Function f : N →Z; f(n) = {cases} n/2 & {n-even} \\ -n-1/2 & {n-odd} {cases} is________.

Correct Answer is : one-one and onto

Function f : N →Z; f(n) = {cases} n/2 & {n-even} \\ -n-1/2 & {n-odd} {cases} is________.

Correct Answer is : one-one and onto

Test Index

GUJCET 2018 Math Solved Paper

Question : 36 of 40

Marks: +1, -0

f(n) = \begin{cases} \frac{n}{2} & \text{n-even} \\ -\frac{n-1}{2} & \text{n-odd} \end{cases}

is________.

not one-one and not onto
one-one but not onto
not one-one but onto
one-one and onto

Validate

Solution:

f(n) =

\begin{cases} \frac{n}{2} \text{;} & \text{n-even} \\ -\frac{n-1}{2} \text{;} & \text{n-odd} \end{cases}

when

n=2,4,6

, 8.1....

f(n)=\{1,2,3 \dots\}

when

n=1,3,5 \dots

f(n)=\{0,-1,-2,-3 \dots\}

\therefore

Function is one-one and onto

Clealy f is one-one function as for each even natural number we have different image as positive integer and for odd natural number image set is set of non-positive integer. also their are no any integer for which pre image does not exists. Henceone-one onto

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