Let `y=sin^(2)x+cos^(4)x`. Then, for all real x
(a) the maximum value of y is 2
(b) the minimum value of y is `3/4`

Options

A. the maximum value of y is 2
B. the minimum value of y is `3/4`
C. `y le 1`
D. `y ge 1/4`

Answer

To solve the problem, we need to analyze the expression \( y = \sin^2 x + \cos^4 x \). We will find both the maximum and minimum values of \( y \).

### Step 1: Rewrite \( \cos^4 x \)
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 \( y \) as:
\[ y = \sin^2 x + (1 - \sin^2 x)^2 \]

### Step 2: Expand the expression
Now, we expand \( (1 - \sin^2 x)^2 \):
\[ (1 - \sin^2 x)^2 = 1 - 2\sin^2 x + \sin^4 x \]
Substituting this back into the expression for \( y \):
\[ y = \sin^2 x + 1 - 2\sin^2 x + \sin^4 x \]
This simplifies to:
\[ y = \sin^4 x - \sin^2 x + 1 \]

### Step 3: Let \( t = \sin^2 x \)
Let \( t = \sin^2 x \). Since \( \sin^2 x \) can take values from 0 to 1, we have:
\[ y = t^2 - t + 1 \]

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

### Step 5: Evaluate \( y \) at the critical point
Now we evaluate \( y \) at \( t = \frac{1}{2} \):
\[ y\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{1}{4} - \frac{2}{4} + \frac{4}{4} = \frac{3}{4} \]

### Step 6: Check the endpoints
Next, we check the endpoints:
- When \( t = 0 \):
\[ y(0) = 0^2 - 0 + 1 = 1 \]
- When \( t = 1 \):
\[ y(1) = 1^2 - 1 + 1 = 1 \]

### Conclusion
Thus, the minimum value of \( y \) is \( \frac{3}{4} \) (at \( t = \frac{1}{2} \)), and the maximum value of \( y \) is \( 1 \) (at both \( t = 0 \) and \( t = 1 \)). Therefore, we conclude:
- The maximum value of \( y \) is \( 1 \).
- The minimum value of \( y \) is \( \frac{3}{4} \).

### Final Answers:
(a) The maximum value of \( y \) is **not** 2.
(b) The minimum value of \( y \) is \( \frac{3}{4} \).