Q2 of 107 Page 10

Check the injectivity and surjectivity of the following functions:

f : R → R given by f (x) = x2

It is given that f : R → R given by f (x) = x2

We can see that f(-1) = f(1) = 1, but -1 ≠ 1


⇒ f is not injective.


Now, let -2 ϵ R. But, we can see that there does not exists any x in R such that


f(x) = x2 = -2


⇒ f is not surjective.


Therefore, function f is neither injective nor surjective.


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