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JEE Main 201910 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
10 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2019 · 10 January · Evening Shift · Mathematics · Functions

Let N be the set of natural numbers and two functions f and g be...

Let NN be the set of natural numbers and two functions fandf andgbedefinedasbe defined asf, g: N \to Nsuchthat<br/><br/>such that<br /><br />f\left(n\right) = \left\{n+12,ifnisoddn2,ifniseven\begin{array}{ll}\dfrac{n + 1}{2}, ifnisodd \\ \dfrac{n}{2}, ifniseven\end{array}\right)

and g(n)=n−(−1)n.g\left(n\right) = n - {\left(- 1\right)}^{n} . Then fogfog 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)=g\left(n\right) =\left\{n+1,ifnoddn−1,ifneven\begin{array}{ll}n + 1, ifnodd \\ n - 1, ifneven\end{array}\right)

f(g(1))=f(2)=1f\left(g\left(1\right)\right) = f ( 2 ) = 1

f(g(2))=f(1)=1f\left(g\left(2\right)\right) = f ( 1 ) = 1

∴f(g(x))∴ f\left(g\left(x\right)\right) is many one.

f(g(2k))=f(2k−1)=kf\left(g\left(2k\right)\right) = f ( 2k - 1 ) = k

f(g(2k+1))=f(2k+2)=k+1f\left(g\left(2k + 1\right)\right) = f\left(2k + 2\right) = k + 1

∴f(g(x))∴ f ( g ( x ) ) is onto.
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