Vardaan Learning Institute
In set theory, the order of elements within a set does not matter (e.g. $\{1, 2\} = \{2, 1\}$). However, in many mathematical applications, the order of elements is critical. This motivates the concept of an Ordered Pair.
An ordered pair consists of two objects or elements in a given fixed order, denoted as $(a, b)$, where $a$ is the first element (first coordinate/abscissa) and $b$ is the second element (second coordinate/ordinate).
Equality of Ordered Pairs: Two ordered pairs $(a, b)$ and $(c, d)$ are equal if and only if their corresponding first and second elements are equal:
$$(a, b) = (c, d) \iff a = c \quad\text{and}\quad b = d$$Given two non-empty sets $A$ and $B$, the Cartesian product $A \times B$ (read as "$A$ cross $B$") is the set of all ordered pairs $(a, b)$ such that $a \in A$ and $b \in B$:
$$A \times B = \{(a, b) : a \in A \text{ and } b \in B\}$$If either $A = \emptyset$ or $B = \emptyset$, then $A \times B = \emptyset$.
Cardinality Principle: If $n(A) = p$ and $n(B) = q$, then:
$$n(A \times B) = n(A) \times n(B) = p \times q$$Ordered Triplet & $\mathbb{R}^3$: $A \times A \times A = \{(a, b, c) : a, b, c \in A\}$. In particular, $\mathbb{R} \times \mathbb{R} = \mathbb{R}^2$ represents the 2D Cartesian plane, and $\mathbb{R} \times \mathbb{R} \times \mathbb{R} = \mathbb{R}^3$ represents 3-dimensional Euclidean space.
If $\left(\dfrac{x}{3} + 1, \; y - \dfrac{2}{3}\right) = \left(\dfrac{5}{3}, \; \dfrac{1}{3}\right)$, find the values of $x$ and $y$.
1. Equate corresponding coordinates:
$\dfrac{x}{3} + 1 = \dfrac{5}{3} \implies \dfrac{x}{3} = \dfrac{5}{3} - 1 = \dfrac{2}{3} \implies \mathbf{x = 2}$.
$y - \dfrac{2}{3} = \dfrac{1}{3} \implies y = \dfrac{1}{3} + \dfrac{2}{3} = \dfrac{3}{3} \implies \mathbf{y = 1}$.
2. Result: $\mathbf{x = 2, \; y = 1}$.
The Cartesian product $A \times A$ has $9$ elements among which are found $(-1, 0)$ and $(0, 1)$. Find the set $A$ and the remaining elements of $A \times A$.
1. $n(A \times A) = 9 \implies [n(A)]^2 = 9 \implies n(A) = 3$.
2. Since $(-1, 0) \in A \times A$ and $(0, 1) \in A \times A$, all elements $-1, 0, 1$ must belong to $A$.
Since $n(A) = 3$, we have $\mathbf{A = \{-1, 0, 1\}}$.
3. $A \times A = \{(-1,-1), (-1,0), (-1,1), (0,-1), (0,0), (0,1), (1,-1), (1,0), (1,1)\}$.
4. Remaining 7 elements: $\{(-1,-1), (-1,1), (0,-1), (0,0), (1,-1), (1,0), (1,1)\}$.
In everyday life, a relation connects pairs of objects (e.g. "is brother of", "is greater than"). Mathematically, a relation formalises the connection between elements of two sets.
A relation $R$ from a non-empty set $A$ to a non-empty set $B$ is a subset of the Cartesian product $A \times B$:
$$R \subseteq A \times B$$If $(a, b) \in R$, we write $a \,R\, b$ (read as "$a$ is related to $b$ by relation $R$"). The element $b$ is called the image of $a$ under $R$, and $a$ is called a pre-image of $b$.
Total Number of Relations: If $n(A) = p$ and $n(B) = q$, then $n(A \times B) = pq$. Since any subset of $A \times B$ is a relation, the total number of relations from $A$ to $B$ is:
$$\text{Total Relations} = 2^{pq}$$The total number of non-empty relations is $2^{pq} - 1$.
Let $A = \{1, 2, 3, \dots, 14\}$. Define a relation $R$ from $A$ to $A$ by $R = \{(x, y) : 3x - y = 0, \text{ where } x, y \in A\}$. Write down its domain, codomain, and range.
