(C) Testing for one-one: $f(1) = \frac{1+1}{2} = 1$ and $f(2) = \frac{2}{2} = 1$. Since $f(1) = f(2)$ for $1 \neq 2$, the function is not one-one (it

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(C) Testing for one-one: $f(1) = \frac{1+1}{2} = 1$ and $f(2) = \frac{2}{2} = 1$. Since $f(1) = f(2)$ for $1 \neq 2$, the function is not one-one (it is many-one).Testing for onto: For any $y \in N$, if $y$ is odd, we can have $n = 2y-1$, then $f(2y-1) = \frac{(2y-1)+1}{2} = y$. If $y$ is even, we can have $n = 2y$, then $f(2y) = \frac{2y}{2} = y$. Thus, for every element in the codomain, there is a pre-image in the domain. Therefore, the function is onto.

Class 12 Mathematics Relation and Function Type of Functions based on Mapping

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Let $f: N \rightarrow N$ be defined by $f(n) = \begin{cases} \frac{n+1}{2}; & \text{if } n \text{ is odd} \\ \frac{n}{2}; & \text{if } n \text{ is even} \end{cases}$, for all $n \in N$ then $f$ is $\dots \dots \dots$

GUJCET 2026 All GUJCET

A
One-one but not onto
B
One-one and onto
C
Many-one and onto
D
Neither one-one nor onto

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Class 12

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Relation and Function

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Type of Functions based on Mapping

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GUJCET 2026 Paper All GUJCET years →

Show that the Signum Function $f: R \rightarrow R$,given by $f(x) = \begin{cases} 1, & \text{if } x > 0 \\ 0, & \text{if } x = 0 \\ -1, & \text{if } x < 0 \end{cases}$ is neither one-one nor onto.

Medium

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If $f : R \to R$ is defined by $f(x) = 2x + \cos x$,then $f$ is

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Let $A \subseteq R, B \subseteq R$ and $f: A \rightarrow B$ be defined by $f(x)=x^2-3x+2$. If $f$ is a bijection,then

MediumAP EAMCET 2023

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If $f: R \to R$ is a continuous function such that $|f(x) - f(y)| \geqslant |e^x - e^y|$ for all $x, y \in R$,then $f(x)$ is:

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Let $A = \{0, 1, 2, 3, 4, 5, 6, 7\}$. Then the number of bijective functions $f: A \rightarrow A$ such that $f(1) + f(2) = 3 - f(3)$ is equal to $.....$

DifficultJEE MAIN 2021

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Let $f: N \rightarrow N$ be defined by $f(n) = \begin{cases} \frac{n+1}{2}; & \text{if } n \text{ is odd} \\ \frac{n}{2}; & \text{if } n \text{ is even} \end{cases}$, for all $n \in N$ then $f$ is $\dots \dots \dots$

Similar Questions

Show that the Signum Function $f: R \rightarrow R$,given by $f(x) = \begin{cases} 1, & \text{if } x > 0 \\ 0, & \text{if } x = 0 \\ -1, & \text{if } x < 0 \end{cases}$ is neither one-one nor onto.

If $f : R \to R$ is defined by $f(x) = 2x + \cos x$,then $f$ is

Let $A \subseteq R, B \subseteq R$ and $f: A \rightarrow B$ be defined by $f(x)=x^2-3x+2$. If $f$ is a bijection,then

If $f: R \to R$ is a continuous function such that $|f(x) - f(y)| \geqslant |e^x - e^y|$ for all $x, y \in R$,then $f(x)$ is:

Let $A = \{0, 1, 2, 3, 4, 5, 6, 7\}$. Then the number of bijective functions $f: A \rightarrow A$ such that $f(1) + f(2) = 3 - f(3)$ is equal to $.....$

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