Question

Let $f:R\to R$ be given by $f\left(x\right)=\left[x^2\right]+\left[x+1\right]-3$ where [x] denotes the greatest integer less than or equal to x. Then, f(x) is
(a) many-one and onto
(b) many-one and into
(c) one-one and into
(d) one-one and onto

Answer 1

(b) many-one and into

$f:R\to R$

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

It is many one function because in this case for two different values of x
we would get the same value of f(x) .
$$\begin{gathered} \mathrm{For} \\ x=1.1,1.2\in R \\ f(1.1)=\left[{\left(1.1\right)}^2\right]+\left[1.1+1\right]-3 \\ =\left[1.21\right]+\left[2.1\right]-3 \\ =1+2-3 \\ =0 \\ f(1.1)=\left[{\left(1.2\right)}^2\right]+\left[1.2+1\right]-3 \\ =\left[1.44\right]+\left[2.2\right]-3 \\ =1+2-3 \\ =0 \end{gathered}$$

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$\ne$Range
Hence, the given function is into function.

Therefore, f(x) is many one and into