1. Given function

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

We need the number of points in $(-2,2)$ where $f(x)$ is not continuous.

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2. Where can discontinuity occur?

The functions $|x^2-1|$ and $\sin(\cdot)$ are continuous wherever defined.

The only possible discontinuities come from the greatest integer terms:

• $[x]$ changes value at integers,
• $[x+1]$ also changes value at integers.

In the interval $(-2,2)$, the integers are:

$-1,\ 0,\ 1$

So we only need to check continuity at $x=-1,0,1$.

For non-integer $x$, both $[x]$ and $[x+1]$ are constant on each interval, so $f$ is continuous there.

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3. Compute $f(x)$ on each interval

(i) For $x\in(-2,-1)$

Then

$[x]=-2, \qquad [x+1]=-1$

Hence

$$f(x)=-2|x^2-1|+\sin\left(\frac{\pi}{-2+3}\right)-(-1)$$ $$= -2|x^2-1|+\sin(\pi)+1$$ $$= -2|x^2-1|+1$$

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(ii) For $x\in(-1,0)$

Then

$[x]=-1, \qquad [x+1]=0$

So

$$f(x)=-|x^2-1|+\sin\left(\frac{\pi}{-1+3}\right)-0$$ $$= -|x^2-1|+\sin\left(\frac{\pi}{2}\right)$$ $$= -|x^2-1|+1$$

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(iii) For $x\in(0,1)$

Then

$[x]=0, \qquad [x+1]=1$

Thus

$$f(x)=0\cdot |x^2-1|+\sin\left(\frac{\pi}{3}\right)-1$$ $$= \frac{\sqrt3}{2}-1$$

So it is constant on $(0,1)$.

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(iv) For $x\in(1,2)$

Then

$[x]=1, \qquad [x+1]=2$

So

$$f(x)=|x^2-1|+\sin\left(\frac{\pi}{4}\right)-2$$ $$= |x^2-1|+\frac{\sqrt2}{2}-2$$

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4. Check continuity at the candidate points

At $x=-1$

First compute the function value:

$[-1]=-1, \qquad [-1+1]=[0]=0$

So

$$f(-1)=(-1)|(-1)^2-1|+\sin\left(\frac{\pi}{2}\right)-0$$ $$= (-1)\cdot 0+1=1$$

Left-hand limit

From $(-2,-1)$:

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

since $|x^2-1|\to 0$.

Right-hand limit

From $(-1,0)$:

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

Thus

$$\lim_{x\to-1^-}f(x)=\lim_{x\to-1^+}f(x)=f(-1)=1$$

So $f$ is continuous at $x=-1$.

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At $x=0$

Function value:

$[0]=0, \qquad [1]=1$

Hence

$$f(0)=0\cdot|0^2-1|+\sin\left(\frac{\pi}{3}\right)-1=\frac{\sqrt3}{2}-1$$

Left-hand limit

From $(-1,0)$:

For $x\in(-1,0)$, $x^2<1$, so

$|x^2-1|=1-x^2$

Therefore

$$f(x)=-(1-x^2)+1=x^2$$

So

$$\lim_{x\to 0^-}f(x)=0$$

Right-hand limit

From $(0,1)$:

$$\lim_{x\to 0^+}f(x)=\frac{\sqrt3}{2}-1$$

Since

$$0 \ne \frac{\sqrt3}{2}-1,$$

$f$ is not continuous at $x=0$.

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At $x=1$

Function value:

$[1]=1, \qquad [2]=2$

So

$$f(1)=1\cdot |1^2-1|+\sin\left(\frac{\pi}{4}\right)-2=\frac{\sqrt2}{2}-2$$

Left-hand limit

From $(0,1)$:

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

Right-hand limit

From $(1,2)$:

As $x\to1^+$, $|x^2-1|\to0$, so

$$\lim_{x\to1^+}f(x)=\frac{\sqrt2}{2}-2$$

These are unequal, hence $f$ is not continuous at $x=1$.

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5. Conclusion

The function is discontinuous at exactly:

$x=0,\ 1$

So the number of points of discontinuity is

$\boxed{2}$

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6. Comparison with stored answer

Stored correct answer: $2$

Our derived answer: $2$

They match.