CBSE Class 11 Mathematics Relations Functions Worksheet Set 03

Case Based MCQs

Ordered Pairs The ordered pair of two elements a and b is denoted by (a, b ) : a is first element (or first component) and b is second element (or second component). Two ordered pairs are equal if their corresponding elements are equal. i.e. (a, b ) = (c , d ) Þ a = c and b = d Cartesian Product of Two Sets For two non-empty sets A and B, the cartesian product A × B is the set of all ordered pairs of elements from sets A and B. In symbolic form, it can be written as A × B = {(a, b ) : a ∈A, b ∈B}

Based on the above topics, answer the following questions.

Question. If (a − 3, b + 7) = (3, 7), then the value of a and b are
(a) 6, 0
(b) 3, 7
(c) 7, 0
(d) 3, −7
Answer : B

Question. If (x + 6, y − 2) = (0, 6), then the value of x and y are
(a) 6, 8
(b) − 6,−8
(c) −6,8
(d) 6,−8
Answer : A

Question. If (x + 2, 4) = (5, 2x + y ), then the value of x and y are
(a) −3 2 ,
(b) 3, 2
(c) −3,−2
(d) 3,−2
Answer : C

Question. Let A and B be two sets such that A × B consists of 6 elements. If three elements of A × B are (1, 4), (2, 6) and (3, 6), then
(a) (A × B) = (B × A)
(b) (A × B) ¹ (B × A)
(c) A × B = {(1,4), (1, 6), (2,4)}
(d) None of the above
Answer : A

Question. If n(A × B ) = 45, then n(A) cannot be
(a) 15
(b) 17
(c) 5
(d) 9
Answer : B

Representation of a Relation A relation can be represented algebraically by roster form or by set-builder form and visually it can be represented by an arrow diagram which are given below
(i) Roster form In this form, we represent the relation by the set of all ordered pairs belongs to R.
(ii) Set-builder form In this form, we represent the relation R from set A to set B as R = {(a, b ) : a ∈ A, b ∈B and the rule which relate the elements of A and B}.
(iii) Arrow diagram To represent a relation by an arrow diagram, we draw arrows from first element to second element of all ordered pairs belonging to relation R.

Based on the above topics, answer the following questions.

Question. Expression of R = {(a,b ): 2a +b =5; a,b ∈ W} as the set of ordered pairs (in roster form) is
(a) R = {(5, 0), (3, 1), (1, 2)}
(b) R = {(0, 5), (1, 3), (1, 2)}
(c) R = {(0, 5), (1, 3), (2, 1)}
(d) None of the above
Answer : C

Question. The relation given in (ii) can be written in set-builder form as
(a) R ={(x, y) :x ∈ P,y ∈ Q and x is the square of y}
(b) R ={(x, y) :x ∈ P,y ∈ Q and y is the square of x}
(c) R ={(x, y) :x ∈ P,y ∈ Q and x =± y}
(d) None of the above
Answer : A

Question. If A = {a, b} and B = {2, 3}, then the number of relations from A to B is
(a) 4
(b) 8
(c) 6
(d) 16
Answer : D

Question. If n (A) = 3 and B = {2, 3, 4, 6, 7, 8}, then the number of relations from A to B is
(a) 2³
(b) 2⁶
(c) 2¹⁸
(d) 2⁹
Answer : C

Assertion-Reasoning MCQs

Directions Each of these questions contains two statements Assertion (A) and Reason (R). Each of the questions has four alternative choices, any one of the which is the correct answer. You have to select one of the codes (a), (b), (c) and
(d) given below.
(a) A is true, R is true; R is a correct explanation of A.
(b) A is true, R is true; R is not a correct explanation of A.
(c) A is true; R is false.
(d) A is false; R is true.