1. Condition $3x - y = 0 \iff y = 3x$. Since $x, y \in A = \{1, 2, \dots, 14\}$:
• For $x = 1 \implies y = 3 \in A \implies (1, 3) \in R$
• For $x = 2 \implies y = 6 \in A \implies (2, 6) \in R$
• For $x = 3 \implies y = 9 \in A \implies (3, 9) \in R$
• For $x = 4 \implies y = 12 \in A \implies (4, 12) \in R$
• For $x = 5 \implies y = 15 \notin A$ (stops here).
2. Roster Form: $R = \{(1, 3), (2, 6), (3, 9), (4, 12)\}$.
3. $\mathbf{\text{Domain}(R) = \{1, 2, 3, 4\}}$.
4. $\mathbf{\text{Range}(R) = \{3, 6, 9, 12\}}$.
5. $\mathbf{\text{Codomain}(R) = A = \{1, 2, 3, \dots, 14\}}$.
Let $A = \{x, y, z\}$ and $B = \{1, 2\}$. Find the total number of relations from $A$ to $B$.
1. Here $n(A) = 3$ and $n(B) = 2$.
2. $n(A \times B) = 3 \times 2 = 6$.
3. Total number of subsets of $A \times B = 2^6 = \mathbf{64}$.
4. Hence, there are $\mathbf{64}$ relations from $A$ to $B$.
A function (or mapping) is a special kind of relation in which every input has a uniquely determined output.
A relation $f$ from a non-empty set $A$ to a non-empty set $B$ is said to be a function if:
We denote a function as $f: A \to B$, and write $y = f(x)$, where $y$ is the image of $x$ under $f$, and $x$ is the pre-image of $y$.
Total Functions: If $n(A) = p$ and $n(B) = q$, each of the $p$ elements of $A$ can choose any of the $q$ images in $B$. Thus, the total number of functions from $A$ to $B$ is:
$$\text{Total Functions} = q^p = [n(B)]^{n(A)}$$Examine whether the following relations are functions or not. Give reasons:
(i) $R_1 = \{(2, 1), (3, 1), (4, 2)\}$
(ii) $R_2 = \{(2, 2), (2, 4), (3, 3), (4, 4)\}$
(iii) $R_3 = \{(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7)\}$
1. $R_1$: Elements $2, 3, 4$ have unique images ($1, 1, 2$). Even though $2$ and $3$ map to the same element $1$, each input has exactly one image. Hence, $R_1$ is a function.
2. $R_2$: Element $2$ appears twice as the first coordinate with two different images ($2$ and $4$). Hence, $R_2$ is NOT a function.
3. $R_3$: Every element $1, 2, 3, 4, 5, 6$ has a distinct and unique image. Hence, $R_3$ is a function.
A function $f: A \to B$ is called a real-valued function if $B \subseteq \mathbb{R}$. If both $A \subseteq \mathbb{R}$ and $B \subseteq \mathbb{R}$, it is called a real function.
| Function Name | Equation $f(x)$ | Domain | Range | Graph Characteristics |
|---|---|---|---|---|
| Identity Function | $f(x) = x$ | $\mathbb{R}$ | $\mathbb{R}$ | Straight line passing through the origin at $45^\circ$ angle. |
| Constant Function | $f(x) = c$ | $\mathbb{R}$ | $\{c\}$ | Horizontal line parallel to $x$-axis at distance $|c|$. |
| Square Function | $f(x) = x^2$ | $\mathbb{R}$ | $[0, \infty)$ | Upward-opening parabola with vertex at $(0, 0)$. |
| Cubic Function | $f(x) = x^3$ | $\mathbb{R}$ | $\mathbb{R}$ | S-shaped curve symmetric about the origin. |
| Reciprocal Function | $f(x) = \dfrac{1}{x}$ | $\mathbb{R} \setminus \{0\}$ | $\mathbb{R} \setminus \{0\}$ | Rectangular hyperbola in Quadrants I and III. |
• Domain: $\mathbb{R}$ • Range: $[0, \infty)$ (Non-negative real numbers)
• Graph: V-shaped, symmetric about the $y$-axis with sharp corner at $(0, 0)$.
• Domain: $\mathbb{R}$ • Range: $\{-1, 0, 1\}$ (A finite 3-element set!)
