Question

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

• A. 49
• B. 50
• C. 51
• D. 61

Answer 1

Answer: (B)

50

Answer 2

The function is $f(x) = \{\frac{x}{5}\} + [\frac{x}{2}]$.

The fractional part function $\{\frac{x}{5}\}$ is discontinuous when $\frac{x}{5}$ is an integer, i.e., $x = 5k$ for $k \in \mathbb{Z}$. In $[0, 100]$, these points are $\{0, 5, 10, \dots, 100\}$, which are $21$ points.

The greatest integer function $[\frac{x}{2}]$ is discontinuous when $\frac{x}{2}$ is an integer, i.e., $x = 2m$ for $m \in \mathbb{Z}$. In $[0, 100]$, these points are $\{0, 2, 4, \dots, 100\}$, which are $51$ points.

The potential points of discontinuity for $f(x)$ are the union of these two sets: $D = \{x \in [0, 100] \mid \frac{x}{5} \in \mathbb{Z}\} \cup \{x \in [0, 100] \mid \frac{x}{2} \in \mathbb{Z}\}$.

Let's analyze the continuity at points $x = 5k$ for $k \in \{1, 2, \dots, 19\}$. If $k$ is odd ($k=2j+1$), $x = 10j+5$. $f(10j+5) = \{2j+1\} + [\frac{10j+5}{2}] = 0 + [5j+2.5] = 5j+2$. $\lim_{x \to (10j+5)^-} f(x) = \lim_{x \to (10j+5)^-} \{\frac{x}{5}\} + \lim_{x \to (10j+5)^-} [\frac{x}{2}] = 1 + [\frac{10j+5-\epsilon}{2}] = 1 + [5j+2.5-\epsilon] = 1 + 5j+2 = 5j+3$. Since $f(10j+5) \neq \lim_{x \to (10j+5)^-} f(x)$, these $10$ points ($5, 15, \dots, 95$) are discontinuities.

If $k$ is even ($k=2j$), $x = 10j$. $f(10j) = \{2j\} + [\frac{10j}{2}] = 0 + 5j = 5j$. $\lim_{x \to (10j)^-} f(x) = 1 + [\frac{10j-\epsilon}{2}] = 1 + [5j-\epsilon] = 1 + 5j-1 = 5j$. $\lim_{x \to (10j)^+} f(x) = 0 + [\frac{10j+\epsilon}{2}] = 0 + [5j+\epsilon] = 5j$. These $10$ points ($10, 20, \dots, 100$) are points of continuity.

Now consider points $x = 2m$ where $x$ is not a multiple of 5. These are even numbers not ending in 0. For $x=2m$, where $m$ is not a multiple of 5: $f(2m) = \{\frac{2m}{5}\} + [\frac{2m}{2}] = \{\frac{2m}{5}\} + m$. $\lim_{x \to (2m)^-} f(x) = \lim_{x \to (2m)^-} \{\frac{x}{5}\} + \lim_{x \to (2m)^-} [\frac{x}{2}] = \{\frac{2m}{5}\} + [m-\epsilon] = \{\frac{2m}{5}\} + m-1$. Since $f(2m) \neq \lim_{x \to (2m)^-} f(x)$, these points are discontinuities. The even numbers in $(0, 100)$ are $2, 4, \dots, 98$ (49 points). The multiples of 10 in $(0, 100)$ are $10, 20, \dots, 90$ (9 points). The number of even numbers not divisible by 5 is $49 - 9 = 40$. These are discontinuities.

The endpoints $x=0$ and $x=100$ are points of continuity.

Total number of discontinuities = (discontinuities at $x=10j+5$) + (discontinuities at $x=2m$ not divisible by 5) Total = $10 + 40 = 50$.