Question. Let A = {1, 2, 3, 4, 6}. If R is the relation on A defined by {(a, b) : a, b ∈ A, b is exactly divisible by a}.
Assertion (A) The relation R in Roster form is {(6, 3), (6, 2), (4, 2)}.
Reason (R) The domain and range of R is {1, 2, 3, 4, 6}.
Answer : D

Question. Let R be a relation defined by R = {(x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}} Then, consider the following
Assertion (A) The domain of R is {0, 1, 2, 3, 4, 5}.
Reason (R) The range of R is {0, 1, 2, 3, 4, 5}.
Answer : C

Question. Assertion (A) The domain of the relation R = {(x + 2, x + 4) : x ∈ N, x < 8} is {3, 4, 5, 6, 7, 8, 9}.
Reason (R) The range of the relation R = {(x + 2, x + 4) : x ∈ N, x < 8} is {1, 2, 3, 4, 5, 6, 7}.
Answer : C

Question. Assertion (A) The range of the function f (x ) = 2 − 3x, x ∈ R, x > 0 is R.
Reason (R) The range of the function f(x ) = x² + 2 is [2, ∝).
Answer : D

Question. Assertion (A) Let A = {1, 2, 3, 5},
B = {4, 6, 9} and R = {(x, y) : |x − y | is odd, x ∈ A, y ∈ B}. Then, domain of R is {1, 2, 3, 5}.
Reason (R) |x |is always positive ∀ x ∈ R.
Answer : B
 

Question 1. If A = {2, 3}, B = {4, 5}, C = {5, 6} Find (i) A \times B (ii) A \times (B \cup C) (iii) A \times (B \cap C) Find (i) A \times (B \cup C) = (A \times B) \cup (A \times C) (ii) A \times (B - C) = (A \times B) - (A \times C)
Answer: Given sets are \(A = \{2, 3\}\), \(B = \{4, 5\}\), and \(C = \{5, 6\}\).
(i) \(A \times B = \{2, 3\} \times \{4, 5\} = \{(2, 4), (2, 5), (3, 4), (3, 5)\}\)
(ii) First find \(B \cup C = \{4, 5\} \cup \{5, 6\} = \{4, 5, 6\}\).
Now, \(A \times (B \cup C) = \{2, 3\} \times \{4, 5, 6\} = \{(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\}\)
(iii) First find \(B \cap C = \{4, 5\} \cap \{5, 6\} = \{5\}\).
Now, \(A \times (B \cap C) = \{2, 3\} \times \{5\} = \{(2, 5), (3, 5)\}\)
Verification of properties:
(i) Left Hand Side: \(A \times (B \cup C) = \{(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\}\)
Right Hand Side: \(A \times B = \{(2, 4), (2, 5), (3, 4), (3, 5)\}\)
\(A \times C = \{(2, 5), (2, 6), (3, 5), (3, 6)\}\)
\((A \times B) \cup (A \times C) = \{(2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\}\)
Since LHS = RHS, the property is verified.
(ii) First find \(B - C = \{4, 5\} - \{5, 6\} = \{4\}\).
Left Hand Side: \(A \times (B - C) = \{2, 3\} \times \{4\} = \{(2, 4), (3, 4)\}\)
Right Hand Side: \((A \times B) - (A \times C) = \{(2, 4), (2, 5), (3, 4), (3, 5)\} - \{(2, 5), (2, 6), (3, 5), (3, 6)\} = \{(2, 4), (3, 4)\}\)
Since LHS = RHS, the property is verified.
In simple words: Cartesian product of sets pairs each element of the first set with every element of the second set.

Exam Tip: Always compute set operations inside parentheses first before finding the Cartesian product.

Question 2. If the ordered pairs (x, -1) and (5, y) belong to the set \(\{ (a, b) : b = 2a - 3 \}\), find the values of x and y.
Answer: The relation is given by \(b = 2a - 3\).
Since \((x, -1)\) belongs to the set, substitute \(a = x\) and \(b = -1\):
\(-1 = 2x - 3\)
\(2x = 3 - 1\)
\(2x = 2 \implies x = 1\)
Since \((5, y)\) belongs to the set, substitute \(a = 5\) and \(b = y\):
\(y = 2(5) - 3\)
\(y = 10 - 3 = 7\)
Thus, \(x = 1\) and \(y = 7\).
In simple words: Substitute the coordinates of the given ordered pairs into the given equation to find the unknown variables.

