The sum of the maximum and minimum values of the function `f(x)=1/(1+(2cosx-4sinx)^2)i s`

Options

A. (a) `22/21`
B. (b) `21/20`
C. (c) `22/20`
D. (d) `21/11`

Answer

To solve the problem, we need to find the sum of the maximum and minimum values of the function \( f(x) = \frac{1}{1 + (2 \cos x - 4 \sin x)^2} \).

### Step-by-Step Solution:

1. **Identify the variable part of the function**:
The function can be rewritten as:
\[ f(x) = \frac{1}{1 + g(x)} \]
where \( g(x) = (2 \cos x - 4 \sin x)^2 \).

2. **Find the maximum and minimum values of \( g(x) \)**:
We need to analyze the expression \( 2 \cos x - 4 \sin x \). We can use the formula for the maximum and minimum values of a linear combination of sine and cosine:
\[ a \cos x + b \sin x \text{ has a maximum value of } \sqrt{a^2 + b^2} \text{ and a minimum value of } -\sqrt{a^2 + b^2}. \]
Here, \( a = 2 \) and \( b = -4 \).

3. **Calculate the maximum and minimum values**:
\[ \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}. \]
Therefore,
\[ -\sqrt{20} \leq 2 \cos x - 4 \sin x \leq \sqrt{20}. \]
This implies:
\[ -2\sqrt{5} \leq 2 \cos x - 4 \sin x \leq 2\sqrt{5}. \]

4. **Square the expression**:
Since \( g(x) = (2 \cos x - 4 \sin x)^2 \), we can find the minimum and maximum values of \( g(x) \):
- The minimum value occurs when \( 2 \cos x - 4 \sin x = 0 \), thus \( g(x) = 0 \).
- The maximum value occurs when \( 2 \cos x - 4 \sin x = 2\sqrt{5} \), thus:
\[ g(x) = (2\sqrt{5})^2 = 20. \]

5. **Substitute back to find \( f(x) \)**:
- The maximum value of \( f(x) \) occurs when \( g(x) \) is at its minimum:
\[ f_{\text{max}} = \frac{1}{1 + 0} = 1. \]
- The minimum value of \( f(x) \) occurs when \( g(x) \) is at its maximum:
\[ f_{\text{min}} = \frac{1}{1 + 20} = \frac{1}{21}. \]

6. **Calculate the sum of maximum and minimum values**:
\[ f_{\text{max}} + f_{\text{min}} = 1 + \frac{1}{21} = \frac{21}{21} + \frac{1}{21} = \frac{22}{21}. \]

### Final Answer:
The sum of the maximum and minimum values of the function \( f(x) \) is:
\[ \frac{22}{21}. \]