Let's analyze each option carefully based on the definitions given for $f(x) = \arcsin(x)$ and $g(x) = \sin(x)$.

Step-by-Step Analysis

1. Option (a): $\exists x \in \mathbb{R} \text{ such that } f(g(x)) \text{ is not defined}$

   To evaluate this, let’s look at $f(g(x)) = \arcsin(\sin(x))$.

   • The function $g(x) = \sin(x)$ outputs values in the range $[-1, 1]$, which are all valid inputs for $f(x) = \arcsin(x)$, as $\arcsin$ is defined for inputs in $[-1, 1]$.
   • Therefore, there are no values of $x \in \mathbb{R}$ for which $f(g(x))$ is undefined.

   Conclusion: Option (a) is incorrect.

2. Option (b): $f(g(x)) = x, \, \forall x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

   • In this interval, $f(g(x)) = \arcsin(\sin(x)) = x$ holds true because $\arcsin$ and $\sin$ are inverse functions on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
   • Hence, $f(g(x)) = x$ is indeed valid for all $x$ in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

   Conclusion: Option (b) is correct.

3. Option (c): $f(g(x)) = x, \, \forall x \in \mathbb{R}$

   • This statement implies that $\arcsin(\sin(x)) = x$ for all $x \in \mathbb{R}$.
   • However, this is only true within the principal branch of the inverse sine function, $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. For values of $x$ outside this interval, $\arcsin(\sin(x))$ will not return $x$ directly, as $\arcsin$ will return values within $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

   Conclusion: Option (c) is incorrect.

4. Option (d): $\forall x \in \mathbb{R}, f(g(x)) = g(f(x))$

   • Let's examine $f(g(x))$ and $g(f(x))$:
     • $f(g(x)) = \arcsin(\sin(x))$.
     • $g(f(x)) = \sin(\arcsin(x)) = x$ for $x \in [-1, 1]$.
   • However, $f(g(x))$ is not necessarily equal to $g(f(x))$ for all $x \in \mathbb{R}$ since they are defined differently depending on the values outside the principal branch.

   Conclusion: Option (d) is incorrect.

5. Option (e): $f(g(x)) = \arcsin(x) \cdot \sin(x)$

   • This option suggests that $\arcsin(\sin(x)) = \arcsin(x) \cdot \sin(x)$.
   • This is not a correct identity in trigonometry. The correct identity is $\arcsin(\sin(x)) = x$ for $x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, without involving multiplication.

   Conclusion: Option (e) is incorrect.

Final Answer

The correct answer is:

• (b) $f(g(x)) = x, \, \forall x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

Would you like more details on any specific part of the solution?

Related Questions:

1. What is the domain and range of the $\arcsin(x)$ function?
2. How does the composition of trigonometric functions work outside their principal domains?
3. Can we generalize the identity $f(g(x)) = x$ for other inverse functions?
4. What are the properties of the sine function over its extended domain?
5. How would this analysis change if we considered the $\arccos(x)$ function?

Tip: When dealing with compositions of inverse trigonometric functions, always consider the domain and range restrictions carefully to determine where the identity $f(g(x)) = x$ holds true.