Consider ` f(x) = sin^(-1) [2x] + cos^(-1) ( [ x] - 1)` ( where `[.]` denotes greatest integer function .) If domain of `f(x) " is " [a,b) ` and the range of ` f(x) ` is `{c,d}" then " a + b + (2d)/c ` is equal to ( where ` c lt d`)

Answer

To solve the problem, we need to analyze the function \( f(x) = \sin^{-1}(2x) + \cos^{-1}(x - 1) \) where the greatest integer function is applied. We will find the domain and range of this function step by step.

### Step 1: Determine the domain of \( f(x) \)

1. **Analyze \( \sin^{-1}(2x) \)**:
- The function \( \sin^{-1}(y) \) is defined for \( -1 \leq y \leq 1 \).
- Therefore, for \( \sin^{-1}(2x) \), we have:
\[ -1 \leq 2x \leq 1 \implies -\frac{1}{2} \leq x \leq \frac{1}{2} \]

2. **Analyze \( \cos^{-1}(x - 1) \)**:
- The function \( \cos^{-1}(y) \) is defined for \( -1 \leq y \leq 1 \).
- Therefore, for \( \cos^{-1}(x - 1) \), we have:
\[ -1 \leq x - 1 \leq 1 \implies 0 \leq x \leq 2 \]

3. **Combine the domains**:
- The domain of \( f(x) \) is the intersection of the two intervals:
\[ -\frac{1}{2} \leq x \leq \frac{1}{2} \quad \text{and} \quad 0 \leq x \leq 2 \]
- Thus, the domain is:
\[ [0, \frac{1}{2}] \]

### Step 2: Determine the range of \( f(x) \)

1. **Evaluate \( f(x) \) at the endpoints of the domain**:
- At \( x = 0 \):
\[ f(0) = \sin^{-1}(0) + \cos^{-1}(-1) = 0 + \pi = \pi \]
- At \( x = \frac{1}{2} \):
\[ f\left(\frac{1}{2}\right) = \sin^{-1}(1) + \cos^{-1}\left(-\frac{1}{2}\right) = \frac{\pi}{2} + \frac{2\pi}{3} = \frac{3\pi}{6} + \frac{4\pi}{6} = \frac{7\pi}{6} \]

2. **Analyze the behavior of \( f(x) \) in the interval**:
- The function is continuous and increases from \( \pi \) to \( \frac{7\pi}{6} \) as \( x \) moves from \( 0 \) to \( \frac{1}{2} \).

3. **Determine the range**:
- Thus, the range of \( f(x) \) is:
\[ [\pi, \frac{7\pi}{6}] \]

### Step 3: Identify \( a, b, c, d \)

From the domain and range:
- \( a = 0 \)
- \( b = \frac{1}{2} \)
- \( c = \pi \)
- \( d = \frac{7\pi}{6} \)

### Step 4: Calculate \( a + b + \frac{2d}{c} \)

1. **Calculate \( \frac{2d}{c} \)**:
\[ \frac{2d}{c} = \frac{2 \cdot \frac{7\pi}{6}}{\pi} = \frac{14\pi/6}{\pi} = \frac{14}{6} = \frac{7}{3} \]

2. **Calculate \( a + b + \frac{2d}{c} \)**:
\[ a + b + \frac{2d}{c} = 0 + \frac{1}{2} + \frac{7}{3} \]
- Convert \( \frac{1}{2} \) to a common denominator of 6:
\[ \frac{1}{2} = \frac{3}{6}, \quad \frac{7}{3} = \frac{14}{6} \]
- Therefore:
\[ a + b + \frac{2d}{c} = \frac{3}{6} + \frac{14}{6} = \frac{17}{6} \]

### Final Result

Thus, the final answer is:
\[ \frac{17}{6} \]