Let `f(x)=a+2b cos^(-1)x, b gt0`. If domain and range of f(x) are the same set, then `(b-a)` is equal to :

Options

A. `1-(1)/(pi)`
B. `(2)/(pi)`
C. `(2)/(pi)+1`
D. `1+(1)/(pi)`

Answer

To solve the problem, we need to analyze the function \( f(x) = a + 2b \cos^{-1} x \) with the condition that the domain and range of \( f(x) \) are the same set. Let's go through the steps to find \( b - a \).

### Step 1: Identify the Domain of \( f(x) \)
The function \( \cos^{-1} x \) is defined for \( x \) in the interval \([-1, 1]\). Therefore, the domain of \( f(x) \) is:
\[ \text{Domain of } f(x) = [-1, 1] \]

### Step 2: Determine the Range of \( f(x) \)
Next, we need to find the range of \( f(x) \). The minimum and maximum values of \( \cos^{-1} x \) occur at the endpoints of the domain:
- At \( x = -1 \):
\[ f(-1) = a + 2b \cos^{-1}(-1) = a + 2b \cdot \pi \]
- At \( x = 1 \):
\[ f(1) = a + 2b \cos^{-1}(1) = a + 2b \cdot 0 = a \]

Thus, the range of \( f(x) \) is:
\[ \text{Range of } f(x) = [a, a + 2b\pi] \]

### Step 3: Set Domain Equal to Range
Since the domain and range must be the same, we set:
\[ [-1, 1] = [a, a + 2b\pi] \]

This gives us two equations:
1. \( a = -1 \)
2. \( a + 2b\pi = 1 \)

### Step 4: Solve for \( b \)
Substituting \( a = -1 \) into the second equation:
\[ -1 + 2b\pi = 1 \]
\[ 2b\pi = 1 + 1 = 2 \]
\[ b = \frac{2}{2\pi} = \frac{1}{\pi} \]

### Step 5: Calculate \( b - a \)
Now we can find \( b - a \):
\[ b - a = \frac{1}{\pi} - (-1) = \frac{1}{\pi} + 1 \]

### Final Answer
Thus, the value of \( b - a \) is:
\[ \boxed{\frac{1}{\pi} + 1} \]

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