Let $[x]$ be the greatest integer less than or equals to $x$. Then, at which of the following point(s) the function $f(x) = x \cos(\pi(x + [x]))$ is discontinuous?

$[x] = \lfloor x \rfloor = \text{greatest integer } \le x$

$f(x) = x\cos(\pi(x+[x]))$

$\text{test } x = -1,\ 0,\ 1,\ 2$

$\text{To find: point(s) of discontinuity}$

$[x] \text{ jumps only at integers; } x,\ \cos \text{ are continuous everywhere}$ $1.\ f(n^+),\ f(n^-)$ $2.\ f(n)$ $3.\ \text{compare all three}$ $x=n+t\ (t\to0^+):\ [x]=n$ $\cos(\pi(2n+t)) = \cos(2\pi n + \pi t) = \cos(\pi t) \to 1$ $f(n^+) = n\cdot 1 = n$ $x=n-t\ (t\to0^+):\ [x]=n-1$ $\cos(\pi(2n-1-t)) = \cos(\pi+\pi t) = -\cos(\pi t) \to -1$ $f(n^-) = n\cdot(-1) = -n$ $f(n) = n\cos(2\pi n) = n$ $f(n^-)=f(n^+)=f(n) \iff -n = n \iff n=0$ $\text{continuous only at } x=0;\ \text{discontinuous at all other integers}$ $x=-1:\ \text{disc.};\quad x=0:\ \text{cont.};\quad x=1:\ \text{disc.};\quad x=2:\ \text{disc.}$ $\boxed{\text{discontinuous at } x=-1,\ 1,\ 2 \Rightarrow (1),(3),(4)}$