The function `f: R-> R : f(x)=sinx` is

Options

A. One One
B. Onto
C. Into
D. None Of these

Answer

C

Explanation

We know that `0,pi in R`
and f(0) = sin 0 = 0
and f(`pi`)= sin `pi` = 0
`:. f(0)= f (pi)`
Therefore two different elements 0 and `pi` have the same image in co-domain R, so f is not one-one.
Again we know that `-1le sin xle 1`
Therefore , all values except in`[-1, 1]` have no preimages in R.
Therefore, f is not onto.
So, f is neither one-one nor onto. Hence Proved.