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Question
The function
Options
injective but not surjective
surjective but not injective
injective as well as surjective
neither injective nor surjective
Solution
Injectivity:
Let x and y be any two elements in the domain (R), such that f(x) = f(y). Then,
\[ \Rightarrow x = \pm y\]
Surjectivity:
\[\text{and} f\left( 1 \right) = 1^2 = 1, \]
\[f\left( - 1 \right) = f\left( 1 \right)\]
So, the answer is (d) .
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