If `f(x)'{{:(sin((a-x)/(2))tan[(pix)/(2a)],"for",xgta),(([cos((pix)/(2a))])/(a-x),"for",x lta):}`
(where [x] is the greatest integer function of x) and a gt 0, then

Options

A. `f(a^(-))lt0`
B. f has a removable discontinuity at x = a
C. f has an irremovable discontinuity at x = a
D. `f(a^(+))lt0`

Explanation

To solve the problem, we need to analyze the function \( f(x) \) defined in two parts depending on the value of \( x \). The function is given as:

\[ f(x) = \begin{cases} \sin\left(\frac{a-x}{2}\right) \tan\left(\frac{\pi x}{2a}\right) & \text{for } x > a \\ \frac{\cos\left(\frac{\pi x}{2a}\right)}{a-x} & \text{for } x < a \end{cases} \]

We need to check the continuity of \( f(x) \) at \( x = a \).

### Step 1: Find \( f(a^-) \) (the left-hand limit as \( x \) approaches \( a \))

For \( x < a \), we use the second part of the function:

\[ f(a^-) = \lim_{h \to 0} f(a - h) = \lim_{h \to 0} \frac{\cos\left(\frac{\pi (a-h)}{2a}\right)}{a - (a - h)} = \lim_{h \to 0} \frac{\cos\left(\frac{\pi (a-h)}{2a}\right)}{h} \]

As \( h \to 0 \):

\[ f(a^-) = \lim_{h \to 0} \frac{\cos\left(\frac{\pi a}{2a} - \frac{\pi h}{2a}\right)}{h} = \lim_{h \to 0} \frac{\cos\left(\frac{\pi}{2} - \frac{\pi h}{2a}\right)}{h} \]

Using the fact that \( \cos\left(\frac{\pi}{2} - x\right) = \sin(x) \):

\[ f(a^-) = \lim_{h \to 0} \frac{\sin\left(\frac{\pi h}{2a}\right)}{h} \]

Using the limit property \( \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \):

\[ f(a^-) = \lim_{h \to 0} \frac{\sin\left(\frac{\pi h}{2a}\right)}{h} = \lim_{h \to 0} \frac{\frac{\pi h}{2a}}{h} = \frac{\pi}{2a} \]

### Step 2: Find \( f(a^+) \) (the right-hand limit as \( x \) approaches \( a \))

For \( x > a \), we use the first part of the function:

\[ f(a^+) = \lim_{h \to 0} f(a + h) = \lim_{h \to 0} \sin\left(\frac{a - (a + h)}{2}\right) \tan\left(\frac{\pi (a + h)}{2a}\right) \]

This simplifies to:

\[ f(a^+) = \lim_{h \to 0} \sin\left(-\frac{h}{2}\right) \tan\left(\frac{\pi a}{2a} + \frac{\pi h}{2a}\right) \]

Using \( \sin(-x) = -\sin(x) \):

\[ f(a^+) = -\lim_{h \to 0} \sin\left(\frac{h}{2}\right) \tan\left(\frac{\pi}{2} + \frac{\pi h}{2a}\right) \]

Since \( \tan\left(\frac{\pi}{2} + x\right) = -\cot(x) \):

\[ f(a^+) = -\lim_{h \to 0} \sin\left(\frac{h}{2}\right) \left(-\cot\left(\frac{\pi h}{2a}\right)\right) = \lim_{h \to 0} \sin\left(\frac{h}{2}\right) \cot\left(\frac{\pi h}{2a}\right) \]

Using \( \cot(x) = \frac{\cos(x)}{\sin(x)} \):

\[ f(a^+) = \lim_{h \to 0} \sin\left(\frac{h}{2}\right) \cdot \frac{\cos\left(\frac{\pi h}{2a}\right)}{\sin\left(\frac{\pi h}{2a}\right)} \]

As \( h \to 0 \):

\[ f(a^+) = \lim_{h \to 0} \frac{\sin\left(\frac{h}{2}\right)}{\frac{\pi h}{2a}} \cdot \cos\left(\frac{\pi h}{2a}\right) \cdot \frac{\pi h}{2a} \]

Using the limit property again:

\[ f(a^+) = \frac{2a}{\pi} \cdot \cos(0) = \frac{2a}{\pi} \]

### Step 3: Check for continuity at \( x = a \)

For continuity at \( x = a \), we need \( f(a^-) = f(a^+) \):

\[ \frac{\pi}{2a} \neq \frac{2a}{\pi} \]

Thus, \( f(a^-) \neq f(a^+) \), indicating that \( f(x) \) has a removable discontinuity at \( x = a \).

### Conclusion

The function \( f(x) \) has a removable discontinuity at \( x = a \).