Chapter 1: Relations and Functions
Welcome to the first chapter of Grade 12 Mathematics. This chapter explores relations between elements of sets and the functions that assign elements from one set to another.
Types of Relations
Given a set $A$, a relation $R$ on $A$ is a subset of $A \times A$. A relation $R$ in a set $A$ is called:
- Reflexive: if $(a, a) \in R$, for every $a \in A$.
- Symmetric: if $(a, b) \in R$ implies that $(b, a) \in R$, for all $a, b \in A$.
- Transitive: if $(a, b) \in R$ and $(b, c) \in R$ implies that $(a, c) \in R$, for all $a, b, c \in A$.
A relation $R$ in a set $A$ is said to be an equivalence relation if $R$ is reflexive, symmetric and transitive.
Fig 1. Visual representation of relation properties
Types of Functions
A function $f: X \rightarrow Y$ is defined as:
- One-One (Injective): If the images of distinct elements of $X$ under $f$ are distinct, i.e., for every $x_1, x_2 \in X$, $f(x_1) = f(x_2)$ implies $x_1 = x_2$.
- Onto (Surjective): If every element of $Y$ is the image of some element of $X$ under $f$, i.e., for every $y \in Y$, there exists an element $x$ in $X$ such that $f(x) = y$.
- Bijective: If $f$ is both one-one and onto.
Fig 2. One-One but not Onto Function
Competency-Based Questions
1. [CBSE 2021] Show that the relation $R$ in the set $A = \{1, 2, 3, 4, 5\}$ given by $R = \{(a, b) : |a - b| \text{ is even}\}$ is an equivalence relation.
Solution:
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Reflexive: For any element $a \in A$, $|a - a| = 0$, which is even. So, $(a, a) \in R$ for all $a \in A$. Therefore, $R$ is reflexive.
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Symmetric: Let $(a, b) \in R$. This means $|a - b|$ is even. Since $|a - b| = |b - a|$, $|b - a|$ is also even. So, $(b, a) \in R$. Therefore, $R$ is symmetric.
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Transitive: Let $(a, b) \in R$ and $(b, c) \in R$. This means $|a - b|$ is even and $|b - c|$ is even. Let $a - b = 2k_1$ and $b - c = 2k_2$ for some integers $k_1, k_2$. Then $a - c = (a - b) + (b - c) = 2(k_1 + k_2)$, which is even. So, $|a - c|$ is even. Thus, $(a, c) \in R$. Therefore, $R$ is transitive.
Since $R$ is reflexive, symmetric, and transitive, it is an equivalence relation.
2. [Sample Paper 2023] Check whether the function $f : \mathbb{R} \rightarrow \mathbb{R}$ defined by $f(x) = 1 + x^2$ is one-one, onto or bijective. Justify your answer.
Solution:
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One-One Check: Let $x_1, x_2 \in \mathbb{R}$ such that $f(x_1) = f(x_2)$. Then $1 + x_1^2 = 1 + x_2^2 \implies x_1^2 = x_2^2 \implies x_1 = \pm x_2$. Since $x_1 \neq x_2$ is possible (e.g., $f(1) = 2$ and $f(-1) = 2$), $f$ is not one-one.
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Onto Check: Let $y \in \mathbb{R}$ be in the codomain. We want to find $x \in \mathbb{R}$ such that $f(x) = y$. $1 + x^2 = y \implies x^2 = y - 1$. For $x$ to be a real number, we must have $y - 1 \ge 0 \implies y \ge 1$. If we take $y = 0 \in \mathbb{R}$ (codomain), there is no real $x$ such that $f(x) = 0$. Thus, $f$ is not onto.
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Since it is neither one-one nor onto, it is not bijective.
3. [CBSE 2019] Let $f : \mathbb{N} \rightarrow \mathbb{Y}$ be a function defined as $f(x) = 4x + 3$, where $Y = \{y \in \mathbb{N} : y = 4x + 3 \text{ for some } x \in \mathbb{N}\}$. Show that $f$ is invertible. Find the inverse.
Solution:
To show $f$ is invertible, we can show it is a bijection. However, given the definition of $Y$, $f$ is clearly onto $Y$. Let’s check one-one: $f(x_1) = f(x_2) \implies 4x_1 + 3 = 4x_2 + 3 \implies 4x_1 = 4x_2 \implies x_1 = x_2$. So $f$ is one-one. Since $f$ is a bijection, it is invertible.
To find the inverse, let $y = 4x + 3$. Then $4x = y - 3 \implies x = \frac{y - 3}{4}$. So, $f^{-1}(y) = \frac{y - 3}{4}$.