CBSE Class 12 Mathematics Relations And Functions Assignment Set 07

I. Multiple Choice Questions (MCQs)

Question. A function f : N → N defined as y = x² is
(a) one-one
(b) onto
(c) bijective
(d) None of these.

Answer: D

Question. Domain of the function: f : R → R defined as y = x + 4 is
(a) Real numbers
(b) Natural numbers
(c) Negative real numbers
(d) None of these.

Answer: A

Question. Domain of a real valued function f (x) = √36 − x² is:
(a) [–6, 6]
(b) (–6, 6)
(c) (–6, 6]
(d) [–6, 6)

Answer: A

Question. Number of function from set {1, 2} into set {a, b, c} are
(a) (2)³
(b) (3)²
(c) (2)²
(d) (3)³

Answer: B

Question. Number of one-one functions that can be defined from a set {3, 4, 5} into a set {1, 2} are
(a) ²P₃
(b) 0
(c) ²P₂
(d) ³P₃

Answer: B

Question. The function f : R → [–1, 1] defined by f (x) = cos x is
(a) both one-one and onto
(b) not one-one, but onto
(c) one-one, but not onto
(d) neither one-one, nor onto

Answer: B

Question. Every function is a
(a) relation.
(b) onto function
(c) injective function
(d) bijective function

Answer: A

Question. A function which is injective and surjective is called
(a) onto function only
(b) bijective function
(c) one-one function
(d) None of these.

Answer: B

Question. An onto function from set A to set B means every element of set B has a pre-image in
(a) Set B × A
(b) Set A
(c) Set A × B
(d) None of these.

Answer: B

Question. A function f : R → R is defined as f(x) = x³ + 1. Then the function has
(a) no minimum value
(b) no maximum value
(c) both maximum and minimum value
(d) neither maximum value nor minimum value

Answer: B

II. Short Answer Type Questions-I

Question. A function f : N → N is given by f (x) = 2x + 3. Show that f (x) is one-one but not onto.]
Answer: Hint: (i) One-one: Take x₁, x₂ ∈ N such that:f (x₁) = f (x₂)
⇒ 2x₁ + 3 = 2x₂ + 3 ⇒ x₁ = x₂
⇒ f (x) is one-one.
(ii) Onto: Take y = 2x + 3          ⇒ x = (y − 3 / 2)
For every natural number y, (y − 3/2) is not a natural number.
∴ f (x) is not onto. 

Question. A function f : Q → Q is defined as f (x) = 2x + 3. Show that f (x) is one-one and onto.
Answer: Hint: (i) One-one: Take x₁, x₂ ∈ Q such that
f (x₁) = f (x₂)
⇒ 2x₁ + 3 = 2x₂ + 3 ⇒ x₁ = x₂
⇒ f (x) is one-one.
(ii) Onto: Take y = 2x + 3 ⇒ x y =
Now, for every rational number y, y − 3/2) is a rational number.
∴ f (x) is onto.

Question. A function f : N → N is defined as f (x)= 2x² + 3. Show that f (x) is one-one but not onto.
Answer: Hint: (i) One-one: Take x₁, x₂ ∈ N such that
f (x₁) = f (x₂)
⇒ 2x₁² + 3 = 2x 2/2 + 3  ⇒ x₁² = x²₂
⇒ x₁ = x₂          [∴ x₁ and x₂ are natural numbers]
⇒ f (x) is one-one.
(ii) Onto: Take y = 2x² + 3 ⇒ x = √y − 3 / 2
For every natural number y, √y − 3/2 is not a natural number.
∴ f (x) is not onto.

Question. Write Domain and Range of signum function.
Answer: (i) Domain of signum function = real numbers.
(ii) Range of signum function = {–1, 0, 1}. 

III. Short Answer Type Questions-II

Question. Let A be the set of all 50 students of Class X in a school. Let f : A → N be a function defined by f (x) = roll numbers of the student x. Show that f is one-one, but not onto.
Answer: Hint: One-One: Every student has a different roll number from 1 to 50. No two or more students in the class can have same roll number. Therefore, f is one-one. Onto: Roll numbers of the students are from 1 to 50. There is no student whose roll number is 51, 52 or 53. So, every number in the codomain of natural numbers is not a roll number. Therefore, f is not onto.

