The left hand derivative of `f(x)=[x]sin(pix)` at `x = k`, `k in Z`, is

Options

A. `(-1)^(k) (k - 1)pi`
B. `(-1)^(k-1)(k-1)pi`
C. `(-1)^(k) kpi`
D. `(-1)^(k-1) kpi`

Answer

To find the left-hand derivative of the function \( f(x) = \lfloor x \rfloor \sin(\pi x) \) at \( x = k \), where \( k \) is an integer, we will follow these steps:

### Step 1: Define the left-hand derivative
The left-hand derivative (LHD) of a function at a point is defined as:
\[ f'_-(k) = \lim_{h \to 0^-} \frac{f(k + h) - f(k)}{h} \]

### Step 2: Substitute the function into the limit
For our function, we have:
\[ f(k) = \lfloor k \rfloor \sin(\pi k) = k \cdot 0 = 0 \quad (\text{since } \sin(\pi k) = 0 \text{ for integer } k) \]
Now, we need to evaluate \( f(k + h) \):
\[ f(k + h) = \lfloor k + h \rfloor \sin(\pi (k + h)) \]
Since \( h \) is approaching 0 from the left (negative side), \( \lfloor k + h \rfloor = k - 1 \) when \( h \) is a small negative number.

### Step 3: Rewrite the limit expression
Now, substituting into the limit expression:
\[ f'_-(k) = \lim_{h \to 0^-} \frac{(k - 1) \sin(\pi (k + h)) - 0}{h} = \lim_{h \to 0^-} \frac{(k - 1) \sin(\pi (k + h))}{h} \]

### Step 4: Simplify the sine term
Using the sine addition formula:
\[ \sin(\pi (k + h)) = \sin(\pi k + \pi h) = \sin(\pi k) \cos(\pi h) + \cos(\pi k) \sin(\pi h) \]
Since \( \sin(\pi k) = 0 \) (for integer \( k \)), we have:
\[ \sin(\pi (k + h)) = \cos(\pi k) \sin(\pi h) \]
And since \( \cos(\pi k) = (-1)^k \):
\[ \sin(\pi (k + h)) = (-1)^k \sin(\pi h) \]

### Step 5: Substitute back into the limit
Now substituting back into our limit:
\[ f'_-(k) = \lim_{h \to 0^-} \frac{(k - 1)(-1)^k \sin(\pi h)}{h} \]

### Step 6: Apply the limit property
Using the property \( \lim_{h \to 0} \frac{\sin(h)}{h} = 1 \):
\[ f'_-(k) = (k - 1)(-1)^k \cdot \pi \cdot 1 = (k - 1)(-1)^k \pi \]

### Final Answer
Thus, the left-hand derivative of \( f(x) \) at \( x = k \) is:
\[ f'_-(k) = (k - 1)(-1)^k \pi \]

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