Let `f(x)=ab sin x+bsqrt(1-a^(2))cosx+c`, where `|a|lt1,bgt0` then

Options

A. maximum value of `f(x)=-cos^(-1)a`
B. `f(x)=c` if `x=cos^(-1)alpha`
C. `f(x)=c` if `x=-cos^(-1)alpha`
D. `f(x)=c` if `x=cos^(-1)alpha`.

Explanation

To solve the problem step by step, we will analyze the function given and derive the necessary expressions to find the maximum and minimum values of \( f(x) \).

### Step 1: Write down the function
The function is given as:
\[ f(x) = ab \sin x + b \sqrt{1 - a^2} \cos x + c \]

### Step 2: Factor out \( b \)
We can factor \( b \) from the first two terms:
\[ f(x) = b \left( a \sin x + \sqrt{1 - a^2} \cos x \right) + c \]

### Step 3: Rewrite the sine and cosine terms
We can express \( a \sin x + \sqrt{1 - a^2} \cos x \) in the form of \( R \sin(x + \alpha) \) where:
\[ R = \sqrt{a^2 + (1 - a^2)} = 1 \]
and
\[ \tan \alpha = \frac{\sqrt{1 - a^2}}{a} \]
Thus, we rewrite:
\[ f(x) = b \sin(x + \alpha) + c \]

### Step 4: Determine the maximum and minimum values of \( f(x) \)
The maximum value of \( \sin(x + \alpha) \) is 1 and the minimum value is -1. Therefore:
- Maximum of \( f(x) \):
\[ f_{\text{max}} = b \cdot 1 + c = b + c \]
- Minimum of \( f(x) \):
\[ f_{\text{min}} = b \cdot (-1) + c = -b + c \]

### Step 5: Calculate the difference between the maximum and minimum values
Now, we find the difference:
\[ f_{\text{max}} - f_{\text{min}} = (b + c) - (-b + c) = b + c + b - c = 2b \]

### Step 6: Conclusion
Thus, the difference between the maximum and minimum values of \( f(x) \) is \( 2b \).