Let`f(x) = sin^(-1) (x-3)/(2) - log _(10) (4-x)`
Statement 1. the domain of ` f(x) ` is [1,3]
statement 2. `sin^(-1)` x is defined for `|x| le 1 ` and ` log_(10) x ` is defined for ` x gt 0`

Answer

To solve the problem, we need to determine the domain of the function \( f(x) = \sin^{-1}\left(\frac{x-3}{2}\right) - \log_{10}(4-x) \).

### Step 1: Determine the domain of \( \sin^{-1}\left(\frac{x-3}{2}\right) \)

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

\[ -\frac{1}{2} \leq \frac{x-3}{2} \leq 1 \]

**Hint:** To solve this compound inequality, multiply all parts by 2 to eliminate the fraction.

### Step 2: Solve the inequalities

1. From \( -\frac{1}{2} \leq \frac{x-3}{2} \):
\[ -1 \leq x - 3 \implies x \geq 2 \]

2. From \( \frac{x-3}{2} \leq 1 \):
\[ x - 3 \leq 2 \implies x \leq 5 \]

Combining these results, we find:
\[ 2 \leq x \leq 5 \]
Thus, the domain from this part is \( [2, 5] \).

**Hint:** Remember to check the endpoints when determining the final interval.

### Step 3: Determine the domain of \( -\log_{10}(4-x) \)

The logarithmic function \( \log_{10}(z) \) is defined for \( z > 0 \). Therefore, we need to solve:

\[ 4 - x > 0 \implies x < 4 \]

**Hint:** This inequality indicates that \( x \) must be less than 4.

### Step 4: Combine the domains

Now we have two conditions for the domain of \( f(x) \):
1. From \( \sin^{-1} \): \( [2, 5] \)
2. From \( -\log_{10} \): \( (-\infty, 4) \)

To find the overall domain, we take the intersection of these two intervals:

\[ [2, 5] \cap (-\infty, 4) = [2, 4) \]

**Hint:** The intersection of two intervals gives us the common values that satisfy both conditions.

### Conclusion

The domain of \( f(x) \) is \( [2, 4) \).

### Evaluating the Statements

- **Statement 1** claims the domain is \( [1, 3] \), which is **false**.
- **Statement 2** correctly states the conditions for \( \sin^{-1} \) and \( \log_{10} \), which is **true**.

Thus, the final conclusion is:
- Statement 1: False
- Statement 2: True