Question
A function $f: R \rightarrow R$ is defined such that $f(x)=2+$ $x^2$ then $f$ is :

Answer

(D) Neither one-one nor onto.
because $-1 \neq 1$
but $\quad f(-1)=f(1)=3$
and $-1 \in R$ (co-domain) then - 1 has not any pre image in R (domain).

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