A) surJective but not injecrive
B) injective but not surjective
C) bijective
D) neither suqective nor injective
Correct Answer: B
Solution :
Now, assume f is injective (ie, one-one), since \[{{x}_{1}},{{x}_{2}}\in R\]and \[{{x}_{1}}\ne {{x}_{2}}.\] \[\Rightarrow \]\[{{e}^{{{x}_{1}}}}\ne {{e}^{{{x}_{2}}}}\]\[\Rightarrow \]\[f({{x}_{1}})\ne f({{x}_{2}})\] \[\therefore \]It is ture. Again, assume f is not suriective (onto), since \[{{e}^{x}}>0\] for all x and so negative real number can not be the image of any real number. For example, there is no real x such that\[f(x)=-2.\] Hence, f is injective but not suriective.
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