Knowee
Questions
Features
Study Tools

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
🧐 Not the exact question you are looking for?Go ask a question

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

Similar Questions

Consider the function f : R2 → R2 defined by the formula f (x, y) = (xy, x3). Is f injective? Is itsurjective? Bijective? Explain

If f A —> B and g : B —> C are injectivefunction, then gof: A —> C is an injectivefunction. Prove or disprove.

If f:A→B is a bijective function and n(A)=6 then which of the following is not possible*Number of elements in range of f is 6n(A)=n(B)n(B)=6n(B)=8

7. Determine whether each of these functions is a bijection from R to R.a) f (x) = −3x + 4 b) f (x) = −3x2 + 7 c) f (x) = (x + 1)/(x + 2) d) f (x) = x5 + 1

Which of the following function f: Z X Z → Z is not onto?a.f(a, B) = a – bb.f(a, B) = |b|c.f(a, B) = a + bd.f(a, B) = a

1/1

Upgrade your grade with Knowee

Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.