If the mapping `f(x)=ax+b,a lt0` and maps [-1, 1] onto [0, 2] , then for all values of `theta, A=cos^(2) theta + sin^(4) theta` is such that

Options

A. `f(1/4) le A le f(0)`
B. `f(0) le A le f(-2)`
C. `f(1/3) le A le f(0)`
D. `f(-1) lt A le f (-2)`

Answer

To solve the problem, we need to analyze the function \( f(x) = ax + b \) where \( a < 0 \) and it maps the interval \([-1, 1]\) onto \([0, 2]\). We will also explore the expression \( A = \cos^2 \theta + \sin^4 \theta \).

### Step 1: Determine the values of \( a \) and \( b \)

Given that \( f(x) \) maps \([-1, 1]\) onto \([0, 2]\), we can set up the following equations based on the endpoints of the intervals:

1. When \( x = -1 \), \( f(-1) = 0 \):
\[ 0 = -a + b \quad \text{(1)} \]

2. When \( x = 1 \), \( f(1) = 2 \):
\[ 2 = a + b \quad \text{(2)} \]

### Step 2: Solve the system of equations

From equation (1), we can express \( b \) in terms of \( a \):
\[ b = a \quad \text{(from 1)} \]

Substituting \( b = a \) into equation (2):
\[ 2 = a + a \\ 2 = 2a \\ a = 1 \]

However, since \( a < 0 \), we must have made an error in interpreting the conditions. Let's correct this:

From equation (1):
\[ b = a \quad \text{(1)} \]

Substituting \( b \) into equation (2):
\[ 2 = a + a \\ 2 = 2a \\ a = 1 \quad \text{(not valid since \( a < 0 \))} \]

Instead, we should have:
1. From (1): \( b = a + 0 \)
2. From (2): \( b = 2 - a \)

Setting these equal gives:
\[ a = 2 - a \\ 2a = 2 \\ a = -1 \]

Substituting \( a = -1 \) back into \( b = -a \):
\[ b = 1 \]

Thus, the function is:
\[ f(x) = -x + 1 \]

### Step 3: Analyze \( A = \cos^2 \theta + \sin^4 \theta \)

Now we need to analyze the expression \( A = \cos^2 \theta + \sin^4 \theta \).

Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \):
\[ A = (1 - \sin^2 \theta) + \sin^4 \theta \\ A = 1 - \sin^2 \theta + \sin^4 \theta \]

Let \( y = \sin^2 \theta \), then:
\[ A = 1 - y + y^2 \]

### Step 4: Find the maximum and minimum values of \( A \)

To find the maximum and minimum values of \( A \), we can differentiate:
\[ \frac{dA}{dy} = -1 + 2y \]
Setting the derivative to zero:
\[ -1 + 2y = 0 \\ 2y = 1 \\ y = \frac{1}{2} \]

Now, we evaluate \( A \) at \( y = 0 \), \( y = 1 \), and \( y = \frac{1}{2} \):
1. \( A(0) = 1 - 0 + 0 = 1 \)
2. \( A(1) = 1 - 1 + 1 = 1 \)
3. \( A\left(\frac{1}{2}\right) = 1 - \frac{1}{2} + \left(\frac{1}{2}\right)^2 = 1 - \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \)

### Conclusion

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