Let $$f(x) = x \cdot \left[\frac{x}{2}\right]$$, for $$-10 < x < 10$$, where $$[t]$$ denotes the greatest integer function. Then the number of points of discontinuity of $$f(x)$$ is equal to

We are given the function

$$f(x)=x\,[\,\frac{x}{2}\,],\qquad -10<x<10,$$

where $$[t]$$ denotes the greatest-integer (floor) function. Our task is to count the points inside the open interval $$(-10,10)$$ where $$f(x)$$ is not continuous.

First recall a basic fact about the greatest-integer function:

Whenever the argument of $$[\,\cdot\,]$$ crosses an integer, the value of the greatest-integer function jumps by $$1$$. In symbols, $$[t]$$ is discontinuous exactly at all integer values of $$t$$.

In our expression the argument of the floor function is $$\dfrac{x}{2}$$, so the jump (and therefore the possible discontinuity) occurs whenever

$$\frac{x}{2}=k\quad\text{with }k\in\mathbb Z.$$

Solving for $$x$$ we obtain

$$x=2k,\quad k\in\mathbb Z.$$

Thus the only potential points of discontinuity for $$f(x)$$ in the interval $$(-10,10)$$ are the even integers lying strictly between $$-10$$ and $$10$$:

$$x=-8,-6,-4,-2,0,2,4,6,8.$$

There are nine such points. We must, however, check each of them because the product by the factor $$x$$ might cancel the jump at some specific point.

Let us analyse each candidate point.

1. The point $$x=0$$.

For $$x>0$$ and sufficiently close to $$0$$ we have $$0<x<2$$, so $$\dfrac{x}{2}\in(0,1)$$ and hence $$\bigl[\dfrac{x}{2}\bigr]=0$$. Therefore

$$f(x)=x\cdot 0=0 \quad\text{for }0<x<2.$$

For $$x<0$$ and sufficiently close to $$0$$ we have $$-2<x<0$$, so $$\dfrac{x}{2}\in(-1,0)$$ and hence $$\bigl[\dfrac{x}{2}\bigr]=-1$$. Therefore

$$f(x)=x\cdot(-1)=-x \quad\text{for }-2<x<0.$$

Taking the limit from the right:

$$\lim_{x\to0^{+}}f(x)=\lim_{x\to0^{+}}0=0.$$

Taking the limit from the left:

$$\lim_{x\to0^{-}}f(x)=\lim_{x\to0^{-}}(-x)=0.$$

Both one-sided limits are equal and equal to zero. The actual value of the function at $$x=0$$ is

$$f(0)=0\cdot\Bigl[\frac{0}{2}\Bigr]=0\cdot0=0.$$

Hence the function is continuous at $$x=0$$. Therefore $$x=0$$ does not contribute to the count of discontinuities.

2. The other even integers: $$x=\pm2,\pm4,\pm6,\pm8$$.

We illustrate the reasoning for a representative point, say $$x=2k$$ with $$k\neq0$$, because the algebra is identical for every such $$k$$.

Pick $$x_{0}=2k$$ with $$k\neq0$$. Then $$\dfrac{x_{0}}{2}=k$$ is an integer. Setting $$x=x_{0}+h$$ with a tiny positive $$h$$ (approach from the right) we have

$$\frac{x}{2}=k+\frac{h}{2},\qquad 0<h\ll1,$$

so

$$\Bigl[\frac{x}{2}\Bigr]=k \quad\text{for }x\to x_{0}^{+}.$$

Therefore the right-hand limit is

$$\lim_{x\to x_{0}^{+}}f(x)=\lim_{h\to0^{+}}(x_{0}+h)\,k \;=\;x_{0}\,k.$$

Now approach from the left, i.e. take $$x=x_{0}-h$$ with $$0<h\ll1$$:

$$\frac{x}{2}=k-\frac{h}{2}\quad\Longrightarrow\quad \Bigl[\frac{x}{2}\Bigr]=k-1.$$

Thus the left-hand limit is

$$\lim_{x\to x_{0}^{-}}f(x)=\lim_{h\to0^{+}}(x_{0}-h)\,(k-1) \;=\;x_{0}\,(k-1).$$

Compute the difference between the two one-sided limits:

$$x_{0}\,k - x_{0}\,(k-1)=x_{0}.$$

Since $$x_{0}=2k\neq0$$, this difference is non-zero. Therefore the two one-sided limits are different, which means the function is discontinuous at $$x_{0}=2k.$$

The same calculation applies verbatim for every non-zero even integer in our list. Consequently $$f(x)$$ is discontinuous at each of

$$x=-8,-6,-4,-2,2,4,6,8.$$

Counting the discontinuities

We found discontinuities at eight points and confirmed continuity at $$x=0$$. No other points inside $$(-10,10)$$ cause trouble, so the total number of discontinuities of $$f(x)$$ in the given interval is

$$8.$$

So, the answer is $$8$$.