Define $f: R \rightarrow R$ by $f(x)=\max \{x+1,1-x, 2\}$. Then, $f$ is

Define $f: R \rightarrow R$ by $f(x)=\max \{x+1,1-x, 2\}$. Then, $f$ is
  1. one-one but not onto
  2. onto but not one-one
  3. neither one-one nor onto
  4. both one-one and onto

Solution

$f: R \rightarrow R$ $f(x)=\max \{x+1,1-x, 2\}$ Since, $f(x)$ contains 2 in range when domain is $[-1,1]$ Hence, it is not one-one. Since, $f(x)$ does not take any value less than 2 in it's domain. Hence, Codomain $\neq$ Range $\Rightarrow f(x)$ is not onto. Hence, $f(x)$ is neither one-one nor onto.

Asked in: AP EAMCET 2021 (23 Aug Shift 2)

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