Let N denote the set of all natural numbers, and Z denote the set of all integers. Consider the functions f:Z→Z and g:Z→N defined by f(n) = ((n+1))/2 if n is odd, ((4-n))/2 if n is even, and g(n) = 3+2n if n ≥ 0, -2n if n < 0. Define (g ∘ f)(n) = g(f(n)) for all n ∈ N, and (f ∘ g)(n) = f(g(n)) for all n ∈ Z. Then which of the following statements is (are) TRUE?

Let N denote the set of all natural numbers, and Z denote the set of all integers. Consider the functions f:Z→Z and g:Z→N defined by f(n) = ((n+1))/2 if n is odd, ((4-n))/2 if n is even, and g(n) = 3+2n if n ≥ 0, -2n if n < 0. Define (g ° f)(n) = g(f(n)) for all n ∈ N, and (f ° g)(n) = f(g(n)) for all n f:Z→Z, f(n) = (n+1)/2 n odd [2pt] (4-n)/2 n even g:Z→N, g(n) = 3+2n n ≥ 0 [2pt] -2n n < 0 (g ∘ f)(n) = g(f(n)), n ∈ N; (f ∘ g)(n) = f(g(n)), n ∈ Z Test one-one and onto for f, g, g∘ f, f∘ g f(1) = (1+1)/2 = 1, f(2) = (4-2)/2 = 1 f(1) = f(2) ⇒ f is NOT one-one For any m ∈ Z, n = 2m-1 (odd): f(n) = (2m-1+1)/2 = m ⇒ f is onto Z ⇒ (4) TRUE n ≥ 0: g = 3+2n ∈ 3,5,7,… (odd) n < 0: g = -2n ∈ 2,4,6,… (even) disjoint ranges, each injective ⇒ g is one-one Image(g) = 2,3,4,5,… ⇒ 1 ∉ Image(g) g is NOT onto ⇒ (3) FALSE (g ∘ f)(1) = g(f(1)) = g(1) = 3 + 2(1) = 5 (g ∘ f)(2) = g(f(2)) = g(1) = 5 ⇒ g ∘ f NOT one-one Image(g ∘ f) = 2,3,4,5,… ⇒ 1 never attained ⇒ (1) TRUE: g∘ f neither one-one nor onto n ≥ 0: (f∘ g)(n) = ((3+2n)+1)/2 = n+2 n < 0: (f∘ g)(n) = (4-(-2n))/2 = n+2 (f∘ g)(n) = n+2 ⇒ one-one and onto ⇒ (2) FALSE boxedTRUE statements: (1) and (4) f:onto, not one-one; g:one-one, not onto; g∘ f:neither; f∘ g = n+2

Let $\mathbb{N}$ denote the set of all natural numbers, and $\mathbb{Z}$ denote the set of all integers. Consider the functions $f:\mathbb{Z}\to\mathbb{Z}$ and $g:\mathbb{Z}\to\mathbb{N}$ defined by $f(n) = \begin{cases} \frac{(n+1)}{2} & \text{if } n \text{ is odd}, \\ \frac{(4-n)}{2} & \text{if } n \text{ is even}, \end{cases}$ 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?

Answers: A. $g \circ f$ is NOT one-one and $g \circ f$ is NOT onto · D. $f$ is NOT one-one but $f$ is onto

Relations and Functions (Class 12) Let

\mathbb{N}

denote the set of all natural numbers, and

\mathbb{Z}

denote the set of all integers. Consider the functions

f:\mathbb{Z}\to\mathbb{Z}

and

g:\mathbb{Z}\to\mathbb{N}

defined by

f(n) = \begin{cases} \frac{(n+1)}{2} & \text{if } n \text{ is odd}, \\ \frac{(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?

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Written solution

$f:\mathbb{Z}\to\mathbb{Z}, \quad f(n) = \begin{cases} \dfrac{n+1}{2} & n \text{ odd} \\[2pt] \dfrac{4-n}{2} & n \text{ even} \end{cases}$ $g:\mathbb{Z}\to\mathbb{N}, \quad g(n) = \begin{cases} 3+2n & n \geq 0 \\[2pt] -2n & n < 0 \end{cases}$ $(g \circ f)(n) = g(f(n)),\ n \in \mathbb{N}; \qquad (f \circ g)(n) = f(g(n)),\ n \in \mathbb{Z}$ $\text{Test one-one and onto for } f,\ g,\ g\circ f,\ f\circ g$ $f(1) = \dfrac{1+1}{2} = 1, \qquad f(2) = \dfrac{4-2}{2} = 1$ $f(1) = f(2) \Rightarrow f \text{ is NOT one-one}$ $\text{For any } m \in \mathbb{Z},\ n = 2m-1 \text{ (odd)}:\ f(n) = \dfrac{2m-1+1}{2} = m$ $\Rightarrow f \text{ is onto } \mathbb{Z} \ \Rightarrow (4) \text{ TRUE}$ $n \geq 0:\ g = 3+2n \in \{3,5,7,\ldots\}\ (\text{odd})$ $n < 0:\ g = -2n \in \{2,4,6,\ldots\}\ (\text{even})$ $\text{disjoint ranges, each injective} \Rightarrow g \text{ is one-one}$ $\text{Image}(g) = \{2,3,4,5,\ldots\} \Rightarrow 1 \notin \text{Image}(g)$ $g \text{ is NOT onto} \ \Rightarrow (3) \text{ FALSE}$ $(g \circ f)(1) = g(f(1)) = g(1) = 3 + 2(1) = 5$ $(g \circ f)(2) = g(f(2)) = g(1) = 5 \Rightarrow g \circ f \text{ NOT one-one}$ $\text{Image}(g \circ f) = \{2,3,4,5,\ldots\} \Rightarrow 1 \text{ never attained}$ $\Rightarrow (1) \text{ TRUE: } g\circ f \text{ neither one-one nor onto}$ $n \geq 0:\ (f\circ g)(n) = \dfrac{(3+2n)+1}{2} = n+2$ $n < 0:\ (f\circ g)(n) = \dfrac{4-(-2n)}{2} = n+2$ $(f\circ g)(n) = n+2 \Rightarrow \text{one-one and onto} \Rightarrow (2) \text{ FALSE}$ $\boxed{\text{TRUE statements: } (1) \text{ and } (4)}$ $f:\text{onto, not one-one};\ \ g:\text{one-one, not onto};\ \ g\circ f:\text{neither};\ \ f\circ g = n+2$

Written solution

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