For all real values of `alpha, x = cos^(4) alpha + sin^(2)alpha`, then range of x is:

Options

A. `3/4 le x le 1`
B. `3/4 le x le 13/15`
C. `13/16 le x le 1`
D. `1/2 le x le 2`

Answer

To find the range of \( x = \cos^4 \alpha + \sin^2 \alpha \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ x = \cos^4 \alpha + \sin^2 \alpha \]
We can rewrite \( \cos^4 \alpha \) as \( (\cos^2 \alpha)^2 \). Let \( y = \cos^2 \alpha \). Then, we have:
\[ x = y^2 + (1 - y) \]
since \( \sin^2 \alpha = 1 - \cos^2 \alpha = 1 - y \).

### Step 2: Simplify the expression
Now, we can simplify the expression:
\[ x = y^2 + 1 - y \]
This simplifies to:
\[ x = y^2 - y + 1 \]

### Step 3: Determine the range of \( y \)
Since \( y = \cos^2 \alpha \), the range of \( y \) is from \( 0 \) to \( 1 \) (i.e., \( 0 \leq y \leq 1 \)).

### Step 4: Analyze the quadratic function
The expression \( x = y^2 - y + 1 \) is a quadratic function in terms of \( y \). To find the minimum and maximum values, we can analyze the vertex of the parabola.

The vertex of a quadratic \( ay^2 + by + c \) is given by:
\[ y = -\frac{b}{2a} \]
Here, \( a = 1 \) and \( b = -1 \):
\[ y = -\frac{-1}{2 \cdot 1} = \frac{1}{2} \]

### Step 5: Calculate \( x \) at the vertex
Now, we substitute \( y = \frac{1}{2} \) back into the expression for \( x \):
\[ x = \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{1}{4} - \frac{2}{4} + \frac{4}{4} = \frac{3}{4} \]

### Step 6: Calculate \( x \) at the endpoints
Next, we evaluate \( x \) at the endpoints of the interval for \( y \):
1. When \( y = 0 \):
\[ x = 0^2 - 0 + 1 = 1 \]
2. When \( y = 1 \):
\[ x = 1^2 - 1 + 1 = 1 \]

### Step 7: Conclusion
Thus, the minimum value of \( x \) is \( \frac{3}{4} \) (at \( y = \frac{1}{2} \)), and the maximum value of \( x \) is \( 1 \) (at \( y = 0 \) and \( y = 1 \)). Therefore, the range of \( x \) is:
\[ \left[\frac{3}{4}, 1\right] \]