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The function $f: N \rightarrow N, N$ being the set of of natural numbers. defined by $f(x)=2 x+3$ is
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injective but not surjective
Given $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}$
$\therefore \mathrm{f}(\mathrm{x})=2 \mathrm{x}+3 \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=2>0$
So, $\mathrm{f}(\mathrm{x})$ is increasing, $\forall \mathrm{x} \in \mathrm{N}$.
Hence, $\mathrm{f}(\mathrm{x})$ is injective. Let $\mathrm{f}(\mathrm{x})=\mathrm{y}$
$\Rightarrow \mathrm{y}=2 \mathrm{x}+3 \Rightarrow \mathrm{x}=\frac{\mathrm{y}-3}{2}$
This is injective
$\therefore \mathrm{x}=\frac{1}{2}$
and $\mathrm{y} \in \mathrm{N}$ but $\mathrm{x} \notin \mathrm{N}$
Hence, $\mathrm{f}(\mathrm{x})$ is not surjective.
$\therefore \mathrm{f}(\mathrm{x})=2 \mathrm{x}+3 \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=2>0$
So, $\mathrm{f}(\mathrm{x})$ is increasing, $\forall \mathrm{x} \in \mathrm{N}$.
Hence, $\mathrm{f}(\mathrm{x})$ is injective. Let $\mathrm{f}(\mathrm{x})=\mathrm{y}$
$\Rightarrow \mathrm{y}=2 \mathrm{x}+3 \Rightarrow \mathrm{x}=\frac{\mathrm{y}-3}{2}$
This is injective
$\therefore \mathrm{x}=\frac{1}{2}$
and $\mathrm{y} \in \mathrm{N}$ but $\mathrm{x} \notin \mathrm{N}$
Hence, $\mathrm{f}(\mathrm{x})$ is not surjective.
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