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Question
Check the injectivity and surjectivity of the following function:
f : R → R given by f(x) = x2
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Solution
f : R → R given by f(x) = x2
It is seen that f(–1) = f(1) = 1, but –1 ≠ 1.
∴ f is not injective.
Now, –2 ∈ R, but there does not exist any element x ∈ R such that f(x) = x2 = –2.
∴ f is not surjective.
Hence, function f is neither injective nor surjective.
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