Let $f(x) = \left|\frac{x}{2}\right|$ and $g(x) = \sqrt{3}\cos x + \sin x - 1$, for all real numbers $x$. If $[ \ \ ]$ denotes the greatest integer function, then the range of the function $[f \circ g]$ is:

The function is $g(x) = (\sqrt{3}\cos x + \sin x) - 1$.

The expression $a\cos x + b\sin x$ ranges from $-\sqrt{a^2+b^2}$ to $\sqrt{a^2+b^2}$.

Here, $a = \sqrt{3}$ and $b = 1$.

Maximum value $= \sqrt{(\sqrt{3})^2 + 1^2}$

$= \sqrt{3+1}$

$= 2$

Minimum value $= -2$

So, the range of $(\sqrt{3}\cos x + \sin x)$ is $[-2, 2]$.

Therefore, the range of $g(x)$ is $[-2 - 1, 2 - 1]$, which simplifies to $[-3, 1]$.

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Let $u = g(x)$, so $u \in [-3, 1]$.

The function becomes $f(u) = \left|\frac{u}{2}\right|$.

Analyzing the values of $f(u)$ as $u$ varies from $-3$ to $1$:

If $u = -3$, $f(-3) = |-1.5| = 1.5$ (Maximum value)

If $u = 0$, $f(0) = 0$ (Minimum value)

If $u = 1$, $f(1) = 0.5$

The range of the composite function $(f \circ g)(x)$ is continuous from the minimum to the maximum, which is $[0, 1.5]$.

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For any value $y$ in the interval $[0, 1.5]$, the greatest integer function gives:

For $y \in [0, 1)$, $[y] = 0$

For $y \in [1, 1.5]$, $[y] = 1$

Thus, the possible output values are only the integers $0$ and $1$.

Therefore, the range is $\{0, 1\}$.