1. Cartesian Product of Sets

Ordered Pair

An ordered pair consists of two objects or elements in a given fixed order.

  • Notation: (a,b)(a, b)
  • Equality: Two ordered pairs (a,b)(a, b) and (c,d)(c, d) are equal if and only if a=ca = c and b=db = d.

Definition of Cartesian Product

Given two non-empty sets PP and QQ. The cartesian product P×QP \times Q is the set of all ordered pairs of elements from PP and QQ. P×Q={(p,q):pP,qQ}P \times Q = \{ (p, q) : p \in P, q \in Q \}

Important Note:

  • If n(A)=pn(A) = p and n(B)=qn(B) = q, then n(A×B)=pqn(A \times B) = pq.
  • If at least one of AA or BB is infinite, then A×BA \times B is infinite.
  • A×A×A={(a,b,c):a,b,cA}A \times A \times A = \{(a, b, c) : a, b, c \in A\}. This is called an ordered triplet.

2. Relations

A relation RR from a non-empty set AA to a non-empty set BB is a subset of the cartesian product A×BA \times B.

  • The subset is derived by describing a relationship between the first element (xx) and the second element (yy) of the ordered pair (x,y)(x, y).
  • The second element is called the image of the first element.

Domain and Range

  1. Domain: The set of all first elements of the ordered pairs in a relation RR.
  2. Range: The set of all second elements of the ordered pairs in a relation RR.
  3. Codomain: The whole set BB is called the codomain of the relation RR.
    • Note: Range \subseteq Codomain.

3. Types of Relations (Board Exam Priority)

A relation RR in a set AA is a subset of A×AA \times A.

(A) Empty Relation

A relation RR in a set AA is called empty relation, if no element of AA is related to any element of AA, i.e., R=ϕA×AR = \phi \subset A \times A.

(B) Universal Relation

A relation RR in a set AA is called universal relation, if each element of AA is related to every element of AA, i.e., R=A×AR = A \times A.

(C) Reflexive Relation

A relation RR in a set AA is called reflexive if (a,a)R(a, a) \in R for every aAa \in A.

(D) Symmetric Relation

A relation RR in a set AA is called symmetric if (a,b)R(a, b) \in R implies that (b,a)R(b, a) \in R, for all a,bAa, b \in A.

(E) Transitive Relation

A relation RR in a set AA is called transitive if (a,b)R(a, b) \in R and (b,c)R(b, c) \in R implies that (a,c)R(a, c) \in R, for all a,b,cAa, b, c \in A.

(F) Equivalence Relation

A relation RR in a set AA is said to be an equivalence relation if RR is reflexive, symmetric and transitive.


4. Functions

A relation ff from a set AA to a set BB is said to be a function if every element of set AA has one and only one image in set BB.

Key Conditions:

  1. All elements of AA must be mapped.
  2. Uniqueness: No element in AA can have more than one image in BB.

Notation: If ff is a function from AA to BB and (x,y)f(x, y) \in f, then f(x)=yf(x) = y.

  • yy is the image of xx under ff.
  • xx is the pre-image of yy.

5. Some Real Functions and their Graphs

(A) Identity Function

f:RRf: \mathbb{R} \to \mathbb{R} defined by y=f(x)=xy = f(x) = x for each xRx \in \mathbb{R}.

  • Graph: A straight line passing through origin with slope 1 (4545^\circ).
  • Domain: R\mathbb{R}
  • Range: R\mathbb{R}

(B) Constant Function

f:RRf: \mathbb{R} \to \mathbb{R} defined by y=f(x)=cy = f(x) = c, where cc is a constant.

  • Graph: A line parallel to the x-axis.
  • Domain: R\mathbb{R}
  • Range: {c}\{c\}

(C) Polynomial Function

f:RRf: \mathbb{R} \to \mathbb{R} defined by y=a0+a1x+a2x2++anxny = a_0 + a_1x + a_2x^2 + \dots + a_nx^n.

  • Example: Parabola f(x)=x2f(x) = x^2.
  • Domain: R\mathbb{R}, Range: [0,)[0, \infty).

(D) Rational Function

Functions of the type f(x)g(x)\frac{f(x)}{g(x)}, where f(x)f(x) and g(x)g(x) are polynomial functions and g(x)0g(x) \neq 0.

