A function f from the set of natural numbers to integers defined by f( n ) = \ \beginmatrix n - 1 \over 2,\,when\,n\,is\,odd - n \over 2,\,when\,n\,is\,even \endmatrix . is

Functions PYQ — JEE Main 2003 · Shift 0 · Q67 with Solution

View written solutionFree

Correct answer: D

Given function

The function is defined from natural numbers to integers:

$f: \mathbb{N} \to \mathbb{Z}$

with

\begin{cases} \dfrac{n-1}{2}, & \text{if } n \text{ is odd}\\[4pt] -\dfrac{n}{2}, & \text{if } n \text{ is even} \end{cases}$$ We must check whether it is **one-one** (injective) and/or **onto** (surjective). --- 2. **Understand the values of the function** Let us compute a few values: - For $n=1$ (odd): $$f(1)=\frac{1-1}{2}=0$$ - For $n=2$ (even): $$f(2)=-\frac{2}{2}=-1$$ - For $n=3$ (odd): $$f(3)=\frac{3-1}{2}=1$$ - For $n=4$ (even): $$f(4)=-\frac{4}{2}=-2$$ - For $n=5$ (odd): $$f(5)=\frac{5-1}{2}=2$$ - For $n=6$ (even): $$f(6)=-\frac{6}{2}=-3$$ So the outputs are: $$0,-1,1,-2,2,-3,3,\dots$$ This suggests that every integer appears exactly once. --- 3. **Check one-one (injective)** We show that different natural numbers give different integer values. ### Case 1: Both numbers are odd Let $n=2a+1$ and $m=2b+1$. Then $$f(n)=\frac{(2a+1)-1}{2}=a, \qquad f(m)=\frac{(2b+1)-1}{2}=b$$ If $f(n)=f(m)$, then $a=b$, hence $$n=2a+1=2b+1=m$$ So no two distinct odd numbers have the same image. ### Case 2: Both numbers are even Let $n=2a$ and $m=2b$. Then $$f(n)=-a, \qquad f(m)=-b$$ If $f(n)=f(m)$, then $$-a=-b \implies a=b \implies n=m$$ So no two distinct even numbers have the same image. ### Case 3: One odd and one even Let $n$ be odd and $m$ be even. Then $$f(n)=\frac{n-1}{2}$$ is a **non-negative integer**, while $$f(m)=-\frac{m}{2}$$ is a **negative integer**. A non-negative integer can never equal a negative integer. Hence $$f(n) \ne f(m)$$ So an odd input and an even input can never have the same image. Therefore, the function is **one-one**. --- 4. **Check onto (surjective)** We must show that for every integer $z \in \mathbb{Z}$, there exists some natural number $n$ such that $f(n)=z$. Take any integer $z$. ### If $z \ge 0$ Choose $$n=2z+1$$ Then $n$ is odd, and $$f(n)=\frac{(2z+1)-1}{2}=\frac{2z}{2}=z$$ ### If $z<0$ Write $z=-k$ where $k\in \mathbb{N}$. Choose $$n=2k$$ Then $n$ is even, and $$f(n)=-\frac{2k}{2}=-k=z$$ Thus every integer has a preimage in $\mathbb{N}$. So the function is **onto**. --- 5. **Conclusion** The function is both **one-one** and **onto**. Hence the correct option is: $$\boxed{\text{D: one-one and onto both}}$$ --- 6. **Comparison with stored correct answer** Stored correct answer: $\boxed{D}$ Our derived answer is also $\boxed{D}$, so they agree.

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View written solutionFree

Correct answer: D

Given function

The function is defined from natural numbers to integers:

$f: \mathbb{N} \to \mathbb{Z}$

with

\begin{cases} \dfrac{n-1}{2}, & \text{if } n \text{ is odd}\\[4pt] -\dfrac{n}{2}, & \text{if } n \text{ is even} \end{cases}$$ We must check whether it is **one-one** (injective) and/or **onto** (surjective). --- 2. **Understand the values of the function** Let us compute a few values: - For $n=1$ (odd): $$f(1)=\frac{1-1}{2}=0$$ - For $n=2$ (even): $$f(2)=-\frac{2}{2}=-1$$ - For $n=3$ (odd): $$f(3)=\frac{3-1}{2}=1$$ - For $n=4$ (even): $$f(4)=-\frac{4}{2}=-2$$ - For $n=5$ (odd): $$f(5)=\frac{5-1}{2}=2$$ - For $n=6$ (even): $$f(6)=-\frac{6}{2}=-3$$ So the outputs are: $$0,-1,1,-2,2,-3,3,\dots$$ This suggests that every integer appears exactly once. --- 3. **Check one-one (injective)** We show that different natural numbers give different integer values. ### Case 1: Both numbers are odd Let $n=2a+1$ and $m=2b+1$. Then $$f(n)=\frac{(2a+1)-1}{2}=a, \qquad f(m)=\frac{(2b+1)-1}{2}=b$$ If $f(n)=f(m)$, then $a=b$, hence $$n=2a+1=2b+1=m$$ So no two distinct odd numbers have the same image. ### Case 2: Both numbers are even Let $n=2a$ and $m=2b$. Then $$f(n)=-a, \qquad f(m)=-b$$ If $f(n)=f(m)$, then $$-a=-b \implies a=b \implies n=m$$ So no two distinct even numbers have the same image. ### Case 3: One odd and one even Let $n$ be odd and $m$ be even. Then $$f(n)=\frac{n-1}{2}$$ is a **non-negative integer**, while $$f(m)=-\frac{m}{2}$$ is a **negative integer**. A non-negative integer can never equal a negative integer. Hence $$f(n) \ne f(m)$$ So an odd input and an even input can never have the same image. Therefore, the function is **one-one**. --- 4. **Check onto (surjective)** We must show that for every integer $z \in \mathbb{Z}$, there exists some natural number $n$ such that $f(n)=z$. Take any integer $z$. ### If $z \ge 0$ Choose $$n=2z+1$$ Then $n$ is odd, and $$f(n)=\frac{(2z+1)-1}{2}=\frac{2z}{2}=z$$ ### If $z<0$ Write $z=-k$ where $k\in \mathbb{N}$. Choose $$n=2k$$ Then $n$ is even, and $$f(n)=-\frac{2k}{2}=-k=z$$ Thus every integer has a preimage in $\mathbb{N}$. So the function is **onto**. --- 5. **Conclusion** The function is both **one-one** and **onto**. Hence the correct option is: $$\boxed{\text{D: one-one and onto both}}$$ --- 6. **Comparison with stored correct answer** Stored correct answer: $\boxed{D}$ Our derived answer is also $\boxed{D}$, so they agree.

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