Let `f(x)` be a function such that `f '(x)= log _(1//3) (log _(3) (sin x+ a)).` The complete set of values of 'a' for which `f (x)` is strictly decreasing for all real values of x is:

Options

A. `[4,oo)`
B. `[3,4]`
C. `(-oo,4)`
D. `[2,oo)`

Answer

To solve the problem, we need to analyze the function \( f'(x) = \log_{1/3}(\log_3(\sin x + a)) \) and determine the conditions under which \( f(x) \) is strictly decreasing for all real values of \( x \).

### Step-by-Step Solution:

1. **Understanding the Condition for Decreasing Function:**
A function \( f(x) \) is strictly decreasing if its derivative \( f'(x) < 0 \) for all \( x \).

2. **Setting Up the Inequality:**
We start with the given derivative:
\[ f'(x) = \log_{1/3}(\log_3(\sin x + a)) \]
For \( f(x) \) to be strictly decreasing, we need:
\[ \log_{1/3}(\log_3(\sin x + a)) < 0 \]

3. **Using Properties of Logarithms:**
The logarithm \( \log_{1/3}(y) < 0 \) implies that \( y < 1 \) because the base \( \frac{1}{3} < 1 \). Therefore, we have:
\[ \log_3(\sin x + a) < 1 \]

4. **Exponentiating Both Sides:**
To eliminate the logarithm, we exponentiate:
\[ \sin x + a < 3^1 \]
Simplifying gives:
\[ \sin x + a < 3 \]
Thus:
\[ a < 3 - \sin x \]

5. **Finding the Range of \( \sin x \):**
The function \( \sin x \) oscillates between -1 and 1. Therefore, we need to consider the maximum and minimum values of \( \sin x \):
- When \( \sin x = 1 \):
\[ a < 3 - 1 \implies a < 2 \]
- When \( \sin x = -1 \):
\[ a < 3 - (-1) \implies a < 4 \]

6. **Combining the Results:**
Since \( a \) must satisfy both conditions, we take the more restrictive condition:
\[ a < 2 \]

7. **Conclusion:**
The complete set of values of \( a \) for which \( f(x) \) is strictly decreasing for all real values of \( x \) is:
\[ a < 2 \]

### Final Answer:
The complete set of values of \( a \) for which \( f(x) \) is strictly decreasing for all real values of \( x \) is \( a < 2 \).