Exam Tip: An ordered pair \((a, b)\) satisfying a relation means substituting \(a\) for the first coordinate and \(b\) for the second coordinate makes the equation true.

Question 3. If A and B are two sets having 3 elements in common. If n(A) = 5, n(B) = 4, find n(A \times B) and n((A \times B) \cap (B \times A))
Answer: Given \(n(A) = 5\), \(n(B) = 4\), and number of common elements in \(A\) and \(B\) is \(n(A \cap B) = 3\).
(i) \(n(A \times B) = n(A) \times n(B) = 5 \times 4 = 20\).
(ii) We know that \((A \times B) \cap (B \times A) = (A \cap B) \times (B \cap A)\).
Since \(A \cap B = B \cap A\), we have \(n((A \times B) \cap (B \times A)) = n((A \cap B) \times (A \cap B)) = n(A \cap B) \times n(A \cap B) = 3 \times 3 = 9\).
In simple words: The number of elements in a Cartesian product is the product of the number of elements in each individual set.

Question 4. Determine the domain and range of the following relations:-(i) \(R_1 = \left\{ \left(x, \frac{1}{x}\right) : 0 < x < 6, x \in N \right\}\) (ii) \(R_2 = \{ (x, x^3) : x \text{ is a prime number less than } 10 \}\)
Answer: (i) Given \(0 < x < 6\) and \(x \in N\), the possible values of \(x\) are \(1, 2, 3, 4, 5\).
Relation \(R_1 = \left\{ \left(1, 1\right), \left(2, \frac{1}{2}\right), \left(3, \frac{1}{3}\right), \left(4, \frac{1}{4}\right), \left(5, \frac{1}{5}\right) \right\}\)
Domain of \(R_1 = \{1, 2, 3, 4, 5\}\)
Range of \(R_1 = \left\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}\right\}\)
(ii) Prime numbers less than 10 are \(x \in \{2, 3, 5, 7\}\).
\(R_2 = \{ (2, 8), (3, 27), (5, 125), (7, 343) \}\)
Domain of \(R_2 = \{2, 3, 5, 7\}\)
Range of \(R_2 = \{8, 27, 125, 343\}\)
In simple words: Domain is the set of all first elements, and range is the set of all second elements in the ordered pairs of the relation.

Exam Tip: Carefully list out the elements of the relation first when the domain is a finite discrete set before extracting domain and range.

Question 5. Write all possible relation from A = \{1, 2\} to B = \{0\}.
Answer: Given \(A = \{1, 2\}\) and \(B = \{0\}\).
Number of elements in \(A\) is \(n(A) = 2\) and in \(B\) is \(n(B) = 1\).
Number of elements in the Cartesian product \(A \times B = n(A) \times n(B) = 2 \times 1 = 2\).
\(A \times B = \{(1, 0), (2, 0)\}\)
Total number of relations from \(A\) to \(B\) is \(2^{n(A \times B)} = 2^2 = 4\).
The possible relations are:
1. Empty relation: \(R_1 = \emptyset\)
2. \(R_2 = \{(1, 0)\}\)
3. \(R_3 = \{(2, 0)\}\)
4. Universal relation: \(R_4 = \{(1, 0), (2, 0)\}\)
In simple words: Any subset of the Cartesian product forms a valid relation. Since there are 2 elements in the product, there are 4 possible subsets.

Exam Tip: Total relations from set \(A\) to set \(B\) is given by \(2^{n(A) \cdot n(B)}\).

Question 6. How many relations are possible from a set A of n elements to another set B of n elements ?
Answer: Given number of elements in set \(A\) is \(n(A) = n\) and in set \(B\) is \(n(B) = n\).
The number of elements in the Cartesian product \(A \times B\) is \(n(A) \times n(B) = n \times n = n^2\).
A relation from \(A\) to \(B\) is a subset of \(A \times B\).
Therefore, the total number of possible relations is \(2^{n^2}\).
In simple words: A set with \(k\) elements has \(2^k\) subsets. Here \(k = n^2\).

