The question concerns the function $f : \mathbb{R} \to \mathbb{R}$, defined as $f(x) = a \arctan\left(\frac{x}{b}\right)$, where $a > 0$ and $b > 0$. The task is to determine the domain of the inverse function $f^{-1}$.

Solution:

1. The function $f(x) = a \arctan\left(\frac{x}{b}\right)$ maps $x \in \mathbb{R}$ to values constrained by the range of $\arctan$, which is $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

2. Since $\arctan\left(\frac{x}{b}\right)$ takes values in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, multiplying by $a$ scales the range to: $\left(-\frac{\pi}{2}a, \frac{\pi}{2}a\right).$

3. The domain of the inverse function $f^{-1}$ is therefore the range of $f(x)$, which is: $\left(-\frac{\pi}{2}a, \frac{\pi}{2}a\right).$

Answer:

The correct choice is (D): $\left(-\frac{\pi}{2}a, \frac{\pi}{2}a\right).$

Would you like a more detailed explanation or clarification? Here are some related questions for deeper understanding:

1. What are the general properties of inverse functions and their domains and ranges?
2. How does the range of $\arctan(x)$ affect transformations of this function?
3. Why does $a > 0$ and $b > 0$ ensure the function is invertible?
4. How would the result change if $a < 0$?
5. Can you derive the explicit formula for the inverse $f^{-1}(x)$?

Tip: For inverse functions, always remember that the domain of the inverse is the range of the original function!