Question 6

MCQMEDIUM

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

(A)

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

(B)

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

(C)

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

(D)

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

Detailed Solution

Analyze function $f$ : $f(1) = 1$ and $f(2) = 1$ , so $f$ is not one-one.

For any $k \in \mathbb{Z}$ , if $k > 0$ , $n = 2k-1$ (odd $\in \mathbb{N}$ ) maps to $k$ ; if $k \leq 0$ , $n = 4-2k$ (even $\in \mathbb{N}$ ) maps to $k$ .

Thus $f$ is onto, making (D) true. Analyze function $g$ : for $n \geq 0$ , range is odd numbers $\geq 3$ ; for $n < 0$ , range is even numbers $\geq 2$ .

The value $1$ is never attained, so $g$ is not onto, making (C) false.

For $(g \circ f)$ : since $f(1)=f(2)$ , $g(f(1))=g(f(2))$ , so it is not one-one.

Since $1$ is not in the range of $g$ , it is not in the range of $g \circ f$ , so it is not onto. Thus (A) is true.

Question 6

Detailed Solution

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