Let $f(x)=x \cdot\left[\frac{x}{2}\right]$, for $-10

Question

Let $f(x)=x \cdot\left[\frac{x}{2}\right]$, for $-10<x<10$, where $[t]$ denotes the greatest integer function. Then the number of points of discontinuity of $f$ is equal to

Professor PC Thomas Classes · Answer

(8) : We have, $x \in(-10,10)$\ $\Rightarrow \quad \frac{x}{2} \in(-5,5)$\ $\Rightarrow\left[\frac{x}{2}ight]$ can be $-4,-3,-2,-1,0,1,2,3,4$\ Also, we have $f(x)=x .\left[\frac{x}{2}ight]$\ Now, at $x=0, f(0)=f\left(0^{+}ight)=f\left(0^{-}ight)$\ $\Rightarrow \quad f(x)$ is continuous at $x=0$.\ Hence, $f(x)$ will be discontinuous at integral point i.e., $\pm 4, \pm 3, \pm 2, \pm 1$ i.e., at 8 points $f(x)$ will be discontinuous.\