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A discontinuous function $y=f(x)$ satisfying $x^2+y^2=4$ is given by $f(x)$=_________

$\begin{array}{1 1}(a)\;\left\{\begin{array}{1 1}\sqrt{4-x^2},&-2\leq x\leq 0\\-\sqrt{4-x^2}&0\leq x\leq 2\end{array}\right.&(b)\;\left\{\begin{array}{1 1}\sqrt{x^2-4},&-2\leq x\leq 0\\-\sqrt{4-x^2}&0\leq x\leq 2\end{array}\right.\\(c)\;\left\{\begin{array}{1 1}\sqrt{4-x^2},&2\leq x\leq 0\\-\sqrt{4-x^2}&0\leq x\leq 2\end{array}\right.&(d)\;\left\{\begin{array}{1 1}\sqrt{4-x^2},&-4\leq x\leq 0\\-\sqrt{4-x^2}&0\leq x\leq 2\end{array}\right.\end{array}$

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$f(x)=x^2+y^2=4$
$\quad\;\;\;=\sqrt{4-x^2}\qquad-2\leq x\leq 0$
$\quad\;\;\;=-\sqrt{4-x^2}\qquad 0\leq x\leq 2$
By choosing any area of circle $x^2-y^2=4$.We can define a discontinuous function of which is
$f(x)=\left\{\begin{array}{1 1}\sqrt{4-x^2},&-2\leq x\leq 0\\-\sqrt{4-x^2}&0\leq x\leq 2\end{array}\right.$
Hence (a) is the correct answer.
answered Dec 20, 2013 by sreemathi.v
 

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