If `A = "sin"^(2)x + cos^(4)x` then for all realx :

Options

A. `(3)/(4)leAle1`
B. `(13)/(16)leAle1`
C. `1leAle2`
D. `(3)/(4)leAle(13)/(16)`

Answer

To find the maximum and minimum values of \( A = \sin^2 x + \cos^4 x \), we will follow these steps:

### Step 1: Rewrite \( A \)
We start with the expression:
\[ A = \sin^2 x + \cos^4 x \]
Using the identity \( \sin^2 x + \cos^2 x = 1 \), we can express \( \cos^4 x \) in terms of \( \sin^2 x \):
\[ \cos^4 x = (\cos^2 x)^2 = (1 - \sin^2 x)^2 \]
Thus, we can rewrite \( A \) as:
\[ A = \sin^2 x + (1 - \sin^2 x)^2 \]

### Step 2: Expand \( A \)
Now, we expand the square:
\[ A = \sin^2 x + (1 - 2\sin^2 x + \sin^4 x) \]
This simplifies to:
\[ A = \sin^4 x - \sin^2 x + 1 \]

### Step 3: Substitute \( y = \sin^2 x \)
Let \( y = \sin^2 x \). Then \( y \) ranges from 0 to 1. We can rewrite \( A \) in terms of \( y \):
\[ A = y^2 - y + 1 \]

### Step 4: Find the critical points
To find the maximum and minimum values of \( A \), we differentiate \( A \) with respect to \( y \):
\[ \frac{dA}{dy} = 2y - 1 \]
Setting the derivative equal to zero to find critical points:
\[ 2y - 1 = 0 \implies y = \frac{1}{2} \]

### Step 5: Evaluate \( A \) at critical points and endpoints
Now we evaluate \( A \) at \( y = 0 \), \( y = 1 \), and \( y = \frac{1}{2} \):

1. For \( y = 0 \):
\[ A(0) = 0^2 - 0 + 1 = 1 \]

2. For \( y = 1 \):
\[ A(1) = 1^2 - 1 + 1 = 1 \]

3. For \( y = \frac{1}{2} \):
\[ A\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \frac{1}{2} + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{1}{4} - \frac{2}{4} + \frac{4}{4} = \frac{3}{4} \]

### Step 6: Determine maximum and minimum values
From the evaluations:
- \( A(0) = 1 \)
- \( A(1) = 1 \)
- \( A\left(\frac{1}{2}\right) = \frac{3}{4} \)

Thus, the minimum value of \( A \) is \( \frac{3}{4} \) and the maximum value is \( 1 \).

### Final Result
We conclude that:
\[ \frac{3}{4} \leq A \leq 1 \]