$f(x) = [x]$ represents the greatest integer less than or equal to $x$.
$$[2.7] = 2, \quad [5] = 5, \quad [0.8] = 0, \quad [-1.3] = -2, \quad [-0.4] = -1$$• Domain: $\mathbb{R}$ • Range: $\mathbb{Z}$ (Set of all integers)
Core Properties:
Find the domain and range of the real function $f(x) = -|x|$.
1. Domain: Since $|x|$ is defined for all real numbers $x \in \mathbb{R}$, $\mathbf{\text{Domain} = \mathbb{R}}$.
2. Range: For any $x \in \mathbb{R}$, $|x| \ge 0$. Multiplying by $-1$ reverses the inequality:
$-|x| \le 0 \implies f(x) \le 0$.
3. Thus, the range consists of all non-positive real numbers: $\mathbf{\text{Range} = (-\infty, 0]}$.
Solve the equation for $x \in \mathbb{R}$: $2[x] = x + \{x\}$.
1. Recall $x = [x] + \{x\} \implies \{x\} = x - [x]$.
2. Substitute $x = [x] + \{x\}$ into the equation:
$2[x] = ([x] + \{x\}) + \{x\} = [x] + 2\{x\} \implies [x] = 2\{x\} \implies \{x\} = \dfrac{[x]}{2}$.
3. By definition, $0 \le \{x\} < 1$. Therefore:
$0 \le \dfrac{[x]}{2} < 1 \implies 0 \le [x] < 2$.
4. Since $[x]$ must be an integer, the only possible values for $[x]$ are $0$ and $1$:
• If $[x] = 0 \implies \{x\} = 0/2 = 0 \implies x = [x] + \{x\} = 0 + 0 = \mathbf{0}$.
• If $[x] = 1 \implies \{x\} = 1/2 \implies x = [x] + \{x\} = 1 + \dfrac{1}{2} = \mathbf{\dfrac{3}{2}}$.
5. Solutions: $\mathbf{x \in \left\{0, \; \dfrac{3}{2}\right\}}$.
The domain of a real function $f(x)$ is the set of all real numbers $x$ for which $f(x)$ produces a well-defined real number.
1. Direct Bounds: Use known inequalities (e.g. $x^2 \ge 0$, $|x| \ge 0$, $-1 \le \sin x \le 1$, $0 \le \sqrt{g(x)} < \infty$).
2. Express $x$ in terms of $y$: Let $y = f(x)$, solve for $x = \phi(y)$, and find the conditions on $y$ that make $x$ real and in the domain of $f$.
3. Quadratic Vertex: For $y = ax^2 + bx + c$, if $a > 0$, range is $\left[-\dfrac{D}{4a}, \infty\right)$; if $a < 0$, range is $\left(-\infty, -\dfrac{D}{4a}\right]$.
4. Semi-Circle Identity: For $f(x) = \sqrt{a^2 - x^2}$, the domain is $[-a, a]$ and the range is $[0, a]$.
Find the domain and range of the real function $f(x) = \sqrt{9 - x^2}$.
1. Domain: For $f(x)$ to be real, the radicand must be non-negative:
$9 - x^2 \ge 0 \iff x^2 \le 9 \iff -3 \le x \le 3$.
$\mathbf{\text{Domain} = [-3, 3]}$.
2. Range: For any $x \in [-3, 3]$, $0 \le x^2 \le 9$.
Multiplying by $-1$: $-9 \le -x^2 \le 0$.
Adding $9$: $0 \le 9 - x^2 \le 9$.
Taking principal square root: $0 \le \sqrt{9 - x^2} \le 3$.
$\mathbf{\text{Range} = [0, 3]}$.
Find the range of the function $f(x) = \dfrac{x^2}{1 + x^2}$ for $x \in \mathbb{R}$.
1. Let $y = \dfrac{x^2}{1 + x^2}$. Since $x^2 \ge 0$ and $1 + x^2 > 0$, we have $y \ge 0$.
2. Cross-multiply: $y(1 + x^2) = x^2 \implies y + y x^2 = x^2 \implies x^2(1 - y) = y \implies x^2 = \dfrac{y}{1 - y}$.
3. Since $x$ is real, $x^2 \ge 0$. Thus:
$\dfrac{y}{1 - y} \ge 0 \iff \dfrac{y}{y - 1} \le 0 \iff 0 \le y < 1$ (note $y \neq 1$ as denominator cannot be zero).
4. Therefore: $\mathbf{\text{Range} = [0, 1)}$.
Let $f: X \to \mathbb{R}$ and $g: X \to \mathbb{R}$ be two real functions with domains $D_f$ and $D_g$ respectively.
