Check the nature of the following function.
(i) `f(x)=sin x, x in R " (ii) " f(x) =sin x, x in N`

Explanation

(i) `f(x) =sinx, x in R`
Let `f(x_(1))=f(x_(2))`
`implies sinx_(1)=sinx_(2)`
`implies x_(1)=n pi+(-1)^(n)x_(2),n in Z.`
There are infinite pair `(x_(1),x_(2)), x_(1) ne x_(2)` which satisfy above equation.
i.e., `f(0)=f(pi)=0`
So, f(x) is many-one .
(ii) `f(x) = sinx, x in N`
Let `f(x_(1))=f(x_(2))`
`implies sin x_(1)=sin x_(2)`
`implies x_(1)=n pi+(-1)^(n)x_(2),` where `x_(1),x_(2) in N " and "n in Z.`
This is equation is possible only for `n=0`, for which `x_(1)=x_(2).`
So, f(x) is one-one.