Let N be the set of natural numbers and two functions f and g be defined as f,g:N→N such that fn= (n+1)/2, if n is odd n/2, if n is even and gn=n--1^n. Then fog is:

Correct answer: (A) onto but not one-one. We have, gn= n+1, if n odd n-1, if n even fg1=f(2)=1 fg2=f(1)=1 ∴fgx is many one. fg2k=f(2k-1)=k fg2k+1=f2k+2=k+1 ∴f(g(x)) is onto.

Q59 / 90 · JEE Main 2019 (10 Jan Shift 2) · Mathematics

Let $N$ be the set of natural numbers and two functions $f$ and $g$ be defined as $f, g : N \to N$ such that

$f (n) = \{\begin{matrix} \frac{n + 1}{2}, i f n i s o d d \\ \frac{n}{2}, i f n i s e v e n \end{matrix}$

and $g (n) = n - {(- 1)}^{n} .$ Then $f o g$ is:

Correct answer
A
onto but not one-one
B
Both one-one and onto
C
One-one but not onto
D
Neither one-one nor onto

Solution

We have,

g (n) = \{\begin{matrix} n + 1, i f n o d d \\ n - 1, i f n e v e n \end{matrix}

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.

Hence, the correct answer is option: 'A'.

Let N be the set of natural numbers and two functions f and g be defined as f,g:N→N such thatfn=n+12, if n is odd n2, if n is evenand gn=n--1n. Then fog is:

Solution

Relevant Questions

Let $N$ be the set of natural numbers and two functions $f$ and $g$ be defined as $f, g : N \to N$ such that

$f (n) = \{\begin{matrix} \frac{n + 1}{2}, i f n i s o d d \\ \frac{n}{2}, i f n i s e v e n \end{matrix}$

and $g (n) = n - {(- 1)}^{n} .$ Then $f o g$ is: