**Hint:** Analyze the value of \( P_n \) at \( \theta = \frac{\pi}{4} \) to see how it compares to the endpoints.

### Step 5: Conclusion
From our evaluations:
- \( P_n(0) = 1 \)
- \( P_n\left(\frac{\pi}{2}\right) = 1 \)
- \( P_n\left(\frac{\pi}{4}\right) \) approaches infinity as \( n \) decreases.

Thus, the minimum value of \( P_n \) is:
\[ \text{Minimum of } P_n = 1 \]

### Final Answer:
The minimum of \( P_n \) is \( \boxed{1} \).

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