1. Addition: $(f + g)(x) = f(x) + g(x)$, $\text{Domain} = D_f \cap D_g$.
2. Subtraction: $(f - g)(x) = f(x) - g(x)$, $\text{Domain} = D_f \cap D_g$.
3. Multiplication: $(f \cdot g)(x) = f(x) \cdot g(x)$, $\text{Domain} = D_f \cap D_g$.
4. Scalar Multiple: $(c \cdot f)(x) = c \cdot f(x)$, $\text{Domain} = D_f$.
5. Quotient: $\left(\dfrac{f}{g}\right)(x) = \dfrac{f(x)}{g(x)}$, $\mathbf{\text{Domain} = (D_f \cap D_g) \setminus \{x : g(x) = 0\}}$.
Let $f(x) = \sqrt{x}$ and $g(x) = x$. Find $(f+g)(x)$, $(f-g)(x)$, $(fg)(x)$, and $\left(\dfrac{f}{g}\right)(x)$ along with their domains.
1. Domains: $D_f = [0, \infty)$ and $D_g = \mathbb{R}$. Thus $D_f \cap D_g = [0, \infty)$.
• $(f+g)(x) = \sqrt{x} + x, \quad \text{Domain} = [0, \infty)$.
• $(f-g)(x) = \sqrt{x} - x, \quad \text{Domain} = [0, \infty)$.
• $(fg)(x) = x\sqrt{x} = x^{3/2}, \quad \text{Domain} = [0, \infty)$.
• $\left(\dfrac{f}{g}\right)(x) = \dfrac{\sqrt{x}}{x} = \dfrac{1}{\sqrt{x}}$, where $g(x) = x \neq 0 \implies \mathbf{\text{Domain} = (0, \infty)}$.
Let $f$ be a real function defined on a symmetric domain (i.e. $x \in D_f \iff -x \in D_f$).
A function $f(x)$ is said to be periodic if there exists a positive real number $T$ such that:
$$f(x + T) = f(x) \quad \forall x \in D_f$$The smallest such positive value $T$ is called the fundamental period of $f(x)$.
Standard Periods:
If $f(x)$ is a polynomial function satisfying $f(x)f(1/x) = f(x) + f(1/x)$ and $f(4) = 65$, find $f(6)$.
1. The standard polynomial satisfying $f(x) + f(1/x) = f(x)f(1/x)$ is of the form $f(x) = 1 \pm x^n$.
2. Given $f(4) = 65$: $1 \pm 4^n = 65 \implies \pm 4^n = 64 = 4^3$.
Thus, the sign is positive ($+$) and $n = 3$. Hence, $\mathbf{f(x) = 1 + x^3}$.
3. Now compute $f(6)$: $f(6) = 1 + 6^3 = 1 + 216 = \mathbf{217}$.
Find the domain of $f(x) = \dfrac{1}{\sqrt{|x| - x}}$.
1. The expression under the square root in the denominator must be strictly positive:
$|x| - x > 0 \iff |x| > x$.
2. Recall that $|x| = x$ for all $x \ge 0$, and $|x| > x$ strictly when $x < 0$.
3. Therefore, the inequality holds if and only if $x < 0$.
4. $\mathbf{\text{Domain} = (-\infty, 0)}$.
| Concept | Key Result |
|---|---|
| Cartesian Product Cardinality | $n(A \times B) = n(A) \cdot n(B) = p \cdot q$ |
| Total Relations | $2^{n(A) \cdot n(B)} = 2^{pq}$ |
| Total Functions | $[n(B)]^{n(A)} = q^p$ |
| Modulus Function $|x|$ | Domain: $\mathbb{R}$, Range: $[0, \infty)$ |
| Signum Function $\text{sgn}(x)$ | Domain: $\mathbb{R}$, Range: $\{-1, 0, 1\}$ |
| Greatest Integer $[x]$ | Domain: $\mathbb{R}$, Range: $\mathbb{Z}$ |
| Semi-Circle $\sqrt{a^2-x^2}$ | Domain: $[-a, a]$, Range: $[0, a]$ |
| Algebra of Functions Domain | Domain of $f \pm g, fg$ is $D_f \cap D_g$; for $f/g$ it is $(D_f \cap D_g) \setminus \{x : g(x) = 0\}$ |