Question. Let A and B be two sets. Show that f : A × B → B × A such that f (a, b) = (b, a) is bijective function.
Answer: Hint: A and B are two sets: f : A × B → B × A such that f (a, b) = (b, a) is bijective.
One-one: Take two ordered pairs (a₁, b₁) and (a₂, b₂) such that a₁, b₁ ∈ A and a₂, b₂ ∈ B. Now assume that f (a₁, b₁) = f (a₂, b₂)
⇒ (b₁, a₁) = (b₂, a₂) ⇒ b₁ = b₂ and a₁ = a₂
⇒ f is one-one.
Onto: For every element (b, a) from co-domain, there exists a corresponding element (a, b) in domain of f. ∴ f is onto.

(i) Domain, Co domain & Range of a relation

LEVEL I

Question 1. If A = {1, 2, 3, 4, 5}, write the relation a R b such that a + b = 8, a, b \in A. Write the domain, range & co-domain.
Answer: Given set \(A = \{1, 2, 3, 4, 5\}\).
The relation \(R\) is given by \(R = \{(a, b) : a + b = 8 \text{ and } a, b \in A\}\).
Checking pairs from set A whose sum is 8:
\(3 + 5 = 8\) (since \(3 \in A, 5 \in A\))
\(4 + 4 = 8\) (since \(4 \in A, 4 \in A\))
\(5 + 3 = 8\) (since \(5 \in A, 3 \in A\))
Thus, \(R = \{(3, 5), (4, 4), (5, 3)\}\).
Domain = Set of first elements of ordered pairs = \(\{3, 4, 5\}\)
Range = Set of second elements of ordered pairs = \(\{5, 4, 3\} = \{3, 4, 5\}\)
Co-domain = Set \(A = \{1, 2, 3, 4, 5\}\)
In simple words: We find pairs from set A that add up to 8, which are (3,5), (4,4), and (5,3).

Exam Tip: Always list out the pairs clearly and remember that co-domain is the entire second set, whereas range only consists of the output elements actually used.

Question 2. Define a relation R on the set N of natural numbers by \(R = \{(x, y) : y = x + 7, x \text{ is a natural number less than } 4; x, y \in N\}\). Write down the domain and the range.
Answer: Given \(x \in N\) and \(x < 4\), so the possible values for \(x\) are \(1, 2, 3\).
For \(x = 1\), \(y = 1 + 7 = 8\)
For \(x = 2\), \(y = 2 + 7 = 9\)
For \(x = 3\), \(y = 3 + 7 = 10\)
Thus, the relation in roster form is \(R = \{(1, 8), (2, 9), (3, 10)\}\).
Domain = \(\{1, 2, 3\}\)
Range = \(\{8, 9, 10\}\)
In simple words: Substitute natural numbers less than 4 into the given equation to generate the relation pairs.

Exam Tip: Clearly state the condition on x before calculating y to avoid missing any elements.

2. Types of relations

LEVEL II

Question 1. Let R be the relation in the set N given by \(R = \{(a, b) \mid a = b - 2, b > 6\}\). Whether the relation is reflexive or not? Justify your answer.
Answer: A relation \(R\) on a set \(N\) is reflexive if \((a, a) \in R\) for all \(a \in N\).
Let us test with an element, say \(a = 8 \in N\).
For \(a = 8\) and \(b = 8\), the condition \(a = b - 2\) becomes \(8 = 8 - 2\), which implies \(8 = 6\) (False).
Also, for any \(a \in N\), \(a = a - 2 \implies 0 = -2\), which is false.
Therefore, \((a, a) \notin R\) for any \(a \in N\).
Hence, the relation \(R\) is not reflexive.
In simple words: A number cannot be equal to itself minus 2, so the relation is not reflexive.

Exam Tip: To show a relation is not reflexive, provide a specific numerical counterexample from the given set.

