$\begin{array}{1 1} \text{many to one and onto function.} \\ \text{many to one and into function.} \\\text{one to one and into function.} \\\text{ bijection.} \end{array}$

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- $[x+n]=[x]+n\:\:if\:\:n\in Z$

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

$=[x]^2+[x]+1-3$

$=[x]^2+[x]-2$

Factorising the quadratic expression in $[x]$ we get

$f(x)=([x]+2)([x]-1)$

$f(x)=0\Rightarrow\:[x]+2=0\:\:or\:\:[x]-1=0$

$\Rightarrow\:[x]=-2\:\:or\:\:[x]=1$

$\Rightarrow\:x\in[1,2)\:\:or\:\:x\in [-2,-1)$

$\Rightarrow\:f$ is not one to one but is many to one function.

Also since $f(x)=[x]^2+[x]-2$ and $[x]^2,\:[x]$ take only integer values,

$f(x)$ assumes only integer values.i.e., $f(x)\in Z$.

$\Rightarrow\:$ Range of $f$ is Z

But given that $f:R\rightarrow R$

$\Rightarrow\:f$ is not onto function, but is into function.

$\Rightarrow\:f$ is many to one into function.

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