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The function \mathrm{f(x)}  is discontinuous only at \mathrm{x=0} such that - \mathrm{f(x)=1 \vee x \in R}.The total number of such functions is

Option: 1

2


Option: 2

3


Option: 3

0


Option: 4

None of these


Answers (1)

best_answer

\mathrm{(i) f(x)= \begin{cases}1, & x \leq 0 \\ -1, & x>0\end{cases}}
\mathrm{ \text { (ii) } f(x)= \begin{cases}1, & x<0 \\ -1, & x \geq 0\end{cases} }
\mathrm{ \text { (iii) } f(x)= \begin{cases}-1, & x \leq 0 \\ 1, & x>0\end{cases} }
\mathrm{ \text { (iv) } f(x)= \begin{cases}-1, & x<0 \\ 1, & x \geq 0\end{cases} }
\mathrm{ \text { (v) } f(x)= \begin{cases}1, & x>0 \\ 1, & x<0 \\ -1 & x=0\end{cases} }
\mathrm{ \text { (vi) } f(x)= \begin{cases}-1, & x>0 \\ -1, & x<0 \\ 1, & x=0\end{cases} }

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