If `0 lt x lt pi/2` and `sin^(n) x + cos^(n) x ge 1`, then

Options

A. `n in [ 2, oo)`
B. ` n in (-oo, 2]`
C. `n in [-1,1]`
D. None of these

Answer

To solve the inequality \( \sin^n x + \cos^n x \geq 1 \) for \( 0 < x < \frac{\pi}{2} \), we can follow these steps:

### Step 1: Apply the AM-GM Inequality
We know that for any two non-negative numbers \( a \) and \( b \), the Arithmetic Mean (AM) is greater than or equal to the Geometric Mean (GM). Thus, we can write:

\[ \frac{\sin^n x + \cos^n x}{2} \geq \sqrt{\sin^n x \cos^n x} \]

### Step 2: Rearranging the Inequality
From the AM-GM inequality, we can rearrange it to obtain:

\[ \sin^n x + \cos^n x \geq 2 \sqrt{\sin^n x \cos^n x} \]

### Step 3: Set the Inequality
Since we are given that \( \sin^n x + \cos^n x \geq 1 \), we can combine this with our previous result:

\[ 2 \sqrt{\sin^n x \cos^n x} \geq 1 \]

### Step 4: Square Both Sides
To eliminate the square root, we square both sides of the inequality:

\[ 4 \sin^n x \cos^n x \geq 1 \]

### Step 5: Express in Terms of Sine and Cosine
Using the identity \( \sin x \cos x = \frac{1}{2} \sin 2x \), we can rewrite the left-hand side:

\[ 4 \left(\frac{1}{2} \sin 2x\right)^n \geq 1 \]

This simplifies to:

\[ 2^n \sin^n 2x \geq 1 \]

### Step 6: Solve for \( n \)
From the inequality \( 2^n \sin^n 2x \geq 1 \), we can derive:

\[ \sin^n 2x \geq 2^{-n} \]

### Step 7: Analyze the Range of \( \sin 2x \)
Since \( 0 < x < \frac{\pi}{2} \), it follows that \( 0 < 2x < \pi \). Therefore, \( \sin 2x \) will take values in the range \( (0, 1] \).

### Step 8: Determine Possible Values of \( n \)
To satisfy the inequality for all \( x \) in the given range, we need to analyze the implications of \( n \):

- If \( n < 2 \), then \( 2^{-n} \) becomes larger than 1, which is not possible since \( \sin 2x \) cannot exceed 1.
- If \( n = 2 \), we have \( \sin^2 2x \geq \frac{1}{4} \), which is possible for certain values of \( x \).
- If \( n > 2 \), \( 2^n \) increases, making it harder for \( \sin^n 2x \) to satisfy the inequality.

### Conclusion
Thus, the inequality \( \sin^n x + \cos^n x \geq 1 \) holds for \( n \) in the range:

\[ n \in [2, \infty) \]