  • Example: f(x)=1xf(x) = \frac{1}{x}.
  • Domain: R{0}\mathbb{R} - \{0\}
  • Range: R{0}\mathbb{R} - \{0\}

(E) Modulus Function (Absolute Value)

f(x)=xf(x) = |x|. x={xif x0xif x<0|x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases}

  • Graph: V-shaped starting from origin.
  • Domain: R\mathbb{R}
  • Range: [0,)[0, \infty) (Non-negative real numbers)

(F) Signum Function

f(x)=xxf(x) = \frac{|x|}{x} for x0x \neq 0 and 00 for x=0x=0. f(x)={1if x>00if x=01if x<0f(x) = \begin{cases} 1 & \text{if } x > 0 \\ 0 & \text{if } x = 0 \\ -1 & \text{if } x < 0 \end{cases}

  • Domain: R\mathbb{R}
  • Range: {1,0,1}\{-1, 0, 1\}

(G) Greatest Integer Function

The function f:RRf: \mathbb{R} \to \mathbb{R} defined by f(x)=[x]f(x) = [x], where [x][x] assumes the value of the greatest integer less than or equal to xx.

  • Example: [2.3]=2,[1.4]=2[2.3] = 2, [-1.4] = -2.
  • Graph: Step function.
  • Domain: R\mathbb{R}
  • Range: Z\mathbb{Z} (Integers)

6. Algebra of Real Functions

Let f:XRf: X \to \mathbb{R} and g:XRg: X \to \mathbb{R} be two real functions.

  1. Addition: (f+g)(x)=f(x)+g(x)(f + g)(x) = f(x) + g(x)
  2. Subtraction: (fg)(x)=f(x)g(x)(f - g)(x) = f(x) - g(x)
  3. Multiplication: (fg)(x)=f(x)g(x)(fg)(x) = f(x)g(x)
  4. Quotient: (fg)(x)=f(x)g(x)(\frac{f}{g})(x) = \frac{f(x)}{g(x)}, provided g(x)0g(x) \neq 0.

Solved Examples (15+) — Relations and Functions

Example 1

If (x+1,y2)=(3,1)(x+1, y-2) = (3, 1), find the values of xx and yy.

Solution: Since ordered pairs are equal: x+1=3    x=2x + 1 = 3 \implies x = 2. y2=1    y=3y - 2 = 1 \implies y = 3. Ans: x=2,y=3x=2, y=3.


Example 2

If P={a,b,c}P = \{a, b, c\} and Q={r}Q = \{r\}, form the set P×QP \times Q and Q×PQ \times P.

Solution: P×Q={(a,r),(b,r),(c,r)}P \times Q = \{(a,r), (b,r), (c,r)\}. Q×P={(r,a),(r,b),(r,c)}Q \times P = \{(r,a), (r,b), (r,c)\}. Note that P×QQ×PP \times Q \neq Q \times P.


Example 3

Let A={1,2,3}A = \{1, 2, 3\}. Let R={(1,1),(2,2),(3,3),(1,2),(2,3)}R = \{(1,1), (2,2), (3,3), (1,2), (2,3)\}. Is RR reflexive?

Solution: Yes, because (1,1),(2,2),(3,3)(1,1), (2,2), (3,3) are all present in RR. Every element is related to itself.


Example 4

For the above relation RR, is it symmetric?

Solution: No. We have (1,2)R(1,2) \in R, but (2,1)R(2,1) \notin R.


Example 5

Is the relation R={(a,b):ab}R = \{(a,b): a \le b\} on R\mathbb{R} transitive?

Solution: Let (a,b)R(a,b) \in R and (b,c)R(b,c) \in R. This means aba \le b and bcb \le c. It implies aca \le c, so (a,c)R(a,c) \in R. Yes, it is transitive.


Example 6

Let A={1,2,3,4,5,6}A = \{1, 2, 3, 4, 5, 6\}. Define a relation RR from AA to AA by R={(x,y):y=x+1}R = \{(x, y) : y = x + 1\}. Write down the domain, codomain and range.

Solution: R={(1,2),(2,3),(3,4),(4,5),(5,6)}R = \{(1,2), (2,3), (3,4), (4,5), (5,6)\}.

  • Domain: {1,2,3,4,5}\{1, 2, 3, 4, 5\}.
  • Range: {2,3,4,5,6}\{2, 3, 4, 5, 6\}.
  • Codomain: {1,2,3,4,5,6}\{1, 2, 3, 4, 5, 6\}.