Exam Tip: Do not confuse relations with functions; total relations is \(2^{n(A \times B)}\) whereas total functions from \(A\) to \(B\) is \((n(B))^{n(A)}\).

Question 7. Find the domain for which the functions \(f(x) = 3x^2 - 1\) and \(g(x) = 3 + x\) are equal.
Answer: For the functions \(f(x)\) and \(g(x)\) to be equal, we must have \(f(x) = g(x)\).
\(3x^2 - 1 = 3 + x\)
\(3x^2 - x - 1 - 3 = 0\)
\(3x^2 - x - 4 = 0\)
Solving the quadratic equation by factoring:
\(3x^2 - 4x + 3x - 4 = 0\)
\(x(3x - 4) + 1(3x - 4) = 0\)
\((3x - 4)(x + 1) = 0\)
Thus, \(x = \frac{4}{3}\) or \(x = -1\).
Therefore, the domain for which the functions are equal is \(\left\{ -1, \frac{4}{3} \right\}\).
In simple words: Equate the two function expressions and solve for \(x\) to find the exact points where their outputs match.

Exam Tip: Two functions are equal if they have the same domain and produce identical outputs for every input in that domain.

Question 8. Express the following functions as set of ordered pairs and determine their ranges. (a) \(f : A \to R, f(x) = x^2 + 1\), where \(A = \{-1, 0, 2, 4\}\) (b) \(g : A \to N, g(x) = 2x\), where \(A = \{x : x \in N, x < 10\}\)
Answer: (a) Given \(A = \{-1, 0, 2, 4\}\) and \(f(x) = x^2 + 1\):
For \(x = -1\): \(f(-1) = (-1)^2 + 1 = 1 + 1 = 2\)
For \(x = 0\): \(f(0) = (0)^2 + 1 = 0 + 1 = 1\)
For \(x = 2\): \(f(2) = (2)^2 + 1 = 4 + 1 = 5\)
For \(x = 4\): \(f(4) = (4)^2 + 1 = 16 + 1 = 17\)
Set of ordered pairs for \(f\) = \(\{(-1, 2), (0, 1), (2, 5), (4, 17)\}\)
Range of \(f = \{1, 2, 5, 17\}\)
(b) Given \(A = \{x : x \in N, x < 10\} = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}\) and \(g(x) = 2x\):
For \(x = 1\) to \(9\), values of \(g(x)\) are \(2, 4, 6, 8, 10, 12, 14, 16, 18\).
Set of ordered pairs for \(g\) = \(\{(1, 2), (2, 4), (3, 6), (4, 8), (5, 10), (6, 12), (7, 14), (8, 16), (9, 18)\}\)
Range of \(g = \{2, 4, 6, 8, 10, 12, 14, 16, 18\}\)
In simple words: Substitute each element of the domain into the function formula to generate the second component of each ordered pair.

Exam Tip: Ensure every element of domain \(A\) has a corresponding ordered pair in the function.

Question 9. Let a function f be defined by \(f(x) = x + \frac{1}{x}\), \(x \in R\) Find (i) \(f\left(\frac{1}{x}\right)\) (\(x \neq 0\)) (ii) \(f(2x)\) (iii) \(f(x - 1)\)
Answer: Given \(f(x) = x + \frac{1}{x}\):
(i) Substitute \(x\) with \(\frac{1}{x}\):
\(f\left(\frac{1}{x}\right) = \frac{1}{x} + \frac{1}{\frac{1}{x}} = \frac{1}{x} + x = x + \frac{1}{x} = f(x)\)
(ii) Substitute \(x\) with \(2x\):
\(f(2x) = 2x + \frac{1}{2x}\)
(iii) Substitute \(x\) with \(x - 1\):
\(f(x - 1) = (x - 1) + \frac{1}{x - 1}\)
In simple words: Replace every occurrence of \(x\) in the function definition with the new expression given inside the parentheses.

