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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:Question 2Answera.Injective in b.Surjective in c.Bijective in d.Neither injective nor surjective in

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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.

This problem has been solved

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