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JEE Main201910 Jan 2019Evening ShiftMathematicsFunctionsActual

Let N be the set of natural numbers and two functions f and g be defined as f , g : N → N such that f n = n + 1 2 , i f n i s o d d n 2 , i f n i s e v e n and g n = n - - 1 n . Then f o g is:

Options

  1. Aonto but not one-one
  2. BBoth one-one and onto
  3. COne-one but not onto
  4. DNeither one-one nor onto

Correct answer

A. onto but not one-one

Step-by-step solution

We have, g n = n + 1 ,   i f   n   o d d n - 1 ,   i f   n   e v e n f g 1 = f ( 2 ) = 1 f g 2 = f ( 1 ) = 1 ∴ f g x is many one. f g 2 k = f ( 2 k - 1 ) = k f g 2 k + 1 = f 2 k + 2 = k + 1 ∴ f ( g ( x ) ) is onto.

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