Math JEE Adv. 2025 paper 1

 

To determine which statements are true, we'll analyze the functions $f$, $g$, and their compositions $g\circ f$ and $f\circ g$.

---

Function Definitions:

1. $f:\mathbb{N}\to \mathbb{Z}$:

   $$f(n)=\{\begin{array}{ll}(n+1)\mathrm{/}2& \text{if }n\text{ is odd},\\ (4-n)\mathrm{/}2& \text{if }n\text{ is even}\mathrm{.}\end{array}$$

2. $g:\mathbb{Z}\to \mathbb{N}$:

   $$g(n)=\{\begin{array}{ll}3+2n& \text{if }n\ge 0,\\ -2n& \text{if }n<0.\end{array}$$

---

Analysis of $f$:

• One-One (Injective):

  • For odd $n$, $f(n)=\frac{n+1}{2}$. This is injective since each odd $n$ maps to a unique integer.

  • For even $n$, $f(n)=\frac{4-n}{2}$. This is also injective since each even $n$ maps to a unique integer.

  • However, $f(1)=1$ and $f(4)=0$, but $f(3)=2$ and $f(2)=1$. Here, $f(1)=f(2)=1$, so $f$ is not one-one.

• Onto (Surjective):

  • For any $k\in \mathbb{Z}$, we can find $n\in \mathbb{N}$ such that $f(n)=k$:

    • If $k\ge 1$, take $n=2k-1$ (odd).

    • If $k\le 0$, take $n=4-2k$ (even).

  • Thus, $f$ is onto.

Conclusion for $f$:

• Not one-one (since $f(1)=f(2)=1$).

• Onto (as shown above).

This matches Statement (D).

---

Analysis of $g$:

• One-One (Injective):

  • For $n\ge 0$, $g(n)=3+2n$. This is injective.

  • For $n<0$, $g(n)=-2n$. This is also injective.

  • No overlaps between the two cases because $g(n)\ge 3$ for $n\ge 0$ and $g(n)\ge 2$ for $n<0$, with $g(-1)=2$ and $g(0)=3$. Thus, $g$ is one-one.

• Onto (Surjective):

  • The range of $g$ for $n\ge 0$ is $\{3,5,7,\dots \}$.

  • The range of $g$ for $n<0$ is $\{2,4,6,\dots \}$.

  • The number $1$ is not in the range of $g$, so $g$ is not onto.

Conclusion for $g$:

• One-one (as shown above).

• Not onto (since $1$ is not in the range).

This contradicts Statement (C).

---

Analysis of $g\circ f$:

Compute $g\circ f$ for $n\in \mathbb{N}$:

• If $n$ is odd:

  $$g(f(n))=g\left(\frac{n+1}{2}\right)=3+2\left(\frac{n+1}{2}\right)=n+4.$$

• If $n$ is even:

  $$g(f(n))=g\left(\frac{4-n}{2}\right)\mathrm{.}$$

  • If $\frac{4-n}{2}\ge 0$ (i.e., $n\le 4$):

    $$g\left(\frac{4-n}{2}\right)=3+2\left(\frac{4-n}{2}\right)=7-n\mathrm{.}$$

  • If $\frac{4-n}{2}<0$ (i.e., $n>4$):

    $$g\left(\frac{4-n}{2}\right)=-2\left(\frac{4-n}{2}\right)=n-4.$$

Behavior of $g\circ f$:

• For odd $n$, $g\circ f(n)=n+4$. This is injective but not surjective (e.g., $1\mapsto 5$, $3\mapsto 7$, etc., missing many natural numbers).

• For even $n$, $g\circ f(n)=7-n$ (if $n\le 4$) or $n-4$ (if $n>4$). This is not injective (e.g., $n=2$ and $n=6$ both map to $5$) and not surjective.

Conclusion for $g\circ f$:

• Not one-one (due to overlaps like $g\circ f(2)=g\circ f(6)=5$).

• Not onto (many natural numbers are not in the range).

This matches Statement (A).

---

Analysis of $f\circ g$:

Compute $f\circ g$ for $n\in \mathbb{Z}$:

• If $n\ge 0$:

  $$f(g(n))=f(3+2n)\mathrm{.}$$

  • Since $3+2n$ is odd, $f(3+2n)=\frac{3+2n+1}{2}=n+2$.

• If $n<0$:

  $$f(g(n))=f(-2n)\mathrm{.}$$

  • Since $-2n$ is even, $f(-2n)=\frac{4-(-2n)}{2}=2+n$.

Behavior of $f\circ g$:

• For $n\ge 0$, $f\circ g(n)=n+2$. This is injective.

• For $n<0$, $f\circ g(n)=2+n$. This is also injective.

• The entire range covers all integers $\ge 2$ (from $n\ge 0$) and all integers $\le 1$ (from $n<0$), so $f\circ g$ is onto.

However, $f\circ g(0)=2$ and $f\circ g(-1)=1$, but $f\circ g(-2)=0$, etc. There are no overlaps, so $f\circ g$ is one-one.

Conclusion for $f\circ g$:

• One-one (no overlaps in outputs).

• Onto (covers all integers).

This contradicts Statement (B).

---

Final Answer:

The true statements are (A) and (D).

Answer: $\boxed{{\displaystyle A,D}}$