MCQ
Let a function $f: N \rightarrow N$ be defined by

$f ( n )=\left[\begin{array}{ll}2 n , \,\,\, \,\,\,\,\,\,n =2,4,6,8, \ldots . \\ n -1,\,\,\,  n =3,7,11,15, \ldots . \\ \frac{ n +1}{2}, \,\,\, \,\,\,n =1,5,9,13, \ldots \ldots\end{array}\right.$

then, $f$ is

  • A
    one-one but not onto
  • B
    onto but not one-one
  • C
    neither one-one nor onto
  • one-one and onto

Answer

Correct option: D.
one-one and onto
d
$f ( x )=\left\{\begin{array}{ccc}4 R ;  n =2 R \\ 4 R -2  ;  n =4 R -1 \\ 2 R -1  ;  n =4 R -3\end{array}\right.$

$(R \in N)$

$Note$ that for any element, it will fall into exactly. one of these sets.

$\{y: y=4 R ; y \in N\}$

$\{y: y=4 R-2 ; y \in N\}$

$\{ y : y =2 R -1 ; y \in N \}$

Corresponding to that $y$, we will get exactly one value of $n$.

Thus, $f$ is one - one and onto.

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