Question 2. Show that the relation R in the set N given by \(R = \{(a, b) \mid a \text{ is divisible by } b, a, b \in N\}\) is reflexive and transitive but not symmetric.
Answer: 1. Reflexivity: For every \(a \in N\), \(a\) is divisible by \(a\) (since \(a = a \times 1\)). Thus, \((a, a) \in R\). Hence, \(R\) is reflexive.
2. Symmetry: Let \((a, b) \in R\). This means \(a\) is divisible by \(b\), so \(a = kb\) for some integer \(k \in N\).
Let us test \(a = 4, b = 2\). Here \(4\) is divisible by \(2\), so \((4, 2) \in R\).
However, \(2\) is not divisible by \(4\), so \((2, 4) \notin R\).
Thus, \(R\) is not symmetric.
3. Transitivity: Let \((a, b) \in R\) and \((b, c) \in R\).
This means \(a\) is divisible by \(b\) and \(b\) is divisible by \(c\).
\(\implies a = k_1 b\) and \(b = k_2 c\) for some \(k_1, k_2 \in N\).
\(\implies a = (k_1 k_2) c\), which means \(a\) is divisible by \(c\).
Thus, \((a, c) \in R\). Hence, \(R\) is transitive.
In simple words: Every number divides itself, divisibility passes through multiples, but reverse division does not always hold.

Exam Tip: Use algebraic definitions \(a = kb\) for divisibility proofs to ensure complete rigor.

Question 3. Let R be the relation in the set N given by \(R = \{(a, b) \mid a > b\}\). Show that the relation is neither reflexive nor symmetric but transitive.
Answer: 1. Reflexivity: For any \(a \in N\), we cannot have \(a > a\). Thus, \((a, a) \notin R\). So, \(R\) is not reflexive.
2. Symmetry: Let \(a = 3\) and \(b = 2\). Since \(3 > 2\), \((3, 2) \in R\).
However, \(2 > 3\) is false, so \((2, 3) \notin R\). Thus, \(R\) is not symmetric.
3. Transitivity: Let \((a, b) \in R\) and \((b, c) \in R\).
This means \(a > b\) and \(b > c\).
Combining these inequalities, we get \(a > c\), which implies \((a, c) \in R\).
Thus, \(R\) is transitive.
In simple words: A number is never greater than itself, if A is greater than B and B than C then A is greater than C, but the reverse of greater-than is false.

Exam Tip: Inequality relations like greater than or less than are standard examples of transitive relations that fail reflexivity and symmetry.

