1. Given function

The function is defined on $[-5,5]$ by

We must check continuity at different values of $x$.

2. Idea for continuity

A function is continuous at $x=a$ if

Here, rational and irrational numbers are both dense in $\mathbb{R}$. So when $x\to a$, we can approach $a$ through:

• rational values of $x$, giving $f(x)=x$,
• irrational values of $x$, giving $f(x)=-x$.

If these two approach-values are different, then the limit does not exist, so the function is discontinuous.

3. Check continuity at $x=a\neq 0$

Take any fixed $a\neq 0$.

• Along rational sequence $x_n\to a$,

  $$f(x_n)=x_n \to a.$$

• Along irrational sequence $y_n\to a$,

  $$f(y_n)=-y_n \to -a.$$

Thus the two limiting values are:

Since $a\neq 0$, we have

So the limit $\lim_{x\to a} f(x)$ does not exist.

Hence $f$ is discontinuous at every $a\neq 0$.

4. Check continuity at $x=0$

Now consider $a=0$.

If $x$ is rational, then

If $x$ is irrational, then

In either case,

So as $x\to 0$,

which gives

Also,

because $0$ is rational, so from the definition $f(0)=0$.

Therefore,

so $f$ is continuous at $x=0$.

5. Conclusion

The function is:

• continuous at $x=0$,
• discontinuous at every $x\neq 0$.

So the correct option is

$f(x)$ is discontinuous at every $x$, except $x=0$.

6. Comparison with stored answer

Stored correct answer: $\text{B}$

Our derived answer is also $\text{B}$, so they agree.