Exam Tip: Be careful with algebraic fractions when substituting compound expressions.

Question 10. If \(f(x) = \frac{x - 1}{x + 1}\), \(x \neq -1\), then show that \(f(f(x)) = \frac{-1}{x}\), provided that \(x \neq 0\)
Answer: Given \(f(x) = \frac{x - 1}{x + 1}\).
To find \(f(f(x))\), substitute \(f(x)\) into itself:
\(f(f(x)) = \frac{f(x) - 1}{f(x) + 1}\)
Substitute \(f(x) = \frac{x - 1}{x + 1}\):
\(f(f(x)) = \frac{\frac{x - 1}{x + 1} - 1}{\frac{x - 1}{x + 1} + 1}\)
Taking LCM in numerator and denominator:
Numerator = \(\frac{(x - 1) - (x + 1)}{x + 1} = \frac{x - 1 - x - 1}{x + 1} = \frac{-2}{x + 1}\)
Denominator = \(\frac{(x - 1) + (x + 1)}{x + 1} = \frac{x - 1 + x + 1}{x + 1} = \frac{2x}{x + 1}\)
Dividing numerator by denominator:
\(f(f(x)) = \frac{-2}{x + 1} \times \frac{x + 1}{2x} = \frac{-2}{2x} = \frac{-1}{x}\)
Hence proved.
In simple words: Replace \(x\) in the function formula with the entire formula of \(f(x)\) and simplify the complex fraction.

Exam Tip: Multiply numerator and denominator by \((x + 1)\) to easily eliminate the complex fraction denominators.

Question 11. Let \(f(x) = x^2\) and \(g(x) = 2x + 1\) be two real functions find (i) (f + g)(x) (ii) (f - g)(x) (iii) (fg)(x) (iv) \(\left(\frac{f}{g}\right)(x)\)
Answer: Given \(f(x) = x^2\) and \(g(x) = 2x + 1\).
(i) \((f + g)(x) = f(x) + g(x) = x^2 + 2x + 1\)
(ii) \((f - g)(x) = f(x) - g(x) = x^2 - (2x + 1) = x^2 - 2x - 1\)
(iii) \((fg)(x) = f(x) \cdot g(x) = x^2(2x + 1) = 2x^3 + x^2\)
(iv) \(\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x^2}{2x + 1}\), where \(2x + 1 \neq 0 \implies x \neq -\frac{1}{2}\).
In simple words: Algebra of real functions involves standard arithmetic operations performed directly on their function expressions.

Exam Tip: Always mention the domain restriction for the quotient function where the denominator is zero.

Question 12. If a real function f is defined by \(f(x) = \frac{|x| - x}{2x}\), find its range.
Answer: Given \(f(x) = \frac{|x| - x}{2x}\) for \(x \neq 0\).
Case 1: When \(x > 0\), \(|x| = x\):
\(f(x) = \frac{x - x}{2x} = \frac{0}{2x} = 0\)
Case 2: When \(x < 0\), \(|x| = -x\):
\(f(x) = \frac{-x - x}{2x} = \frac{-2x}{2x} = -1\)
Thus, the function only takes the values \(0\) and \(-1\).
Therefore, the range of \(f\) is \(\{-1, 0\}\).
In simple words: Break the absolute value into positive and negative cases to simplify the fraction for different intervals of \(x\).

Exam Tip: Absolute value definitions \(|x| = x\) if \(x \geq 0\) and \(|x| = -x\) if \(x < 0\) are crucial for simplifying piecewise functions.

