Q. 2 C4.0( 48 Votes )
Check the injecti
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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