Find the domain and range of `f(x) = sin^-1 (log [x]) + log (sin^-1 [x])`, where [.] denotes the greatest integer function.

Answer

To find the domain and range of the function \( f(x) = \sin^{-1}(\log[\lfloor x \rfloor]) + \log(\sin^{-1}[\lfloor x \rfloor]) \), where \([\cdot]\) denotes the greatest integer function, we will break the problem down into manageable steps.

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

#### 1.1: Analyze \( f_1(x) = \sin^{-1}(\log[\lfloor x \rfloor]) \)

The function \( \sin^{-1}(y) \) is defined for \( -1 \leq y \leq 1 \). Therefore, we need:

\[ -1 \leq \log[\lfloor x \rfloor] \leq 1 \]

Taking the exponential (antilog) of the inequalities:

\[ 10^{-1} \leq [\lfloor x \rfloor] \leq 10^1 \]

This simplifies to:

\[ 0.1 \leq [\lfloor x \rfloor] \leq 10 \]

Since \([\lfloor x \rfloor]\) is an integer, the possible values are:

\[ [\lfloor x \rfloor] = 1, 2, \ldots, 10 \]

This means:

\[ 1 \leq \lfloor x \rfloor \leq 10 \]

Thus, the corresponding values of \( x \) are:

\[ 1 \leq x < 11 \]

#### 1.2: Analyze \( f_2(x) = \log(\sin^{-1}[\lfloor x \rfloor]) \)

The function \( \log(y) \) is defined for \( y > 0 \). Therefore, we need:

\[ \sin^{-1}[\lfloor x \rfloor] > 0 \]

Since \( \sin^{-1}(y) \) is positive for \( y > 0 \), we require:

\[ [\lfloor x \rfloor] > 0 \]

This means:

\[ [\lfloor x \rfloor] \geq 1 \]

Thus, the possible values of \([\lfloor x \rfloor]\) remain the same:

\[ [\lfloor x \rfloor] = 1, 2, \ldots, 10 \]

This again leads to:

\[ 1 \leq x < 11 \]

### Step 2: Final Domain of \( f(x) \)

The domain of \( f(x) \) is the intersection of the domains of \( f_1(x) \) and \( f_2(x) \):

\[ \text{Domain of } f(x) = [1, 11) \]

### Step 3: Determine the Range of \( f(x) \)

#### 3.1: Evaluate \( f(x) \) for \( 1 \leq x < 2 \)

For \( 1 \leq x < 2 \), \([\lfloor x \rfloor] = 1\):

\[ f(x) = \sin^{-1}(\log(1)) + \log(\sin^{-1}(1)) = \sin^{-1}(0) + \log\left(\frac{\pi}{2}\right) = 0 + \log\left(\frac{\pi}{2}\right) \]

#### 3.2: Evaluate \( f(x) \) for \( 2 \leq x < 3 \)

For \( 2 \leq x < 3 \), \([\lfloor x \rfloor] = 2\):

\[ f(x) = \sin^{-1}(\log(2)) + \log(\sin^{-1}(2)) \]

However, \(\sin^{-1}(2)\) is not defined, so this case is invalid.

Continuing this way, we find that for \( 1 \leq x < 2 \), the only valid output is:

\[ f(x) = \log\left(\frac{\pi}{2}\right) \]

### Step 4: Final Range of \( f(x) \)

Since the only valid output occurs when \( x \) is in the interval \([1, 2)\):

\[ \text{Range of } f(x) = \left\{ \log\left(\frac{\pi}{2}\right) \right\} \]

### Summary

- **Domain**: \( [1, 11) \)
- **Range**: \( \left\{ \log\left(\frac{\pi}{2}\right) \right\} \)