Mathematics - Functions Question with Solution | TestHub

For one value of $n$ we will get only one corresponding value of $f\left(n\right)$, so $f\left(n\right)$ is one-one

Now, for $n=2,4,6\cdots ; f\left(n\right)=4,8,12...$

for $n=3,7,11\cdots ; f \left(n\right)=2,6,10...$

for $n=1,5,9\cdots ; f\left(n\right)=1,3,5,7...$

So range of $f\left(n\right)$ is $N$

Hence $f\left(n\right)$ is onto