Test Index

Functions Part 1

Section: Mathematics

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Question : 13 of 100

Marks: +1, -0

Let

f, g: \mathbb{N}-\{1\} \rightarrow \mathbb{N}

be functions defined by

f(a)=\alpha

\alpha

is the maximum of the powers of those primes

p

p^\alpha

a

g(a)=a+1

a \in \mathbb{N}-\{1\}

. Then, the function

f+g

[27-Jul-2022-Shift-1]

one-one but not onto
onto but not one-one
both one-one and onto
neither one-one nor onto

Solution:

f, g: N-\{1\} \rightarrow N

defined as

f(a)=\alpha

, where

\alpha

is the maximum power of those primes

p

such that

p^\alpha

divides a.

g(a)=a+1

Now,

f(2)=1, \; g(2)=3 \; \Rightarrow \; (f+g)(2)=4

f(3)=1, \; g(3)=4 \; \Rightarrow \; (f+g)(3)=5

f(4)=2, \; g(4)=5 \Rightarrow (f+g)(4)=7

f(5)=1, \; g(5)=6 \Rightarrow (f+g)(5)=7

\because (f+g)(5)=(f+g)(4)

\therefore f+g

is not one-one

Now,

\because f_{\min}=1, g_{\min}=3

So, there does not exist any

x \in N-\{1\}

such that

(f+g)(x)=1,2,3

\therefore f+g

is not onto

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