Let $f, g: N \rightarrow N$ such that $f(n+1)=f(n)+f(1)$ $\forall \mathrm{n} \in \mathrm{N}$ and $g$ be any arbitrary function. Which of the following statements is NOT true?
Correct Option: , 4
$f(n+1)-f(n)=f(1)$
$\Rightarrow \mathrm{f}(\mathrm{n})=\mathrm{nf}(1)$
$\Rightarrow \mathrm{f}$ is one-one
Now, Let $\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}_{2}\right)\right)=\mathrm{f}\left(\mathrm{g}\left(\mathrm{x}_{1}\right)\right)$
$\Rightarrow \mathrm{g}\left(\mathrm{x}_{2}\right)=\mathrm{g}\left(\mathrm{x}_{1}\right)$ (as $\mathrm{f}$ is one-one)
$\Rightarrow \mathrm{x}_{1}=\mathrm{x}_{2}$ (as fog is one-one)
$\Rightarrow \mathrm{g}$ is one-one
Now, $\mathrm{f}(\mathrm{g}(\mathrm{n}))=\mathrm{g}(\mathrm{n}) \mathrm{f}(\mathrm{l})$
may be many-one if
$\mathrm{g}(\mathrm{n})$ is many-one