Question

$f:N\to N$, then show that $f\left(n\right)=2n+3\forall nϵN.$

• A. many one, into.
• B. one-one, into
• C. one-one, not onto
• D. many-one, onto

Answer 1

Answer: (B), (C)

The correct options are
B one-one, into
C one-one, not onto

Given, $f\left(n\right)=2n+3$
Now for $n_1,n_2ϵN$
If $f\left(n_1\right)=f\left(n_2\right)$
Then $2n_1+3=2n_2+3$
$\Rightarrow n_1=n_2$
Hence for $f\left(n_1\right)=f\left(n_2\right)$, it implies that $n_1=n_2$.
Therefore, it is a one-one function.
Now the smallest natural number is $1$.
$f\left(1\right)=5$.
Thus domain is $N$, while range is $\{5,7,\mathrm{9...2}n+1\}$
Hence, the above function is an into function.