The domain of `sin ^(-1)[log _(3) ((x)/(3)) ]` is

Options

A. `[1,9]`
B. `(-1,0)`
C. ` [-9,1]`
D. `[-9,-1]`

Answer

To find the domain of the function \( y = \sin^{-1} \left( \log_3 \left( \frac{x}{3} \right) \right) \), we need to ensure that the expression inside the inverse sine function is valid. The inverse sine function, \( \sin^{-1}(x) \), is defined for \( x \) in the range \([-1, 1]\). Therefore, we need to find the values of \( x \) for which:

\[ -1 \leq \log_3 \left( \frac{x}{3} \right) \leq 1 \]

### Step 1: Solve the inequality \( \log_3 \left( \frac{x}{3} \right) \leq 1 \)

To solve this, we can rewrite the logarithmic inequality:

\[ \log_3 \left( \frac{x}{3} \right) \leq 1 \]

This implies:

\[ \frac{x}{3} \leq 3^1 \]

\[ \frac{x}{3} \leq 3 \]

Multiplying both sides by 3 gives:

\[ x \leq 9 \]

### Step 2: Solve the inequality \( \log_3 \left( \frac{x}{3} \right) \geq -1 \)

Now we solve the other part of the inequality:

\[ \log_3 \left( \frac{x}{3} \right) \geq -1 \]

This implies:

\[ \frac{x}{3} \geq 3^{-1} \]

\[ \frac{x}{3} \geq \frac{1}{3} \]

Multiplying both sides by 3 gives:

\[ x \geq 1 \]

### Step 3: Combine the results

From the two inequalities we have derived:

1. \( x \leq 9 \)
2. \( x \geq 1 \)

Combining these results, we find that the domain of the function is:

\[ 1 \leq x \leq 9 \]

Thus, the domain of \( \sin^{-1} \left( \log_3 \left( \frac{x}{3} \right) \right) \) is:

\[ [1, 9] \]

### Final Answer:
The domain of \( \sin^{-1} \left( \log_3 \left( \frac{x}{3} \right) \right) \) is \( [1, 9] \).

---