Question 13. Find the domain of each of the following real valued functions :- (i) \(f(x) = \frac{1}{x + 2}\) (ii) \(f(x) = \frac{x - 1}{x - 3}\) (iii) \(f(x) = \frac{x^2 - 3x + 2}{x^2 - 2x - 3}\) (iv) \(f(x) = \frac{x^2 + 3x + 5}{x^2 - 5x + 4}\) (v) \(f(x) = \sqrt{x - 2}\) (vi) \(f(x) = \frac{1}{\sqrt{1 - x}}\) (vii) \(f(x) = \sqrt{4 - x^2}\)
Answer: (i) \(f(x) = \frac{1}{x + 2}\):
Denominator cannot be zero: \(x + 2 \neq 0 \implies x \neq -2\).
Domain = \(R - \{-2\}\).
(ii) \(f(x) = \frac{x - 1}{x - 3}\):
Denominator cannot be zero: \(x - 3 \neq 0 \implies x \neq 3\).
Domain = \(R - \{3\}\).
(iii) \(f(x) = \frac{x^2 - 3x + 2}{x^2 - 2x - 3}\):
Denominator factorizes as \(x^2 - 2x - 3 = (x - 3)(x + 1)\).

Denominator cannot be zero: \((x - 3)(x + 1) \neq 0 \implies x \neq 3, x \neq -1\).
Domain = \(R - \{-1, 3\}\).
(iv) \(f(x) = \frac{x^2 + 3x + 5}{x^2 - 5x + 4}\):
Denominator factorizes as \(x^2 - 5x + 4 = (x - 1)(x - 4)\).
Denominator cannot be zero: \((x - 1)(x - 4) \neq 0 \implies x \neq 1, x \neq 4\).
Domain = \(R - \{1, 4\}\).
(v) \(f(x) = \sqrt{x - 2}\):
Expression under square root must be non-negative: \(x - 2 \geq 0 \implies x \geq 2\).
Domain = \([2, \infty)\).
(vi) \(f(x) = \frac{1}{\sqrt{1 - x}}\):
Expression under square root must be strictly positive: \(1 - x > 0 \implies x < 1\).
Domain = \((-\infty, 1)\).
(vii) \(f(x) = \sqrt{4 - x^2}\):
Expression under square root must be non-negative: \(4 - x^2 \geq 0 \implies x^2 \leq 4 \implies -2 \leq x \leq 2\).
Domain = \([-2, 2]\).
In simple words: Set denominators not equal to zero and quantities inside square roots greater than or equal to zero to find the domain.

Exam Tip: Denominator terms can never be zero, and even-index radical terms must have non-negative arguments.

