Let `f(x)` be a function such that `f^(prime)(x)=(log)_(1/3)[(log)_3(sinx+a)]dot` If `f(x)` is decreasing for all real values of `x ,` then `a in (1,4)` (b) `a in (4,oo)` `a in (2,3)` (d) `a in (2,oo)`

Options

A. `a in (1,4)`
B. `a in (4,oo)`
C. `a in (2,3)`
D. `a in(2,oo)`

Explanation

To solve the problem, we start with the given information about the function \( f(x) \) and its derivative \( f'(x) \).

### Step 1: Understand the condition for \( f(x) \) to be decreasing
Since \( f(x) \) is decreasing for all real values of \( x \), it implies that:
\[ f'(x) < 0 \quad \text{for all } x \]

### Step 2: Analyze the derivative
The derivative is given as:
\[ f'(x) = \log_{1/3}(\log_3(\sin x + a)) \]
To analyze when this is negative, we can rewrite the logarithm:
\[ f'(x) = -\log_3(\log_3(\sin x + a)) \]
This is negative if:
\[ \log_3(\log_3(\sin x + a)) > 0 \]
This means:
\[ \log_3(\sin x + a) > 1 \]

### Step 3: Solve the inequality
From the inequality \( \log_3(\sin x + a) > 1 \), we can exponentiate both sides:
\[ \sin x + a > 3 \]
Thus, we can rearrange this to find:
\[ \sin x > 3 - a \]

### Step 4: Determine the range of \( \sin x \)
The sine function has a range of \([-1, 1]\). Therefore, for the inequality \( \sin x > 3 - a \) to hold for all \( x \), we need:
\[ 3 - a < -1 \]
Solving this gives:
\[ 3 + 1 < a \quad \Rightarrow \quad a > 4 \]

### Step 5: Conclusion
Thus, for \( f(x) \) to be decreasing for all real values of \( x \), \( a \) must satisfy:
\[ a > 4 \]
This corresponds to option (b) \( a \in (4, \infty) \).

### Final Answer:
The correct option is (b) \( a \in (4, \infty) \).