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

Explanation

`{f}: {R} -> {R}`


`f(x)=[x^{2}]+[x+1]-3`


It is many one function since, 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}`


` Rightarrow {f}(1.1) &=[(1.1)^{2}]+[1.1+1]-3 &=[1.21]+[2.1]-3 &=1+2-3 &=0 Rightarrow {f}(1.2) &=[(1.2)^{2}]+[1.2+1]-3 &=[1.44]+[2.2]-3 &=1+2-3 &=0 `


`Rightarrow {f}(1.2)=[(1.2)^{2}]+[1.2+1]-3`


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


`=1+2-3`


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