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$f$ is defined in [-5,5] as $f(x)=\left\{\begin{array}{1 1}x&if\;x\;is\;rational\\-x&if\;x\;is\;irrational\end{array}\right.$. Then

$\begin{array}{1 1}(A)\;\text{f(x) is continuous at every x,except x=0}\\(B)\;\text{f(x) is discontinuous at every x,except x=0}\\(C)\;\text{f(x) is continuous everywhere}\\(D)\;\text{f(x) is discontinuous everywhere}\end{array}$

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Let $a$ is a rational number other than 0 in [-5,5] then
$f(a)=a$ and $\lim\limits_{x\to a}f(x)=-a$
As in the immediate neighborhood of a rational number,we find irrational numbers.
$\therefore f(x)$ is not continuous at any rational number.
If $a$ is irrational number,then
$f(a)=-a$ and $\lim\limits_{x\to a}f(x)=a$
$\therefore f(x)$ is not continuous at any irrational number clearly.
$\lim\limits_{x\to 0}f(x)=f(0)=0$
$\therefore f(x)$ is continuous at $x=0$
Hence (b) is the correct answer.
answered Jan 3, 2014 by sreemathi.v

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