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  • A discontinuous function  satisfying. <img alt="\mathrm{x^2+y^2=4}" src="htt

A discontinuous function \mathrm{y=f(x)} satisfying. \mathrm{x^2+y^2=4} is given by \mathrm{f(x)=} ? for \mathrm{0 \leq x \leq 2}

Option: 1

\mathrm{-\sqrt{\left(4-x^2\right)}}


Option: 2

\mathrm{4 x}


Option: 3

\mathrm{\sqrt{x^2-4}}


Option: 4

\mathrm{ \sqrt{x^2+4}}


Answers (1)

best_answer

By choosing any arc of the circle \mathrm{x^2+y^2=4} we can define a discontinuous function, one of which is
Hence,\mathrm{\begin{aligned} & f(x)=\sqrt{\left(4-x^2\right) ;-2} \leq x \leq 0 \\ & f(x)=\left\{\begin{array}{l} \sqrt{\left(4-x^2\right)} ;-2 \leq x \leq 0 \\ -\sqrt{\left(4-x^2\right)} ; 0 \leq x \leq 2 \end{array}\right. \end{aligned}}

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Ritika Jonwal

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