If `A= cos^(4)theta+sin^(2)theta` , then for all values of `theta` :

Options

A. `1 lt A le2`
B. `(13)/(16) le A le 1`
C. `(3)/(4) le A le (13)/(16)`
D. `(3)/(4) le A le 1`

Answer

To solve the problem, we need to find the range of the expression \( A = \cos^4 \theta + \sin^2 \theta \).

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

### Step 2: Expand the expression
Now, we expand \( (1 - \sin^2 \theta)^2 \):
\[ A = (1 - 2\sin^2 \theta + \sin^4 \theta) + \sin^2 \theta \]
This simplifies to:
\[ A = 1 - 2\sin^2 \theta + \sin^4 \theta + \sin^2 \theta \]
Combining like terms gives:
\[ A = 1 - \sin^2 \theta + \sin^4 \theta \]

### Step 3: Substitute \( x = \sin^2 \theta \)
Let \( x = \sin^2 \theta \). Since \( \sin^2 \theta \) varies from 0 to 1, we have:
\[ A = 1 - x + x^2 \]
Now, we need to find the range of \( A \) as \( x \) varies from 0 to 1.

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

### Step 5: Evaluate \( A \) at critical points and endpoints
Now we evaluate \( A \) at \( x = 0 \), \( x = 1 \), and \( x = \frac{1}{2} \):
1. When \( x = 0 \):
\[ A(0) = 1 - 0 + 0^2 = 1 \]
2. When \( x = 1 \):
\[ A(1) = 1 - 1 + 1^2 = 1 \]
3. When \( x = \frac{1}{2} \):
\[ A\left(\frac{1}{2}\right) = 1 - \frac{1}{2} + \left(\frac{1}{2}\right)^2 = 1 - \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \]

### Step 6: Determine the range of \( A \)
From the evaluations:
- The maximum value of \( A \) is 1.
- The minimum value of \( A \) is \( \frac{3}{4} \).

Thus, the range of \( A \) is:
\[ \frac{3}{4} \leq A \leq 1 \]

### Final Answer
The final answer is:
\[ A \in \left[\frac{3}{4}, 1\right] \]