Question

A mapping to \[N{\text{ }}to{\text{ }}N\] is defined as follows
\[f:{\text{ }}N{\text{ }} - - - > N\]
 \[f(n){\text{ = (n + 5}}{{\text{)}}^2},n\varepsilon N\]
(\[N\] is a set of natural numbers). Then,
A.\[F\] is not one to one
B.\[F\] is onto
C.\[F\] is both one to one and onto
D.\[F\] is one to one but not onto

Answer 1

Hint: - So we have to find the cases of one-one and onto.
To check onto, we check that the inverse of the function and also the codomain exist in that interval.
To check one-one, we have to check that the function does not give the same value for two different points.


Complete step by step solution:
We are given
Now, \[{1^{st}}\] we will check for one-one,
\[f\left( {n\_1} \right) = f\left( {n\_2} \right)\]
Let ${n_1},{n_2}\varepsilon N$
Now, assume
${({n_1} - 5)^2} = {({n_2} - 5)^2}$
\[ = > n\_1 - 5 = n\_2 - 5\]
\[ = > n\_1 = n\_2\]
\[\left( {n\_1} \right){\text{ }} = \left( {n\_2} \right)\] \[\therefore \] \[R\] is one-one
Now, we will check form onto:
${(x - 5)^2} = y$
Now let’s check for the onto condition
$x - 5 = {y^{(1/2)}}$
$x = {y^{(1/2)}} + 5$
Therefore we cannot find the values of \[y\] all natural numbers.
Therefore this is not onto.
\[R\] is not onto.
Hence, f is one-one but not onto
Therefore, the option \[\left( D \right)\] is correct.


Note: - While checking one-one and onto, follow the basic rules that rules \[i.e.\]for one-one different elements should have different values and for onto range should be equal to domain.