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Consider the function f : R → {0, 1} such that \(f(x)=\left\lbrace \begin{matrix} 1 \ \text{if} \ x \ \text{is rational} \\\ 0 \ \text{if} \ x \ \text{is irrational} \end{matrix} \right.\)Which one of the following is correct?1. The function is one-one into2. The function is many-one into3. The function is one-one onto4. The function is many-one onto |
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Answer» Correct Answer - Option 4 : The function is many-one onto Concept: Let f(x) be any function. f (x) is onto if range of f (x ) = Codomain The function f is said to be many-one functions if there exist two or more than two different elements in X having the same image in Y.
Calculations: Given function f : R → {0, 1} such that \(f(x)=\left\lbrace \begin{matrix} 1 \ \text{if} \ x \ \text{is rational} \\\ 0 \ \text{if} \ x \ \text{is irrational} \end{matrix} \right.\) Codomain = {0, 1} Since, on taking a straight line parallel to the x-axis, the group of given function intersect it at many points. ⇒ f (x) is many-one. Range of function is {0, 1} As range of f (x ) = Codomain ⇒ f (x) is onto. Hence, f (x) is many-one onto |
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