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Q9:
Let $f : N \to N$ 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 $n \in N$. State whether the function $f$ is bijective. Justify your answer.

Solution :

Step 1: Analyzing Injectivity (One-to-One Property)

A function $f: A \to B$ is injective if $f(n_1) = f(n_2) \implies n_1 = n_2$. We evaluate this by testing values for $n \in \mathbb{N}$.

  • Consider $n_1 = 1$ (odd): $f(1) = \frac{1+1}{2} = 1$.
  • Consider $n_2 = 2$ (even): $f(2) = \frac{2}{2} = 1$.

[By Definition of Injectivity]: Since $f(1) = f(2) = 1$ but $1 \neq 2$, the function $f$ fails the condition for injectivity. Therefore, $f$ is not one-to-one.

Step 2: Analyzing Surjectivity (Onto Property)

A function $f: A \to B$ is surjective if for every $y \in \text{Codomain}$, there exists an $x \in \text{Domain}$ such that $f(x) = y$. Here, the codomain is $\mathbb{N}$.

Case 1: If $y$ is odd, let $n = 2y - 1$. Since $y \geq 1$, $n$ is an odd natural number.
$f(2y-1) = \frac{(2y-1)+1}{2} = \frac{2y}{2} = y$.

Case 2: If $y$ is even, let $n = 2y$. Since $y \geq 1$, $n$ is an even natural number.
$f(2y) = \frac{2y}{2} = y$.

[By Definition of Surjectivity]: Since every element in the codomain $\mathbb{N}$ has a pre-image in the domain $\mathbb{N}$, the function $f$ is onto.

Visual Representation of Mapping

Domain (N) Codomain (N) 1 2 3 1 2

Step 3: Conclusion on Bijectivity

A function is defined as bijective if and only if it is both injective and surjective.
[Per the Definition of Bijective Functions]: Since $f$ is surjective but not injective, it cannot be bijective.

Final Solution: The function $f$ is not bijective because it is not injective (many-to-one), even though it is surjective.


More Questions from Class 12 Mathematics Relations and Functions EXERCISE 1.2


CBSE Solutions for Class 12 Mathematics Relations and Functions


Chapters in CBSE - Class 12 Mathematics


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