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If $f: R \rightarrow C$ is defined by $f(x)=e^{2 i x}$ for $x \in R$, then $f$ is (where $C$ denotes the set of all complex numbers)
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neither one-one nor onto
Given that, $f(x)=e^{2 i x}$ and $f: R \rightarrow C$. Function $f(x)$ is not one-one, because after some values of $x(i e, \pi)$ it will give the same values.
Also, $f(x)$ is not onto, because it has minimum and maximum values $-1-i$ and $1+i$ respectively.
Hence, option (d) is correct.
Also, $f(x)$ is not onto, because it has minimum and maximum values $-1-i$ and $1+i$ respectively.
Hence, option (d) is correct.
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