Let $N$ be the set of natural numbers and two function $f$ and $g$ be defined as $f,\,g:N \to N$ such that $f\left( n \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{n + 1}}{2};}&{{\rm{if}}\,\,n\,\,{\rm{is}}\,\,{\rm{odd}}}\\ {\frac{n}{2};}&{{\rm{if}}\,\,n\,\,{\rm{is}}\,\,{\rm{even}}} \end{array}} \right.$ and $g\left( n \right) = n - {\left( { - 1} \right)^n}$. Then $fog$ is

JEE MATHEMATICS MCQ: FUNCTIONS

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Analyzing +4 | -1

Let

N

be the set of natural numbers and two function

f

and

g

be defined as

f,\,g:N \to N

such that

f\left( n \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\frac{{n + 1}}{2};}&{\mathrm{if}\,\,n\,\,\mathrm{is}\,\,\mathrm{odd}}\\ {\frac{n}{2};}&{\mathrm{if}\,\,n\,\,\mathrm{is}\,\,\mathrm{even}} \end{array}} \right.

and

g\left( n \right) = n - {\left( { - 1} \right)^n}

. Then

fog

View answer and explanation

Correct option: B

Onto but not one – one

Option B (Onto but not one – one) is correct because the function fog is not one-to-one as f(2) = f(1) = 1, but it is onto as every element in the codomain has a pre-image in the domain.

JEE MATHEMATICS MCQ: FUNCTIONS

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