JEE Advanced 2025 Paper 1 Q06 Mathematics Algebra Sequences & Series Medium

JEE Advanced 2025 Paper 1 · Q06 · Sequences & Series

Let $\mathbb{N} = \{1,2,3,\ldots\}$ (natural numbers) and $\mathbb{Z}$ = integers. Define $f: \mathbb{N} \to \mathbb{Z}$ by $f(n) = \frac{n+1}{2}$ if $n$ is odd, $f(n) = \frac{4-n}{2}$ if $n$ is even. Define $g: \mathbb{Z} \to \mathbb{N}$ by $g(n) = 3 + 2n$ if $n \ge 0$, $g(n) = -2n$ if $n < 0$. Define $(g \circ f)(n) = g(f(n))$ for $n \in \mathbb{N}$, and $(f \circ g)(n) = f(g(n))$ for $n \in \mathbb{Z}$. Which statements are TRUE?

  1. A. $g \circ f$ is NOT one-one and $g \circ f$ is NOT onto
  2. B. $f \circ g$ is NOT one-one but $f \circ g$ is onto
  3. C. $g$ is one-one and $g$ is onto
  4. D. $f$ is NOT one-one but $f$ is onto
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:A, D

Solution

$f: \mathbb{N} \to \mathbb{Z}$. $f(1)=1$, $f(2)=1$, so $f$ is NOT one-one. Range of $f$: odd $n=2k-1$ gives $k$ ($k=1,2,\ldots$) $\to$ all positive integers; even $n=2k$ gives $2-k$ ($k=1,2,\ldots$) $\to$ $1,0,-1,-2,\ldots$ $\to$ all integers $\le 1$. So range $= \mathbb{Z}$. $f$ is onto. (D) TRUE. $g: \mathbb{Z} \to \mathbb{N}$. For $n \ge 0$: $g(n)=3,5,7,9,\ldots$ (odd $\ge 3$). For $n < 0$: $g(n)=2,4,6,\ldots$ (even $\ge 2$). So $g$ hits all natural numbers $\ge 2$; $g$ misses $1$. So $g$ is NOT onto. Also injective on each piece and the ranges $\{\text{odd} \ge 3\}$ and $\{\text{even} \ge 2\}$ are disjoint, so $g$ is one-one. (C) FALSE. $g \circ f$: for $n=1$, $f(1)=1$, $g(1)=3+2=5$; for $n=2$, $f(2)=1$, $g(1)=5$. Same value, so not one-one. Misses $1$ (since $g$ misses $1$). NOT onto. (A) TRUE. $f \circ g$: $g(0)=3$, $g(1)=5$, $g(-1)=2$, $g(-2)=4$. $f(g(0))=f(3)=2$, $f(g(-1))=f(2)=1$, $f(g(1))=f(5)=3$, $f(g(-2))=f(4)=0$. Test injectivity: $g(0)=3$ (odd) gives $f=2$; $g(2)=7$ (odd) gives $f=4$; injective on each piece. Even $g$ and odd $g$ both $\ge 2$, but $f$ maps odd $n=(n+1)/2$ (positive ints), even $n=(4-n)/2$ ($\le 1$). Different ranges from each piece, so $f …[truncated]

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