A) Onto
B) Many-one
C) One-one and into
D) Many one and onto
Correct Answer: C
Solution :
Function \[f:R\to R\] is defined by \[f(x)={{e}^{x}}\]. Let \[{{x}_{1}},\,{{x}_{2}}\in R\] and \[f({{x}_{1}})=f({{x}_{2}})\] or \[{{e}^{{{x}_{1}}}}={{e}^{{{x}_{2}}}}\] or \[{{x}_{1}}={{x}_{2}}\]. Therefore f is one-one. Let \[f(x)={{e}^{x}}=y\]. Taking log on both sides, we get \[x=\log y\]. We know that negative real numbers have no pre-image or the function is not onto and zero is not the image of any real number. Therefore function f is into.
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