The maximum and minimum value of f(x) `=ab sin x +b sqrt(1-a^(2)) cos x + c` lie in the interval (assuming `|a| lt 1, b gt 0`)

Options

A. `[b-c, b+c]`
B. `(b-c, b+c)`
C. `[c-b, b+c]`
D. none of these

Answer

To find the maximum and minimum values of the function \( f(x) = ab \sin x + b \sqrt{1 - a^2} \cos x + c \) under the conditions \( |a| < 1 \) and \( b > 0 \), we can follow these steps:

### Step 1: Identify the structure of the function
The function can be rewritten as:
\[ f(x) = A \sin x + B \cos x + c \]
where \( A = ab \) and \( B = b \sqrt{1 - a^2} \).

### Step 2: Find the maximum and minimum values of \( A \sin x + B \cos x \)
The maximum and minimum values of the expression \( A \sin x + B \cos x \) can be determined using the formula:
\[ \text{Max} = \sqrt{A^2 + B^2}, \quad \text{Min} = -\sqrt{A^2 + B^2} \]

### Step 3: Calculate \( A^2 + B^2 \)
Substituting \( A \) and \( B \):
\[ A^2 = (ab)^2 = a^2b^2 \]
\[ B^2 = (b \sqrt{1 - a^2})^2 = b^2(1 - a^2) = b^2 - a^2b^2 \]
Thus,
\[ A^2 + B^2 = a^2b^2 + (b^2 - a^2b^2) = b^2 \]

### Step 4: Find the maximum and minimum values of \( f(x) \)
Now substituting back into the max and min formulas:
\[ \text{Max}(A \sin x + B \cos x) = \sqrt{b^2} = b \]
\[ \text{Min}(A \sin x + B \cos x) = -\sqrt{b^2} = -b \]
Thus, the maximum and minimum values of \( f(x) \) are:
\[ \text{Max}(f(x)) = b + c \]
\[ \text{Min}(f(x)) = -b + c \]

### Final Result
Therefore, the maximum and minimum values of \( f(x) \) are:
- Maximum value: \( b + c \)
- Minimum value: \( c - b \)