Question

Let $\left[y\right]$ and $\left\{y\right\}$ denote the greatest integer less than or equal to $y$ and fractional part of $y$ respectively. Then the number of points of discontinuity of the function $f\left(x\right)=\left[5x\right]+\left\{3x\right\}$ in $[0,5]$ is

Answer 1

$f\left(x\right)=\left[5x\right]-\left[3x\right]+3x$
We know that the greatest integer function $\left[x\right]$ is discontinuous at integral points.
So, we have to check those points where $\left[5x\right]$ and $\left[3x\right]$ take integral values.
$\left[5x\right]$ is discontinuous in $[0,5]$ at
$x=\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5},\frac{5}{5},\frac{6}{5},\cdots ,\frac{24}{5},\frac{25}{5}$
and $\left[3x\right]$ is discontinuous in $[0,5]$ at
$x=\frac{1}{3},\frac{2}{3},\frac{3}{3},\frac{4}{3},\frac{5}{3},\frac{6}{3},\cdots ,\frac{14}{3},\frac{15}{3}$
Common points are $1,2,3,4,5.$

At integral point $x=n,$
$L.H.L.=\underset{h\to 0}{lim}\left(\right[5(n-h)]-[3(n-h)]+3(n-h\left)\right)$
$=(5n-1)-(3n-1)+3n=5n$
$R.H.L.=\underset{h\to 0}{lim}\left(\right[5(n+h)]-[3(n+h)]+3(n+h\left)\right)$
$=5n-3n+3n=5n$
$f\left(n\right)=5n$
So, $f\left(x\right)$ is continuous at $x=n$
This means $f\left(x\right)$ is continuous at $x=1,2,3,4,5$
$\therefore$ Total points of discontinuity $=20+10=30$