Let the given function be $f(x) = [x] \sin(\pi x)$, where $[x]$ denotes the greatest integer function and $k$ is an integer.

We need to find the left-hand derivative of $f(x)$ at $x = k$. The formula for the left-hand derivative of a function $f(x)$ at a point $x=k$ is given by:

$\qquad f'(k^-) = \lim_{h \to 0^-} \frac{f(k+h) - f(k)}{h}$

First, let's evaluate $f(k)$. Since $k$ is an integer, $[k] = k$.

$\qquad f(k) = [k] \sin(\pi k) = k \sin(\pi k)$

For any integer $k$, $\sin(\pi k) = 0$. Therefore,

$\qquad f(k) = k \times 0 = 0$

Next, let's evaluate $f(k+h)$ for $h < 0$ and $h$ approaching $0$. When $h$ is a small negative number (like $-0.001$), $k+h$ is slightly less than $k$. For an integer $k$, the greatest integer less than $k$ but greater than or equal to $k+h$ is $k-1$. So, for $h < 0$ and $h \to 0^-$, $[k+h] = k-1$.

$\qquad f(k+h) = [k+h] \sin(\pi (k+h)) = (k-1) \sin(\pi k + \pi h)$

Now, substitute $f(k+h)$ and $f(k)$ into the limit formula for the left-hand derivative:

$\qquad f'(k^-) = \lim_{h \to 0^-} \frac{(k-1) \sin(\pi k + \pi h) - 0}{h}$

$\qquad f'(k^-) = \lim_{h \to 0^-} \frac{(k-1) \sin(\pi k + \pi h)}{h}$

Using the trigonometric identity $\sin(A+B) = \sin A \cos B + \cos A \sin B$, with $A = \pi k$ and $B = \pi h$:

$\qquad \sin(\pi k + \pi h) = \sin(\pi k) \cos(\pi h) + \cos(\pi k) \sin(\pi h)$

Since $k$ is an integer, $\sin(\pi k) = 0$ and $\cos(\pi k) = (-1)^k$. Substituting these values:

$\qquad \sin(\pi k + \pi h) = 0 \cdot \cos(\pi h) + (-1)^k \sin(\pi h) = (-1)^k \sin(\pi h)$

Substitute this back into the limit expression for $f'(k^-)$:

$\qquad f'(k^-) = \lim_{h \to 0^-} \frac{(k-1) (-1)^k \sin(\pi h)}{h}$

We can pull the terms that do not depend on $h$ out of the limit:

$\qquad f'(k^-) = (k-1) (-1)^k \lim_{h \to 0^-} \frac{\sin(\pi h)}{h}$

We know the standard limit $\lim_{x \to 0} \frac{\sin(ax)}{x} = a$. In this case, $x = h$ and $a = \pi$. The limit as $h$ approaches $0$ from the left is the same as the two-sided limit for this expression.

$\qquad \lim_{h \to 0^-} \frac{\sin(\pi h)}{h} = \pi$

Substitute this limit back into the expression for $f'(k^-)$:

$\qquad f'(k^-) = (k-1) (-1)^k \pi$

Rearranging the terms, we get:

$\qquad f'(k^-) = (-1)^k (k-1) \pi$

Comparing this result with the given options, we find that it matches the first option.

Revision Table: Key Concepts for Left-Hand Derivative

Additional Information: Derivative and Greatest Integer Function

The greatest integer function $f(x) = [x]$ has discontinuities at every integer value. A function can only be differentiable at a point if it is continuous at that point. However, the function $f(x) = [x] \sin(\pi x)$ is interesting because the $\sin(\pi x)$ term is zero at every integer $k$. This makes the function value $f(k) = [k]\sin(\pi k) = k \cdot 0 = 0$.

Let's consider the right-hand derivative at $x=k$ for comparison:

$\qquad f'(k^+) = \lim_{h \to 0^+} \frac{f(k+h) - f(k)}{h}$

For $h > 0$ and $h$ approaching $0$, $k+h$ is slightly greater than $k$. So, $[k+h] = k$.

$\qquad f(k+h) = [k+h] \sin(\pi (k+h)) = k \sin(\pi k + \pi h)$

$\qquad f'(k^+) = \lim_{h \to 0^+} \frac{k \sin(\pi k + \pi h) - 0}{h}$

Using $\sin(\pi k + \pi h) = (-1)^k \sin(\pi h)$ as derived before:

$\qquad f'(k^+) = \lim_{h \to 0^+} \frac{k (-1)^k \sin(\pi h)}{h}$

$\qquad f'(k^+) = k (-1)^k \lim_{h \to 0^+} \frac{\sin(\pi h)}{h}$

$\qquad f'(k^+) = k (-1)^k \pi = (-1)^k k \pi$

Comparing the left-hand derivative $f'(k^-) = (-1)^k (k-1) \pi$ and the right-hand derivative $f'(k^+) = (-1)^k k \pi$, we see that they are generally not equal (unless $k-1 = k$, which is impossible). This confirms that the function is typically not differentiable at integer points, but the left and right derivatives exist and are different.