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(A) The function is defined as $f(x)=[x]|x^{2}-1|+\sin \left(\frac{\pi}{[x]+3}\right)-[x+1]$ for $x \in (-2, 2)$.
We analyze the points of discontinu

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$3$

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(A) The function is defined as $f(x)=[x]|x^{2}-1|+\sin \left(\frac{\pi}{[x]+3}\right)-[x+1]$ for $x \in (-2, 2)$.
We analyze the points of discontinuity by checking the behavior of $[x]$ and $[x+1]$ at integer values $x \in \{-1, 0, 1\}$.
For $x \in (-2, -1)$,$[x] = -2$ and $[x+1] = -1$. Thus,$f(x) = -2|x^2-1| + \sin(\pi/1) - (-1) = -2|x^2-1| + 1$.
For $x \in [-1, 0)$,$[x] = -1$ and $[x+1] = 0$. Thus,$f(x) = -1|x^2-1| + \sin(\pi/2) - 0 = -|x^2-1| + 1$.
For $x \in [0, 1)$,$[x] = 0$ and $[x+1] = 1$. Thus,$f(x) = 0|x^2-1| + \sin(\pi/3) - 1 = \frac{\sqrt{3}}{2} - 1$.
For $x \in [1, 2)$,$[x] = 1$ and $[x+1] = 2$. Thus,$f(x) = 1|x^2-1| + \sin(\pi/4) - 2 = |x^2-1| + \frac{1}{\sqrt{2}} - 2$.
The function is discontinuous at points where the floor functions change values,which are $x = -1, 0, 1$. Checking the limits at these points confirms discontinuity. Therefore,there are $3$ points of discontinuity.

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

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