Question 14. Find the domain and range of the real function f(x) given by (i) \(f(x) = \frac{2 - x}{x - 3}\) (ii) \(f(x) = \frac{1}{\sqrt{x - 5}}\) (iii) \(f(x) = \sqrt{16 - x^2}\) (iv) \(f(x) = \frac{x}{1 + x^2}\) (v) \(f(x) = \frac{3}{2 - x^2}\) (vi) \(f(x) = \frac{x^2 - 9}{x - 3}\)
Answer: (i) \(f(x) = \frac{2 - x}{x - 3}\):
Domain: Denominator \(x - 3 \neq 0 \implies x \neq 3\), so Domain = \(R - \{3\}\).
Range: Let \(y = \frac{2 - x}{x - 3} \implies y(x - 3) = 2 - x \implies yx - 3y = 2 - x \implies x(y + 1) = 3y + 2 \implies x = \frac{3y + 2}{y + 1}\).
For \(x\) to be real, \(y + 1 \neq 0 \implies y \neq -1\), so Range = \(R - \{-1\}\).
(ii) \(f(x) = \frac{1}{\sqrt{x - 5}}\):
Domain: \(x - 5 > 0 \implies x > 5\), so Domain = \((5, \infty)\).
Range: As \(x \to 5^+\), \(f(x) \to \infty\); as \(x \to \infty\), \(f(x) \to 0\), so Range = \((0, \infty)\).
(iii) \(f(x) = \sqrt{16 - x^2}\):
Domain: \(16 - x^2 \geq 0 \implies x^2 \leq 16 \implies x \in [-4, 4]\).
Range: Minimum value is \(0\) (at \(x = \pm 4\)) and maximum value is \(4\) (at \(x = 0\)), so Range = \([0, 4]\).
(iv) \(f(x) = \frac{x}{1 + x^2}\):
Domain: Denominator \(1 + x^2 > 0\) for all real \(x\), so Domain = \(R\).
Range: Let \(y = \frac{x}{1 + x^2} \implies yx^2 - x + y = 0\). Since \(x\) is real, discriminant \(\Delta = 1 - 4y^2 \geq 0 \implies y^2 \leq \frac{1}{4} \implies -\frac{1}{2} \leq y \leq \frac{1}{2}\). Also at \(x = 0\), \(y = 0\), so Range = \(\left[-\frac{1}{2}, \frac{1}{2}\right] - \{0\}\).
(v) \(f(x) = \frac{3}{2 - x^2}\):
Domain: \(2 - x^2 \neq 0 \implies x^2 \neq 2 \implies x \neq \pm\sqrt{2}\), so Domain = \(R - \{-\sqrt{2}, \sqrt{2}\}\).
Range: Since \(x^2 \geq 0\), \(2 - x^2 \leq 2\). When \(x^2\) passes through \(2\), the denominator changes sign, leading to Range = \((-\infty, 0) \cup \left[\frac{3}{2}, \infty\right]\).
(vi) \(f(x) = \frac{x^2 - 9}{x - 3}\):
Domain: Denominator \(x - 3 \neq 0 \implies x \neq 3\), so Domain = \(R - \{3\}\).
Range: For \(x \neq 3\), \(f(x) = \frac{(x - 3)(x + 3)}{x - 3} = x + 3\). Since \(x \neq 3\), \(f(x) \neq 3 + 3 = 6\). Thus Range = \(R - \{6\}\).
In simple words: To find the range, express \(x\) in terms of \(y\) or analyze the maximum and minimum limits of the function output.

Exam Tip: For rational functions, finding range often involves solving for \(x\) in terms of \(y\) and ensuring the denominator of the resulting expression is non-zero.

Question 15. If \(R = \{(x, y) : x, y \in I, 4x^2 + 8y^2 = 36\}\), then represent R by arrow diagram.
Answer: Given relation equation is \(4x^2 + 8y^2 = 36 \implies x^2 + 2y^2 = 9\), where \(x, y \in I\) (integers).
Let us test integer values for \(y\):
If \(y = 0\): \(x^2 + 2(0) = 9 \implies x^2 = 9 \implies x = \pm 3\). Thus, \((3, 0)\) and \((-3, 0)\) belong to \(R\).
If \(y = \pm 1\): \(x^2 + 2(1)^2 = 9 \implies x^2 + 2 = 9 \implies x^2 = 7\) (not an integer).
If \(y = \pm 2\): \(x^2 + 2(2)^2 = 9 \implies x^2 + 8 = 9 \implies x^2 = 1 \implies x = \pm 1\). Thus, \((1, 2), (1, -2), (-1, 2), (-1, -2)\) belong to \(R\).
If \(y = \pm 3\): \(x^2 + 2(9) = 9 \implies x^2 = -9\) (no real integer solution).
Therefore, the relation in roster form is \(R = \{(-3, 0), (3, 0), (-1, 2), (-1, -2), (1, 2), (1, -2)\}\).
Arrow diagram representation: Draw two ovals representing the domain set \(\{-3, -1, 1, 3\}\) and range set \(\{-2, 0, 2\}\) and draw directed arrows from each \(x\) to its corresponding \(y\).
In simple words: Find all integer pairs satisfying the circle-ellipse equation and connect them with arrows from inputs to outputs.

Exam Tip: Always check that both \(x\) and \(y\) satisfy the integer condition \(I\) given in the problem statement.

