JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Continuity and Differentiability

Since $1$ is an integer, $[x+1]=[x]+1$, and $|x^2-1|$ is continuous, so discontinuities can only occur where $[x]$ jumps: at $x=-1,0,1$. On each interval with $[x]=k$, At $x=-1$ the factor $|x^2-1|\to 0$ and the constant parts from the left ($k=-2$) and right ($k=-1$) both equal $1$, so $f$ is continuous there. At $x=0$ the left value is $-|0-1|+1=0$ while the right value is $\sin \frac{\pi }{3}-1$, so $f$ jumps. At $x=1$ the left value is $\sin \frac{\pi }{3}-1$ while the right value is $\sin \frac{\pi }{4}-2$, so $f$ jumps. Thus $f$ is discontinuous at $2$ points.