Number of points of discontinuity of `f(x)={x/5}+[x/2]` in `x in [0,100]` is/are (where [-] denotes greatest integer function and {:} denotes fractional part function)

Options

A. 50
B. 51
C. 52
D. 61

Answer

To find the number of points of discontinuity of the function \( f(x) = \{ \frac{x}{5} \} + \left[ \frac{x}{2} \right] \) in the interval \( x \in [0, 100] \), we will analyze the two components of the function separately: the fractional part function \( \{ \frac{x}{5} \} \) and the greatest integer function \( \left[ \frac{x}{2} \right] \).

### Step 1: Analyze the fractional part function \( \{ \frac{x}{5} \} \)

The fractional part function \( \{ \frac{x}{5} \} \) is discontinuous at points where \( \frac{x}{5} \) is an integer, which occurs at \( x = 5k \) for \( k = 0, 1, 2, \ldots, 20 \) (since \( 5 \times 20 = 100 \)).

**Points of discontinuity for \( \{ \frac{x}{5} \} \):**
- \( x = 0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100 \)
- Total = 21 points

### Step 2: Analyze the greatest integer function \( \left[ \frac{x}{2} \right] \)

The greatest integer function \( \left[ \frac{x}{2} \right] \) is discontinuous at points where \( \frac{x}{2} \) is an integer, which occurs at \( x = 2k \) for \( k = 0, 1, 2, \ldots, 50 \) (since \( 2 \times 50 = 100 \)).

**Points of discontinuity for \( \left[ \frac{x}{2} \right] \):**
- \( x = 0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100 \)
- Total = 51 points

### Step 3: Identify common points of discontinuity

Now we need to find the common points of discontinuity between the two functions. The common points are those values of \( x \) that are multiples of both 5 and 2, which are multiples of 10.

**Common points of discontinuity:**
- \( x = 0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100 \)
- Total = 11 points

### Step 4: Apply the principle of inclusion-exclusion

To find the total number of points of discontinuity of \( f(x) \), we use the principle of inclusion-exclusion:

\[ \text{Total discontinuities} = (\text{Discontinuities of } \{ \frac{x}{5} \}) + (\text{Discontinuities of } \left[ \frac{x}{2} \right]) - (\text{Common discontinuities}) \]

Substituting the values we found:

\[ \text{Total discontinuities} = 21 + 51 - 11 = 61 \]

### Conclusion

Thus, the number of points of discontinuity of \( f(x) \) in the interval \( [0, 100] \) is **61**.

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