The domain of the function `f(x)= sqrt(sin^(-1)x-(pi)/(4))+log(1-x)` is :

Options

A. `x in [(1)/(sqrt(2)), 1]`
B. `x in ((1)/(sqrt(2)), 1]`
C. `x in [(1)/(sqrt(2)), 1)`
D. `x in ((1)/(sqrt(2)),1)`

Answer

To find the domain of the function \( f(x) = \sqrt{\sin^{-1}(x) - \frac{\pi}{4}} + \log(1 - x) \), we need to ensure that both components of the function are defined and valid.

### Step 1: Determine the domain of \( \sin^{-1}(x) \)

The inverse sine function \( \sin^{-1}(x) \) is defined for \( x \) in the interval \([-1, 1]\). Therefore, we have:
\[ -1 \leq x \leq 1 \]

### Step 2: Ensure the expression inside the square root is non-negative

For the square root to be defined, the expression inside it must be non-negative:
\[ \sin^{-1}(x) - \frac{\pi}{4} \geq 0 \]
This simplifies to:
\[ \sin^{-1}(x) \geq \frac{\pi}{4} \]
Since \( \sin^{-1}(x) \) is an increasing function, we can take the sine of both sides:
\[ x \geq \sin\left(\frac{\pi}{4}\right) \]
Calculating \( \sin\left(\frac{\pi}{4}\right) \):
\[ \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \quad \text{(or } \frac{\sqrt{2}}{2} \text{)} \]
Thus, we have:
\[ x \geq \frac{1}{\sqrt{2}} \quad \text{(or } x \geq 0.7071 \text{)} \]

### Step 3: Determine the domain of the logarithm function

The logarithm \( \log(1 - x) \) is defined when its argument is positive:
\[ 1 - x > 0 \implies x < 1 \]

### Step 4: Combine the conditions

Now we combine the conditions:
1. From \( \sin^{-1}(x) \): \( x \geq \frac{1}{\sqrt{2}} \)
2. From \( \log(1 - x) \): \( x < 1 \)

Thus, the combined domain is:
\[ \frac{1}{\sqrt{2}} \leq x < 1 \]

### Final Domain

The domain of the function \( f(x) \) is:
\[ x \in \left[\frac{1}{\sqrt{2}}, 1\right) \]