Let `f(x)={{:(8^((1)/(x))",",xlt0,),(a[x]",",a inR-{0}",",xge0):}` (where [.] denotes the greatest integer function).
Then f(x) is

Options

A. continuous only at a finite number of points.
B. discontinuous at a finite number of points.
C. discontinuous at an infinite number of points.
D. discontinuous at x = 0.

Explanation

To determine the continuity of the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} 8^{\frac{1}{x}} & \text{if } x < 0 \\ a \cdot [x] & \text{if } x \geq 0 \end{cases} \]

where \( [x] \) denotes the greatest integer function, we need to analyze the function at the point \( x = 0 \) and at other integer points for \( x \geq 0 \).

### Step 1: Check continuity at \( x = 0 \)

To check the continuity at \( x = 0 \), we need to evaluate the left-hand limit, right-hand limit, and the function value at \( x = 0 \).

1. **Left-hand limit as \( x \to 0^- \)**:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 8^{\frac{1}{x}} \]
As \( x \) approaches \( 0 \) from the left, \( \frac{1}{x} \) approaches \( -\infty \), thus:
\[ 8^{\frac{1}{x}} \to 0 \]
So, \( \lim_{x \to 0^-} f(x) = 0 \).

2. **Right-hand limit as \( x \to 0^+ \)**:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} a \cdot [x] \]
Since \( [x] = 0 \) for \( x \) in the interval \( [0, 1) \):
\[ \lim_{x \to 0^+} f(x) = a \cdot 0 = 0 \]

3. **Function value at \( x = 0 \)**:
\[ f(0) = a \cdot [0] = a \cdot 0 = 0 \]

Since all three values are equal:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 0 \]
Thus, \( f(x) \) is continuous at \( x = 0 \).

### Step 2: Check continuity at integer points \( x = n \) where \( n \in \mathbb{Z}^+ \)

Now, we will check the continuity at \( x = 1 \) and \( x = 2 \).

1. **At \( x = 1 \)**:
- **Left-hand limit**:
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} a \cdot [x] = a \cdot [1] = a \cdot 0 = 0 \]
- **Right-hand limit**:
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} a \cdot [x] = a \cdot [1] = a \cdot 1 = a \]
Since \( 0 \neq a \), \( f(x) \) is discontinuous at \( x = 1 \).

2. **At \( x = 2 \)**:
- **Left-hand limit**:
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} a \cdot [x] = a \cdot [2] = a \cdot 1 = a \]
- **Right-hand limit**:
\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} a \cdot [x] = a \cdot [2] = a \cdot 2 = 2a \]
Since \( a \neq 2a \) (unless \( a = 0 \)), \( f(x) \) is discontinuous at \( x = 2 \).

### Conclusion

The function \( f(x) \) is continuous at \( x = 0 \) but discontinuous at every positive integer \( n \) where \( n \in \mathbb{Z}^+ \).