If ` A = sin^(2)x + cos^(4)x`, where x in a real number

Options

A. ` 1 le A le 2`
B. `(3)/(4) le A le (13)/(16)`
C. `(3)/(4) le A le 1`
D. `(13)/(16) le A le 1`

Answer

To solve the problem where \( A = \sin^2 x + \cos^4 x \), we need to find the range of the expression. Let's break it down step by step.

### Step 1: Understand the ranges of \( \sin^2 x \) and \( \cos^4 x \)

1. **Range of \( \sin^2 x \)**:
- Since \( \sin x \) varies between -1 and 1, squaring it gives:
\[ \sin^2 x \in [0, 1] \]

2. **Range of \( \cos^4 x \)**:
- Similarly, \( \cos x \) also varies between -1 and 1. Therefore, raising it to the fourth power (which eliminates negative values) gives:
\[ \cos^4 x \in [0, 1] \]

### Step 2: Combine the ranges

Since both \( \sin^2 x \) and \( \cos^4 x \) are non-negative and each can take values from 0 to 1, we can analyze the sum:
\[ A = \sin^2 x + \cos^4 x \]

### Step 3: Find maximum value of \( A \)

To find the maximum value of \( A \), we can consider the extreme values of \( \sin^2 x \) and \( \cos^4 x \):

- When \( x = 0 \):
\[ A = \sin^2(0) + \cos^4(0) = 0 + 1 = 1 \]

- When \( x = 90^\circ \):
\[ A = \sin^2(90) + \cos^4(90) = 1 + 0 = 1 \]

Thus, the maximum value of \( A \) is:
\[ \text{Maximum} = 1 \]

### Step 4: Find minimum value of \( A \)

Next, we check for the minimum value of \( A \). The minimum occurs when both \( \sin^2 x \) and \( \cos^4 x \) are minimized. This typically occurs when \( x = 45^\circ \):

- When \( x = 45^\circ \):
\[ A = \sin^2(45) + \cos^4(45) \]
\[ = \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^4 \]
\[ = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \]

Thus, the minimum value of \( A \) is:
\[ \text{Minimum} = \frac{3}{4} \]

### Conclusion

The range of \( A \) is:
\[ \left[\frac{3}{4}, 1\right] \]

### Final Answer

The correct option is that the minimum value is \( \frac{3}{4} \) and the maximum value is \( 1 \).

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