Let be a function defined as . Then is:Question 2Answera.Injective in b.Surjective in c.Bijective in d.Neither injective nor surjective in
Question
Let be a function defined as . Then is:
- a. Injective in
- b. Surjective in
- c. Bijective in
- d. Neither injective nor surjective in
Solution
To determine whether the function is injective, we need to check if different inputs yield different outputs.
Let's assume two inputs, and , such that and .
Now, let's evaluate the function for these inputs:
Since the outputs are different, we can conclude that the function is injective.
To determine whether the function is surjective, we need to check if every element in the codomain has a corresponding element in the domain.
Let's consider an arbitrary element in the codomain.
Now, let's solve the equation for :
Since we were able to find an input that maps to the arbitrary element , we can conclude that the function is surjective.
Therefore, the function is both injective and surjective, which means it is bijective.
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