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Show that the Signum Function f : R → R, given by f(x) = \(f(x)=\begin{cases}1 & if \;x > 0\\0& if \;x = 0\\-1 & if \;x < 0\end{cases}\) is neither one-one nor onto. |
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Answer» f (x) = 1 ∀ x ∈ [0, ∞ ) hence f is not one-one Range of f = {-1,0, 1} hence f is not onto. |
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