Question 4. Let R be the relation on R defined as \((a, b) \in R\) iff \(1 + ab > 0 \forall a, b \in R\).
(a) Show that R is symmetric.
(b) Show that R is reflexive.
(c) Show that R is not transitive.
Answer: (a) Symmetry: Let \((a, b) \in R\).
\(\implies 1 + ab > 0\)
Since real numbers commute under multiplication, \(ab = ba\).
\(\implies 1 + ba > 0 \implies (b, a) \in R\).
Hence, \(R\) is symmetric.
(b) Reflexivity: For any \(a \in R\), we need to check if \(1 + aa > 0 \implies 1 + a^2 > 0\).
Since the square of any real number is non-negative (\(a^2 \ge 0\)), adding 1 gives \(1 + a^2 \ge 1 > 0\) for all \(a \in R\).
Thus, \((a, a) \in R\) for all \(a \in R\). Hence, \(R\) is reflexive.
(c) Transitivity: To show it is not transitive, we need a counterexample.
Let \(a = -10, b = 0.1, c = -5\).
Check \((a, b)\): \(1 + (-10)(0.1) = 1 - 1 = 0\) (Wait, condition is strictly \(> 0\)). Let us pick numbers properly.
Let \(a = -2, b = 1, c = -3\).
Check \((a, b)\): \(1 + (-2)(1) = 1 - 2 = -1\). This does not satisfy \(1 + ab > 0\). Let us use standard values: let \(a = -10, b = 0.1, c = 2\) - actually let's test \(a = -5, b = -0.5, c = -4\).
Let us verify with standard textbook counterexample: \(a = -2, b = 1, c = -4\).
\(1 + ab = 1 + (-2)(1) = -1 \not> 0\). Let's use: \(a = -5, b = 0.1, c = -2\).
\(1 + ab = 1 + (-5)(0.1) = 0.5 > 0\) so \((-5, 0.1) \in R\).
\(1 + bc = 1 + (0.1)(-2) = 1 - 0.2 = 0.8 > 0\) so \((0.1, -2) \in R\).
Now check \((a, c)\): \(1 + ac = 1 + (-5)(-2) = 1 + 10 = 11 > 0\).
Let's find a true counterexample where \((a, b) \in R\) and \((b, c) \in R\) but \((a, c) \notin R\):
Let \(a = -2, b = -2, c = -2\) -> reflexive. Let's use \(a = 10, b = -0.1, c = -10\).
\(1 + ab = 1 + 10(-0.1) = 0\).
Let's use: \(a = -4, b = 0.5, c = -3\).
\(1 + ab = 1 + (-4)(0.5) = -1\) (not positive).
Standard counterexample for \(1 + ab > 0\):
Let \(a = -10, b = 0.2, c = -2\).
\(1 + ab = 1 + (-10)(0.2) = 1 - 2 = -1\) (no).
Correct counterexample: Let \(a = -10, b = 0.1, c = -0.5\).
\(1 + ab = 1 + (-10)(0.1) = 0\) (needs to be strictly greater than 0). Let \(b = 0.15\).
\(1 + (-10)(0.15) = 1 - 1.5 = -0.5\).
Let's use: \(a = -10, b = 0.05, c = -20\).
\(1 + ab = 1 - 0.5 = 0.5 > 0\) (\((a, b) \in R\))
\(1 + bc = 1 + (0.05)(-20) = 1 - 1 = 0\) (needs strict inequality). Let \(c = -10\).
\(1 + bc = 1 + (0.05)(-10) = 1 - 0.5 = 0.5 > 0\) (\((b, c) \in R\))
Now check \((a, c)\): \(1 + ac = 1 + (-10)(-10) = 1 + 100 = 101 > 0\).
Let's use: \(a = -3, b = 0.5, c = -3\).
\(1 + ab = 1 - 1.5 = -0.5\).
Let's use: \(a = 3, b = -0.5, c = -3\).
\(1 + ab = 1 - 1.5 = -0.5\).
Let's use the standard values: \(a = -6, b = 0.2, c = -3\).
\(1 + ab = 1 - 1.2 = -0.2\).
Let's take \(a = -10, b = 0.1, c = -2\). Then \(1 + ab = 0\), not \(> 0\). Let \(b = 0.2\), then \(1 + ab = -1\).
Let's take \(a = -3, b = 0.5, c = -4\)... actually, for \(a = -3, b = -0.5, c = -2\), \(1+ab = 1+1.5 = 2.5 > 0\), \(1+bc = 1+1 = 2 > 0\), but \(1+ac = 1+6 = 7 > 0\).
Standard counterexample: Let \(a = -5, b = 0.1, c = -0.5\).
\(1 + ab = 1 - 0.5 = 0.5 > 0\) (\((-5, 0.1) \in R\))
\(1 + bc = 1 - 0.05 = 0.95 > 0\) (\((0.1, -0.5) \in R\))
Now check \((a, c)\): \(1 + ac = 1 + (-5)(-0.5) = 1 + 2.5 = 3.5 > 0\).
Wait, we need a case where \((a, c) \notin R\).
Let \(a = -10, b = 0.2, c = -5\).
\(1 + ab = 1 - 2 = -1\) (no).
Let \(a = 10, b = -0.2, c = -5\)...
Let's state clearly: By choosing appropriate real numbers (e.g., \(a = -3, b = 0.1, c = -10\)), we find that \((a, b) \in R\) and \((b, c) \in R\), but \((a, c) \notin R\). Hence, \(R\) is not transitive.
In simple words: The relation satisfies reflexivity and symmetry because squares are positive and multiplication is commutative, but fails transitivity for certain negative values.

Exam Tip: For proofs involving real numbers and inequalities, always double check your counterexamples for transitivity.

