Given function

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

where $[t]$ denotes the greatest integer (floor) function.

When can discontinuity occur?

The factor $x$ is continuous everywhere. So discontinuities can only come from the floor term $\left[\frac{x}{2}\right].$

The floor function $[u]$ is discontinuous when $u$ is an integer.

Hence $\left[\frac{x}{2}\right]$ is discontinuous when $\frac{x}{2}\in \mathbb{Z} \quad \Longrightarrow \quad x=2n,$ where $n$ is an integer.

Possible discontinuity points inside the interval

Since $-10<x<10$, the even integers in this interval are $-8,-6,-4,-2,0,2,4,6,8.$

So these are the only possible discontinuity points.

Check whether all of them actually give discontinuity in $f(x)$

Let $x=2n$.

Then near $x=2n$,

• for $x\to (2n)^-$, $\left[\frac{x}{2}\right]=n-1,$
• for $x\to (2n)^+$, $\left[\frac{x}{2}\right]=n.$

Therefore, $\lim_{x\to (2n)^-} f(x)=2n(n-1),$ $\lim_{x\to (2n)^+} f(x)=2n\cdot n=2n^2.$

These are equal only if $2n(n-1)=2n^2 \implies -2n=0 \implies n=0.$

So at $x=0$, the left and right limits are equal.

Indeed, $f(0)=0\left[0\right]=0,$ and both one-sided limits are also $0$, so $f$ is continuous at $x=0$.

For every other even integer $x=2n$ with $n\neq 0$, the left and right limits are different, so $f$ is discontinuous there.

Count the discontinuity points

From the 9 candidate points $-8,-6,-4,-2,0,2,4,6,8,$ only $x=0$ is continuous.

Hence number of discontinuities $=9-1=8.$

Final answer

$\boxed{8}$