Question
$$f$$ is defined in $$\left[ { - 5,\,5} \right]$$ as $$f\left( x \right) = x$$ if $$x$$ is rational $$ = - x$$ if $$x$$ is irrational. Then-
A.
$$f\left( x \right)$$ is continuous at every $$x,$$ except $$x =0$$
B.
$$f\left( x \right)$$ is discontinuous at every $$x,$$ except $$x =0$$
C.
$$f\left( x \right)$$ is continuous everywhere
D.
$$f\left( x \right)$$ is discontinuous everywhere
Answer :
$$f\left( x \right)$$ is discontinuous at every $$x,$$ except $$x =0$$
Solution :
Let $$a$$ is a rational number other than 0, in $$\left[ { - 5,\,5} \right],$$ then
$$f\left( a \right) = a\,\,\,{\text{and}}\,\,\,\,\mathop {\lim }\limits_{x \to a} f\left( x \right) = - a$$
[As in the immediate neighborhood of a rational number, we find irrational numbers]
$$\therefore f\left( x \right)$$ is not continuous at any rational number
If $$a$$ is irrational number, then
$$f\left( a \right) = - a\,\,\,{\text{and}}\,\,\,\,\mathop {\lim }\limits_{x \to a} f\left( x \right) = a$$
$$\therefore f\left( x \right)$$ is not continuous at any irrational number clearly
$$\mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right) = 0$$
$$\therefore f\left( x \right)$$ is continuous at $$x =0$$