We have, g(n)={\begin{matrix}n+1, if n odd \\ n-1, if n even\end{matrix}
f(g(1))=f(2)=1
f(g(2))=f(1)=1
∴f(g(x)) is many one.
f(g(2k))=f(2k−1)=k
f(g(2k+1))=f(2k+2)=k+1
∴f(g(x)) is onto.
JEE Main 2019 — Mathematics Algebra
Let N be the set of natural numbers and two functions f and g be defined as f,g:N→N such that
f(n)={\begin{matrix}\frac{n+1}{2}, if n is odd \\ \frac{n}{2}, if n is even\end{matrix}
and g(n)=n−(−1)n. Then fog is:
Held on 10 Jan 2019 · Verified 6 Jul 2026.
onto but not one-one
Both one-one and onto
One-one but not onto
Neither one-one nor onto
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