Let `f (x)= {{:(x [x] , 0 le x lt 2),( (x-1), 2 le x le 3):}` where [x]= greatest integer less than or equal to x, then:
The number of values of x for ` x in [0,3]` where `f (x)` is dicontnous is:

Options

A. 0
B. 1
C. 2
D. 3

Answer

To determine the number of values of \( x \) in the interval \([0, 3]\) where the function \( f(x) \) is discontinuous, we will analyze the function piecewise.

The function is defined as follows:
\[ f(x) = \begin{cases} x \cdot [x] & \text{if } 0 \leq x < 2 \\ x - 1 & \text{if } 2 \leq x \leq 3 \end{cases} \]
where \([x]\) is the greatest integer function (also known as the floor function).

### Step 1: Identify the intervals and points of interest

1. The first piece \( f(x) = x \cdot [x] \) is valid for \( 0 \leq x < 2 \).
2. The second piece \( f(x) = x - 1 \) is valid for \( 2 \leq x \leq 3 \).
3. The points of interest for discontinuity will be at the boundaries of these intervals and at the points where the greatest integer function changes, which are the integers.

### Step 2: Check for discontinuities at the boundaries and integer points

#### At \( x = 0 \):
- \( f(0) = 0 \cdot [0] = 0 \)
- Right-hand limit as \( x \to 0^+ \):
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x \cdot [x] = 0 \cdot 0 = 0 \]
- Left-hand limit does not exist since \( x \) cannot be less than 0.
- Since \( f(0) = 0 \) and the right-hand limit is also 0, \( f(x) \) is continuous at \( x = 0 \).

#### At \( x = 1 \):
- \( f(1) = 1 \cdot [1] = 1 \)
- Left-hand limit as \( x \to 1^- \):
\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} x \cdot [x] = 1 \cdot 1 = 1 \]
- Right-hand limit as \( x \to 1^+ \):
\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} x \cdot [x] = 1 \cdot 1 = 1 \]
- Both limits equal \( f(1) \), so \( f(x) \) is continuous at \( x = 1 \).

#### At \( x = 2 \):
- \( f(2) = 2 - 1 = 1 \)
- Left-hand limit as \( x \to 2^- \):
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} x \cdot [x] = 2 \cdot 2 = 4 \]
- Right-hand limit as \( x \to 2^+ \):
\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x - 1) = 2 - 1 = 1 \]
- Since the left-hand limit (4) does not equal the right-hand limit (1), \( f(x) \) is discontinuous at \( x = 2 \).

#### At \( x = 3 \):
- \( f(3) = 3 - 1 = 2 \)
- Left-hand limit as \( x \to 3^- \):
\[ \lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (x - 1) = 3 - 1 = 2 \]
- Right-hand limit does not exist since \( x \) cannot be greater than 3.
- Since both limits equal \( f(3) \), \( f(x) \) is continuous at \( x = 3 \).

### Conclusion
The function \( f(x) \) is discontinuous at one point, which is \( x = 2 \).

Thus, the number of values of \( x \) in the interval \([0, 3]\) where \( f(x) \) is discontinuous is **1**.