Let \mathbb{N} denote the set of all natural numbers, and \mathbb{Z} denote the set of all integers. Consider the functions f: \mathbb{N} \rightarrow \mathbb{Z} and g: \mathbb{Z} \rightarrow \mathbb{N} defined by f(n)={\\begin{array}{c}(n+1) \/ 2 \\\\text{ if }\\ n \\\\text{ is odd }\\ \ (4-n) \/ 2 \\\\text{ if }\\ n \\\\text{ is even }\\\\end{array} and g(n)={\\begin{array}{cc}3+2 n & \\\\text{ if }\\ n \geq 0 \ -2 n & \\\\text{ if }\\ n&lt\ 0\end{array} Define (g \circ f)(n)=g(f(n)) for all n \in \mathbb{N}, and (f \circ g)(n)=f(g(n)) for all n \in \mathbb{Z}. Then which of the following statements is (are) TRUE? [JEE Adv 2025 P1]

g \circ f is NOT one-one and g \circ f is NOT onto

f \circ g is NOT one-one but f \circ g is onto

Functions

Section: Mathematics

Question : 3 of 30

Marks: +1, -0

Let ℕ denote the set of all natural numbers, and ℤ denote the set of all integers.
Consider the functions f:ℕ⟶ℤ and g:ℤ⟶ℕ defined by

f(n)={

(n+1)∕2‌ if ‌n‌ is odd ‌

(4−n)∕2‌ if ‌n‌ is even ‌

and

g(n)={

3+2n	‌ if ‌n≥0
−2n	‌ if ‌n<0

Define (g∘f)(n)=g(f(n)) for all n∈ℕ, and (f∘g)(n)=f(g(n)) for all n∈ℤ.
Then which of the following statements is (are) TRUE?

[JEE Adv 2025 P1]

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k+1	k∈N∪{0}
2−k,‌	k∈N

3+2f(n),‌	f(n)≥0
−2f(n),‌	f(n)<0

n+4	n=2k+1
7−n,‌	n=2k

3+2n,‌	n≥0
−2n,‌	n<0

n+2,‌	n≥0
n+2,‌	n<0