The number of points where the function
$$f(x) = \begin{cases} |2x^2 - 3x - 7| & \text{if } x \leqslant -1 \\ [4x^2 - 1] & \text{if } -1 < x < 1 \\ |x+1| + |x-2| & \text{if } x \geqslant 1 \end{cases}$$
where $$[t]$$ denotes the greatest integer $$\leqslant t$$, is discontinuous is ______

Given,

$$f(x)= \begin{cases} |2x^2-3x-7|, & x\le-1\\ [4x^2-1], & -1<x<1\\ |x+1|+|x-2|, & x\ge1\end{cases}$$

For $$x\le-1$$ and $$x\ge1,$$ the functions are continuous.

So, discontinuities can occur only:

- inside $$(-1,1)$$ due to GIF
- at $$x=-1,1$$

Now,

$$f(x)=[4x^2-1],\qquad -1<x<1$$

Since

$$-1\le4x^2-1<3$$

possible integer values are

$$-1,0,1,2$$

Case 1:

$$4x^2-1=-1\implies x=0$$

At $$x=0,$$ the GIF does not jump.

Hence, continuous at $$x=0$$.

Case 2:

$$4x^2-1=0\implies x=\pm\frac12$$

Discontinuous at both points.

Case 3:

$$4x^2-1=1\implies x=\pm\frac1{\sqrt2}$$

Discontinuous at both points.

Case 4:

$$4x^2-1=2\implies x=\pm\frac{\sqrt3}{2}$$

Discontinuous at both points.

Thus, discontinuities inside $$(-1,1)$$ are

$$x=\pm\frac12,\ \pm\frac1{\sqrt2},\ \pm\frac{\sqrt3}{2}$$

Total points $$=6$$

Now check boundary points.

At $$x=-1,$$

$$f(-1)=2,\qquad \lim_{x\to-1^-}f(x)=2,\qquad \lim_{x\to-1^+}f(x)=2$$

Hence, continuous at $$x=-1$$.

At $$x=1,$$

$$\lim_{x\to1^-}f(x)=2,\qquad f(1)=3$$

Hence, discontinuous at $$x=1$$.

Therefore, total number of discontinuity points is

$$6+1=7$$

Hence, $$\boxed{7}$$.