Let `f(x)=sin[pi/6sin(pi/2sinx)]` for all `x in RR`

Options

A. Range of `f " is " [-(1)/(2),(1)/(2)]`
B. Range of fog is `[-(1)/(2),(1)/(2)]`
C. `underset(x to 0)(lim)(f(x))/(g(x))=(pi)/(6)`
D. There is an `x in R` such that `(gof)(x)=1`

Explanation

To solve the problem, we need to analyze the function \( f(x) = \sin\left(\frac{\pi}{6} \sin\left(\frac{\pi}{2} \sin x\right)\right) \) and determine its range.

### Step-by-Step Solution:

1. **Understanding the Inner Function**:
- Start with the innermost function: \( \sin x \).
- The sine function oscillates between -1 and 1 for all real numbers \( x \).
- Therefore, \( \sin x \in [-1, 1] \).

**Hint**: Remember that the sine function has a range of \([-1, 1]\).

2. **Analyzing the Next Layer**:
- Next, consider \( \frac{\pi}{2} \sin x \).
- Since \( \sin x \) ranges from -1 to 1, we have:
\[ \frac{\pi}{2} \sin x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \]

**Hint**: Multiplying by a constant scales the range accordingly.

3. **Evaluating the Sine of the Next Layer**:
- Now, we look at \( \sin\left(\frac{\pi}{2} \sin x\right) \).
- Since \( \frac{\pi}{2} \sin x \) ranges from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), we find:
\[ \sin\left(\frac{\pi}{2} \sin x\right) \in [-1, 1] \]

**Hint**: The sine function also has a range of \([-1, 1]\) when its argument is within \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).

4. **Applying the Final Layer**:
- Now we analyze \( \frac{\pi}{6} \sin\left(\frac{\pi}{2} \sin x\right) \).
- Since \( \sin\left(\frac{\pi}{2} \sin x\right) \) ranges from -1 to 1, we have:
\[ \frac{\pi}{6} \sin\left(\frac{\pi}{2} \sin x\right) \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right] \]

**Hint**: Again, multiplying by a constant scales the range.

5. **Finding the Sine of the Final Expression**:
- Finally, we evaluate \( f(x) = \sin\left(\frac{\pi}{6} \sin\left(\frac{\pi}{2} \sin x\right)\right) \).
- Since \( \frac{\pi}{6} \sin\left(\frac{\pi}{2} \sin x\right) \) ranges from \(-\frac{\pi}{6}\) to \(\frac{\pi}{6}\), we find:
\[ f(x) \in \left[\sin\left(-\frac{\pi}{6}\right), \sin\left(\frac{\pi}{6}\right)\right] \]
- Knowing that \( \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \) and \( \sin\left(-\frac{\pi}{6}\right) = -\frac{1}{2} \), we conclude:
\[ f(x) \in \left[-\frac{1}{2}, \frac{1}{2}\right] \]

**Hint**: The sine function is periodic and symmetric, which helps in evaluating its values at specific angles.

### Conclusion:
The range of the function \( f(x) \) is \( [-\frac{1}{2}, \frac{1}{2}] \).