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A Function F from the Set of Natural Numbers to the Set of Integers Defined by (A) Neither One-one Nor onto (B) One-one but Not onto (C) onto but Not One-one (D) One-one and onto - Mathematics

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Question

A function f from the set of natural numbers to the set of integers defined by

\[f\left( n \right)\begin{cases}\frac{n - 1}{2}, & \text{when n is odd} \\ - \frac{n}{2}, & \text{when n is even}\end{cases}\]

 

Options

  • neither one-one nor onto

  • one-one but not onto

  • onto but not one-one

  • one-one and onto

MCQ
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Solution

Injectivity:
Let x and y be any two elements in the domain (N).

\[Case-1: \text{Both  x and y are even}.\] 

\[\text{Let}f\left( x \right) = f\left( y \right)\] 
\[ \Rightarrow \frac{- x}{2} = \frac{- y}{2}\] 
\[ \Rightarrow - x = - y\] 
\[ \Rightarrow x = y\] 
\[Case-2: \text{Both x and y are odd}.\] 
\[\text{Let}f\left( x \right) = f\left( y \right)\] 
\[ \Rightarrow \frac{x - 1}{2} = \frac{y - 1}{2}\] 
\[ \Rightarrow x - 1 = y - 1\] 
\[ \Rightarrow x = y\] 
\[Case-3: \text{ Let  x be even and y be odd}.\] 
\[\text{Then},f\left( x \right) = \frac{- x}{2}\text{and}f\left( y \right) = \frac{y - 1}{2}\] 
\[\text{Then, clearly}\] 
\[x \neq y \] 
\[ \Rightarrow f\left( x \right) \neq f\left( y \right)\] 
\[\text{From all the cases,fis one-one}.\] 
Surjectivity:
\[\text{Co-domain off} = Z = \left\{ . . . , - 3, - 2, - 1, 0, 1, 2, 3, . . . . \right\}\] 

\[\text{Range of f} = \left\{ . . . , \frac{- 3 - 1}{2}, \frac{- \left( - 2 \right)}{2}, \frac{- 1 - 1}{2}, \frac{0}{2}, \frac{1 - 1}{2}, \frac{- 2}{2}, \frac{3 - 1}{2}, . . . \right\}\] 
\[ \Rightarrow \text{Range of f} = \left\{ . . . , - 2, 1, - 1, 0, 0, - 1, 1, . . . \right\}\] 
\[ \Rightarrow\text{Range of f } = \left\{ . . . , - 2, - 1, 0, 1, 2, . . . . \right\}\] 
\[ \Rightarrow \text{Co-domain of f = Range of f}\] 
\[\Rightarrow\] f is onto.
So, the answer is (d).
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Chapter 2: Functions - Exercise 2.6 [Page 77]

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RD Sharma Mathematics [English] Class 12
Chapter 2 Functions
Exercise 2.6 | Q 26 | Page 77

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