Let `f: R->R` be given by `f(x)=[x]^2+[x+1]-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

Options

A. many-one and onto
B. many-one and into
C. one-one and into
D. one-one and onto

Answer

B

Explanation

We have
`f(x)=[x]^(2)+[x+1]-3`
`Rightarrow f(x)=[x]^(2)+[x]+1-3 [therefore [x+n]=[x]+n, "where n "in Z]`
`Rightarrow f(x)=[x]^(2)+[x]-2`
`Rightarrow f(x)=([x]+2) ([x]-1)`
Clearly, f(x)=0 for all `x in [1,2] uu [-2, -1]`
So, f is a many-one function.
Also, f(x) assumes only integral values.
`therefore "Range of f" ne R`.
Hence, f(x) is a many-one into function.