Let f : R→R be given by

Solution:

(b) many-one and into

$f: R \rightarrow R$

$f(x)=\left[x^{2}\right]+[x+1]-3$

It is many one function because in this case for two different values of x
we would get the same value of f(x) .

For

$x=1.1,1.2 \in R$

$f(1.1)=\left[(1.1)^{2}\right]+[1.1+1]-3$

$=[1.21]+[2.1]-3$

$=1+2-3$

$=0$

$f(1.1)=\left[(1.2)^{2}\right]+[1.2+1]-3$

$=[1.44]+[2.2]-3$

$=1+2-3$

$=0$

It is into function because for the given domain we would only get the integral values of
f(x).
but R is the codomain of the given function.
That means , Codomain≠≠Range
Hence, the given function is into function.
Therefore, f(x) is many one and into