JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Continuity and Differentiability

The factor $x$ is continuous everywhere, so $f(x)=x\left[\frac{x}{2}\right]$ can break only where $\left[\frac{x}{2}\right]$ jumps, namely where $\frac{x}{2}$ is an integer, that is at even values of $x$. On $(-10,10)$ these are $-8,-6,-4,-2,0,2,4,6,8$. At $x=0$ the jump in $\left[\frac{x}{2}\right]$ is multiplied by $x=0$, so $f$ remains continuous there, while at the other eight even points it is genuinely discontinuous. That leaves $8$ points of discontinuity.