Search any question & find its solution
Question:
Answered & Verified by Expert
Let $\mathrm{N}$ be the set of natural numbers and $\mathrm{f}: \mathrm{N} \rightarrow \mathrm{N}$, be a function given by $\mathrm{f}(\mathrm{x})=\mathrm{x}+1, \mathrm{x} \in \mathrm{N}$.
Which one of the following is correct ?
Options:
Which one of the following is correct ?
Solution:
1946 Upvotes
Verified Answer
The correct answer is:
$\mathrm{f}$ is one-one but not onto
To show $\mathrm{f}$ is one-one. Let $f(x)=f(y)$ (To show: $x=y)$ $\Rightarrow \quad x+1=y+1$
$\Rightarrow \quad x=y$
Hence, $\mathrm{f}$ is one-one, Now, ' $\mathrm{f}$ ' is not onto because every element of codomain does not have it's pre-image in domain.
$\Rightarrow \quad x=y$
Hence, $\mathrm{f}$ is one-one, Now, ' $\mathrm{f}$ ' is not onto because every element of codomain does not have it's pre-image in domain.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.