JEE Advanced 2025 · Paper 1 · Mathematics Q6

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MathematicsMultiple correct+4 / -2Question 6 of 48

Let $\mathbb{N}$ denote the set of all natural numbers, and $\mathbb{Z}$ denote the set of all integers. Consider the functions $f : \mathbb{N} \to \mathbb{Z}$ and $g : \mathbb{Z} \to \mathbb{N}$ defined by

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

and

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

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?

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

g

is one-one and

g

is onto

f

is NOT one-one but

f

is onto

Answer and worked solution

Official answer: A, D

f. Odd $n$ give $1,2,3,\dots$ ; even $n$ give $1,0,-1,-2,\dots$

$f(1)=f(2)=1$ , so $f$ is not one-one.
Every integer is hit, so $f$ is onto.
(D) is true.

g. $n\ge0$ gives $3,5,7,\dots$ ; $n<0$ gives $2,4,6,\dots$

$g$ is one-one.
$1$ is never hit, so $g$ is not onto.
(C) is false.

g∘f.

$f(1)=f(2)$ , so $g\circ f$ is not one-one.
Its range lies inside the range of $g$ , which misses 1, so it is not onto.
(A) is true.

f∘g.

For $n\ge0$ : $f(3+2n)=n+2$ .
For $n<0$ : $f(-2n)=\frac{4+2n}{2}=n+2$ .
So $f\circ g(n)=n+2$ , a bijection. (B) is false.

Topics:Composition and inverse functions Types of functions

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