Example 7

Check if R={(1,1),(2,2),(3,3),(1,2),(2,1)}R = \{(1,1), (2,2), (3,3), (1,2), (2,1)\} on A={1,2,3}A=\{1,2,3\} is an equivalence relation.

Solution:

  1. Reflexive: Yes, (1,1),(2,2),(3,3)(1,1), (2,2), (3,3) present.
  2. Symmetric: Yes, (1,2)R    (2,1)R(1,2) \in R \implies (2,1) \in R.
  3. Transitive: (1,2)(1,2) and (2,1)R    (1,1)R(2,1) \in R \implies (1,1) \in R. Yes. Since it satisfies all three, it is an equivalence relation.

Example 8

Find the domain of the function f(x)=x2+2x+1x28x+12f(x) = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}.

Solution: Function is defined when denominator 0\neq 0. x28x+12=0    (x2)(x6)=0x^2 - 8x + 12 = 0 \implies (x-2)(x-6) = 0. x=2,6x = 2, 6. Domain = R{2,6}\mathbb{R} - \{2, 6\}.


Example 9

Find the domain and range of f(x)=9x2f(x) = \sqrt{9 - x^2}.

Solution: Domain: 9x20    x29    3x39 - x^2 \ge 0 \implies x^2 \le 9 \implies -3 \le x \le 3. Domain: [3,3][-3, 3]. Range: Let y=9x2y = \sqrt{9-x^2}. Since square root is non-negative, y0y \ge 0. Also max value is at x=0x=0, y=3y=3. Range: [0,3][0, 3].


Example 10

Find the domain of f(x)=1x5f(x) = \frac{1}{\sqrt{x - 5}}.

Solution: For square root in denominator: x5>0x - 5 > 0 (Strictly greater than). x>5x > 5. Domain: (5,)(5, \infty).


Example 11

Evaluate f(x)=x23x+1f(x) = x^2 - 3x + 1 at x=1x = -1.

Solution: f(1)=(1)23(1)+1=1+3+1=5f(-1) = (-1)^2 - 3(-1) + 1 = 1 + 3 + 1 = 5.


Example 12

Let f={(1,1),(2,3),(0,1),(1,3)}f = \{(1,1), (2,3), (0,-1), (-1,-3)\} be a function from Z\mathbb{Z} to Z\mathbb{Z} defined by f(x)=ax+bf(x) = ax + b. Determine aa and bb.

Solution: For (1,1):1=a(1)+b    a+b=1(1,1): 1 = a(1) + b \implies a+b=1. For (0,1):1=a(0)+b    b=1(0,-1): -1 = a(0) + b \implies b=-1. Substitute b=1b=-1 in first eq: a1=1    a=2a - 1 = 1 \implies a = 2. Function is f(x)=2x1f(x) = 2x - 1.


Example 13

Find the range of f(x)=23cosxf(x) = 2 - 3\cos x.

Solution: Range of cosx\cos x is [1,1][-1, 1]. Max value: 23(1)=52 - 3(-1) = 5. Min value: 23(1)=12 - 3(1) = -1. Range: [1,5][-1, 5].


Example 14

Find the range of f(x)=x3f(x) = |x - 3|.

Solution: Modulus function is always non-negative. x30|x - 3| \ge 0. Range: [0,)[0, \infty).


Example 15

Evaluate [4.7]+[4.7][4.7] + [-4.7] where [\] is Greatest Integer Function.

Solution: [4.7]=4[4.7] = 4. [4.7]=5[-4.7] = -5 (Greatest integer less than -4.7). Sum =4+(5)=1= 4 + (-5) = -1.

Questions and Answers (Board Exam)

Q1. Define an Equivalence Relation.

Answer: A relation RR in a set AA is said to be an equivalence relation if RR is reflexive, symmetric, and transitive.


Q2. What is the domain of a rational function f(x)=p(x)q(x)f(x) = \frac{p(x)}{q(x)}?

Answer: The domain is R{x:q(x)=0}\mathbb{R} - \{x : q(x) = 0\}.


Q3. Draw the graph of the Identity Function.

Answer: It is a straight line passing through the origin (0,0)(0,0) and making an angle of 4545^\circ with the positive x-axis (y=xy=x).


Q4. What is the range of the Signum function?

Answer: The range is the set {1,0,1}\{-1, 0, 1\}.


Q5. If n(A)=pn(A) = p and n(B)=qn(B) = q, how many relations are there from AA to BB?

Answer: Total relations = 2pq2^{pq} (Number of subsets of A×BA \times B).