Let denote the set of all natural numbers, and denote the set of all integers. Consider the functions f : and g : defined by f ( n ) = { ( n + 1 ) 2 if n is odd, ( 4 - n ) 2 if n is even, and g ( n ) = { 3 + 2 n if n 0 , - 2 n if n 0

Question

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

$f (n) =$ ${\begin{matrix} \frac{(n + 1)}{2} & if n is odd, \\ \frac{(4 - n)}{2} & if n is even, \end{matrix}$ and

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

Define $(g \circ f) (n) = g (f (n))$ for all $n \in ℕ$ , and $(f \circ g) (n) = f (g (n))$ for all $n \in ℤ$ .

Then which of the following statement(s) is (are) TRUE? [2025]

Smart Achievers · Accepted Answer

(1, 4)

$Given,$ $f (n) =$ ${\begin{matrix} \frac{(n + 1)}{2} & if n is odd \\ \frac{(4 - n)}{2} & if n is even \end{matrix}$

$f (n) = {(1, 1), (2, 1), (3, 2), (4, 0), (5, 3), (6, - 1), \dots}$

$∴$ $f (n) is many-one and onto function.$

$g (n) =$ ${\begin{matrix} 3 + 2 n & if n \geq 0 \\ - 2 n & if n < 0 \end{matrix}$

$g (n) = {(- 3, 6), (- 2, 4), (- 1, 2), (0, 3), (1, 5), (2, 7), (3, 9), (4, 15), \dots}$

$∴$ $g (n) is one-one and into function.$

$f (g (n)) = 2 + n,$ $n \in ℕ$ $\Rightarrow f \circ g is one-one and into$

$g (f (n)) =$ ${\begin{matrix} 4 + n & if n is odd natural number \\ 7 - n & if n = 2, 4 \\ n - 4 & if n is even natural number and n \geq 6 \end{matrix}$

$g (f (2)) = g (f (1)) = 5$

$∴$ $g \circ f is many-one and into$

Solution

1 $g \circ f$ is NOT one-one and $g \circ f$ is NOT onto

2 $f \circ g$ is NOT one-one but $f \circ g$ is onto

3 $g$ is one-one and $g$ is onto

4 $f$ is NOT one-one but $f$ is onto

Want Us to Email you About Special Offers & Updates?

Solution

1 g∘f is NOT one-one and g∘f is NOT onto

2 f∘g is NOT one-one but f∘g is onto

3 g is one-one and g is onto

4 f is NOT one-one but f is onto

Want Us to Email you About Special Offers & Updates?

1 $g \circ f$ is NOT one-one and $g \circ f$ is NOT onto

2 $f \circ g$ is NOT one-one but $f \circ g$ is onto

3 $g$ is one-one and $g$ is onto

4 $f$ is NOT one-one but $f$ is onto