Answer
$f(x)$ is not continuous for any value of $x$.
Work Step by Step
\[F(x)=\left\{\begin{array}{ll}f(x)\;\;\;\;\;\; ,if \; x\; is \; rational\\
g(x)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;,if \; x\; is \; irrational\end{array}\right. \]
$\Rightarrow F(x)$ is continuous at those values of $x$ for which $f(x)=g(x)$.
Here,
\[f(x)=\left\{\begin{array}{ll}0\;\;\;\;\;\; ,if \; x\; is \; rational\\
1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;,if \; x\; is \; irrational\end{array}\right. \]
$\Rightarrow f(x)$ is continuous at those values of $x$ for which $1=0$
Which is not possible.
Hence $f(x)$ is not continuous for any value of $x$.