Question 16. Give A = \{-2, -1, 0, 1, 2\} and B = \{-3, -1, 1, 5\}. List all elements of \(f = \{(x, y) : y = 2x^2 - 3, x \in A, y \in B\}\). Is f a function ?
Answer: Given \(A = \{-2, -1, 0, 1, 2\}\) and \(B = \{-3, -1, 1, 5\}\).
Calculate \(y = 2x^2 - 3\) for each element in \(A\):
For \(x = -2\): \(y = 2(-2)^2 - 3 = 2(4) - 3 = 5\) (Since \(5 \in B\), ordered pair is \((-2, 5)\)).
For \(x = -1\): \(y = 2(-1)^2 - 3 = 2(1) - 3 = -1\) (Since \(-1 \in B\), ordered pair is \((-1, -1)\)).
For \(x = 0\): \(y = 2(0)^2 - 3 = -3\) (Since \(-3 \in B\), ordered pair is \((0, -3)\)).
For \(x = 1\): \(y = 2(1)^2 - 3 = -1\) (Since \(-1 \in B\), ordered pair is \((1, -1)\)).
For \(x = 2\): \(y = 2(2)^2 - 3 = 2(4) - 3 = 5\) (Since \(5 \in B\), ordered pair is \((2, 5)\)).
Listing all elements of \(f\):
\(f = \{(-2, 5), (-1, -1), (0, -3), (1, -1), (2, 5)\}\)
Is \(f\) a function? Yes, because every element in the domain set \(A\) has a unique image in the codomain set \(B\).
In simple words: A relation is a function if every input value has one and only one output value.

Exam Tip: Verify that no two distinct ordered pairs have the same first element with different second elements.

Q.1 Let A = {1,2} and B = {3,4}.Write A×B. How many subsets will A×B have? List them.

Q.2 Let A = {1,2,3,…..,14}. Define a relation R from A to A by R = {(x,y) : 3x-y = 0, where x,y∈A}. Write down its domain, co-domain and range.

Q.3 If f(x) = x2, find f(1.2) - f(1) / 1.2 - 1

Q.4 Find the inverse of the function of x = 3 - 5y / 2y - y

Q.5 If n(A) = 3 and n(B) = 3, then find n(A B). (1 mark)

Q.6 Let R be a relation from Q to Q defined by R = {(a,b):a,b∈Q and a-b∈Z}.Show that (a) (a,a)∈R for all a∈Q (b) (a,b)∈R implies that (b,a)∈R (c) (a,b)∈R and (b,c)∈R implies that (a,c)∈R.

Q.7 Let A = {9, 10, 11, 12, 13} and let f: A → N be defined by f(n) = the highest prime factor of n. Find the range of f.

Q.8 Find the domain of the function f(x) = x² + 2x + 1 / x² - 8x + 12 

Q.9 Let A = {1, 2, 3, 4} and B = {10, 12, 13, 14, 20}. Whether f: A B defined by f(1) = 10, f(2) = 12, f(3) = 13 is a function? 

Q.10 Find the domain and the range of the real function f defined by f (x) = |x – 1|. (2 marks)

Q.11 Examine the relation : R={(2,1),(3,1),(4,1)} and state whether it is a function or not?

Q.12 A function f is defined by f(x) = 3x-4. Write down the value of f(5) and f(-7).

Q.13 Write the domain of the function f(x) = x+1/x²+6x+5

Q.14 Let A = {1, 2, 6, 8} and let R be a relation on A defined by {(a, b): a, b A, b is exactly divisible by a}
a) Write R in roster form.
b) Find the domain of R.
c) Find the range of R.

Q.15 If f and g are two functions such that f(x) = 5x + 2 and g(x) = x2 + 3, then find f + g and f – g. 

Q.16 Write the domain of the function f(x) = x² - 2x + 3 / x² - x - 20

Q.17 A function f is defined by f(x) = 2x – 5. Write down the values of (2 marks)
(i) f(0), (ii) f(7), (iii) f(–3)

Q.18 If f(x) = x² - 1/x². then find the value of : f(x) + f(1/x)

Q.19 Under which condition a relation f from A to B is said to be a function? (1 mark)

Q.20 If A = {a1,a2} and B = {b1,b2,b3}, then write A × B.