Let $$[t]$$ denote the greatest integer $$\leq t$$. The number of points where the function $$f(x) = [x]|x^2 - 1| + \sin\frac{\pi}{[x]+3} - [x+1]$$, $$x \in (-2, 2)$$ is not continuous is _________.

Let us remember the definition: for any real number $$t$$ the symbol $$[t]$$ denotes “the greatest integer less than or equal to $$t$$.” The function to be studied is

$$f(x)= [x]\;|x^{2}-1|+\sin\!\left(\dfrac{\pi}{[x]+3}\right)-[x+1],\qquad x\in(-2,2).$$

The only positions where discontinuity can occur are the points at which at least one of the three component expressions $$[x]$$, $$|x^{2}-1|$$ or $$[x+1]$$ changes its algebraic description. The absolute-value term $$|x^{2}-1|$$ is continuous everywhere, so we only need to watch where the greatest-integer symbols change value, i.e. at integer values of $$x$$. Inside the open interval $$(-2,2)$$ those integers are $$x=-1,\;0,\;1.$$

We therefore study the function separately on the four open sub-intervals determined by these integers.

1. Interval $$(-2,-1)$$

Here we have $$[x]=-2.$$ Therefore $$[x]+3=1$$ and $$\sin\!\left(\dfrac{\pi}{[x]+3}\right)=\sin(\pi)=0.$$ Further, $$x+1\in(-1,0)$$ so $$[x+1]=-1.$$ Substituting these three results gives

$$f(x)=(-2)\,|x^{2}-1|+0-(-1)= -2|x^{2}-1|+1.$$

2. Interval $$(-1,0)$$

Now $$[x]=-1,$$ whence $$[x]+3=2$$ and $$\sin\!\left(\dfrac{\pi}{[x]+3}\right)=\sin\!\left(\dfrac{\pi}{2}\right)=1.$$ Also $$x+1\in(0,1)$$ so $$[x+1]=0.$$ Hence

$$f(x)=(-1)\,|x^{2}-1|+1-0= -|x^{2}-1|+1.$$

3. Interval $$(0,1)$$

Here $$[x]=0,$$ giving $$[x]+3=3$$ and $$\sin\!\left(\dfrac{\pi}{[x]+3}\right)=\sin\!\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt3}{2}.$$ Also $$x+1\in(1,2)$$ so $$[x+1]=1.$$ Therefore

$$f(x)=0\cdot|x^{2}-1|+\dfrac{\sqrt3}{2}-1=\dfrac{\sqrt3}{2}-1,$$

a constant on this whole sub-interval.

4. Interval $$(1,2)$$

Now $$[x]=1,$$ so $$[x]+3=4$$ and $$\sin\!\left(\dfrac{\pi}{[x]+3}\right)=\sin\!\left(\dfrac{\pi}{4}\right)=\dfrac{\sqrt2}{2}.$$ Moreover $$x+1\in(2,3)$$ hence $$[x+1]=2.$$ We obtain

$$f(x)=1\cdot|x^{2}-1|+\dfrac{\sqrt2}{2}-2=|x^{2}-1|+\dfrac{\sqrt2}{2}-2.$$

Within each open sub-interval the formula is built only from continuous pieces (constants, the continuous absolute value, and simple arithmetic), so $$f(x)$$ is continuous there. We now examine the three candidate points one by one.

Point $$x=-1$$

Left-hand limit (using the $$(-2,-1)$$ formula):

$$\lim_{x\to-1^-}f(x)= -2|(-1)^{2}-1|+1=-2\cdot0+1=1.$$

Right-hand limit (using the $$(-1,0)$$ formula):

$$\lim_{x\to-1^+}f(x)= -|(-1)^{2}-1|+1=-0+1=1.$$

Value of the function at $$x=-1$$:

$$f(-1)=(-1)\cdot0+\sin\!\left(\dfrac{\pi}{2}\right)-[0]=0+1-0=1.$$

Both limits equal the function value, hence $$f(x)$$ is continuous at $$x=-1.$$

Point $$x=0$$

Left-hand limit (use $$(-1,0)$$):

$$\lim_{x\to0^-}f(x)= -|0^{2}-1|+1=-(1)+1=0.$$

Right-hand limit (use $$(0,1)$$, which is constant):

$$\lim_{x\to0^+}f(x)=\dfrac{\sqrt3}{2}-1.$$

Since $$0\neq\dfrac{\sqrt3}{2}-1,$$ the two one-sided limits are different, so the function jumps here. Consequently $$f(x)$$ is not continuous at $$x=0$$ (even though $$f(0)=\dfrac{\sqrt3}{2}-1$$ equals the right limit).

Point $$x=1$$

Left-hand limit (use $$(0,1)$$, constant):

$$\lim_{x\to1^-}f(x)=\dfrac{\sqrt3}{2}-1.$$

Right-hand limit (use $$(1,2)$$):

$$\lim_{x\to1^+}f(x)=|1^{2}-1|+\dfrac{\sqrt2}{2}-2=0+\dfrac{\sqrt2}{2}-2.$$

Because $$\dfrac{\sqrt3}{2}-1\neq\dfrac{\sqrt2}{2}-2,$$ the one-sided limits differ, giving another jump. Hence $$f(x)$$ is discontinuous at $$x=1$$ (again the actual value $$f(1)=\dfrac{\sqrt2}{2}-2$$ matches only the right limit).

There are no other integers inside the open interval $$(-2,2),$$ so the complete list of discontinuities is $$\{0,\,1\}.$$ Thus the number of points of discontinuity is $$2.$$

So, the answer is $$2.$$