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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)` isA. many-one and ontoB. many-one and intoC. one-one and intoD. one-one and onto |
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Answer» Correct Answer - B 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. |
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