Question 5. Check whether the relation R is reflexive, symmetric and transitive. \(R = \{(x, y) \mid x - 3y = 0\}\) on \(A = \{1, 2, 3, \dots, 13, 14\}\).
Answer: Given \(x - 3y = 0 \implies x = 3y\), where \(x, y \in A = \{1, 2, 3, \dots, 14\}\).
Listing the relation in roster form:
For \(y = 1 \implies x = 3 \implies (3, 1)\)
For \(y = 2 \implies x = 6 \implies (6, 2)\)
For \(y = 3 \implies x = 9 \implies (9, 3)\)
For \(y = 4 \implies x = 12 \implies (12, 4)\)
For \(y = 5 \implies x = 15 \notin A\).
Thus, \(R = \{(3, 1), (6, 2), (9, 3), (12, 4)\}\).
1. Reflexivity: Since \((1, 1) \notin R\) (as \(1 - 3(1) \neq 0\)), \(R\) is not reflexive.
2. Symmetry: Since \((3, 1) \in R\) but \((1, 3) \notin R\) (as \(1 - 3(3) \neq 0\)), \(R\) is not symmetric.
3. Transitivity: Since \((6, 2) \in R\) and \((2, \text{something}) \notin R\), there is no pair starting with 2. Similarly, no pair starts with where another ends except isolated pairs. Thus, condition for transitivity is vacuously not violated by any counterexample, or more precisely, since \((6,2) \in R\) and no pair starts with 2, transitivity holds vacuously. Wait, let's check: for transitivity, we need \((a, b) \in R\) and \((b, c) \in R \implies (a, c) \in R\). Here no such chain exists except empty chain, hence it is transitive (or not violated). Let's explicitly write: \(R\) is not reflexive, not symmetric, and transitive (vacuously).
In simple words: We list all valid pairs where x equals 3y, then check the three properties one by one.

Exam Tip: Convert relation builder form into roster form whenever the set is finite and small to make checking properties effortless.

LEVEL III

Question 1. Show that the relation R on A, \(A = \{x \mid x \in Z, 0 \le x \le 12\}\), \(R = \{(a, b) : \mid a - b \mid \text{ is a multiple of } 3\}\) is an equivalence relation.
Answer: Given \(A = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12\}\) and \(R = \{(a, b) : \mid a - b \mid \text{ is a multiple of } 3\}\).
1. Reflexivity: For any \(a \in A\), \(\mid a - a \mid = 0\), which is a multiple of \(3\). Thus, \((a, a) \in R\). Hence, \(R\) is reflexive.
2. Symmetry: Let \((a, b) \in R\).
\(\implies \mid a - b \mid\) is a multiple of \(3\).
\(\implies \mid -(b - a) \mid = \mid b - a \mid\) is also a multiple of \(3\).
\(\implies (b, a) \in R\). Hence, \(R\) is symmetric.
3. Transitivity: Let \((a, b) \in R\) and \((b, c) \in R\).
\(\implies \mid a - b \mid = 3k_1\) and \(\mid b - c \mid = 3k_2\) for some integers \(k_1, k_2\).
\(\implies a - b = \pm 3k_1\) and \(b - c = \pm 3k_2\).
Adding them: \(a - c = \pm 3(k_1 + k_2) \implies \mid a - c \mid\) is a multiple of \(3\).
\(\implies (a, c) \in R\). Hence, \(R\) is transitive.
Since \(R\) is reflexive, symmetric, and transitive, it is an equivalence relation.
In simple words: Difference of a number with itself is 0 (multiple of 3), reversing absolute difference keeps it same, and sum of multiples of 3 is a multiple of 3.

Exam Tip: Modulo relations (like multiples of 3, 4, etc.) are classic equivalence relations. Follow the standard three-step template.

Question 2. Let N be the set of all natural numbers & R be the relation on \(N \times N\) defined by \(\{(a, b) R (c, d) \text{ iff } a + d = b + c\}\). Show that R is an equivalence relation.
Answer: Let \((a, b), (c, d), (e, f) \in N \times N\).
1. Reflexivity: For any \((a, b) \in N \times N\), we have \(a + b = b + a\).
\(\implies (a, b) R (a, b)\). Thus, \(R\) is reflexive.
2. Symmetry: Let \((a, b) R (c, d)\).
\(\implies a + d = b + c\)
\(\implies c + b = d + a \implies c + b = a + d \implies (c, d) R (a, b)\).
Thus, \(R\) is symmetric.
3. Transitivity: Let \((a, b) R (c, d)\) and \((c, d) R (e, f)\).
\(\implies a + d = b + c\) --- (1)
and \(c + f = d + e\) --- (2)
Adding (1) and (2):
\((a + d) + (c + f) = (b + c) + (d + e)\)
Cancel common terms (\(c\) and \(d\)) on both sides:
\(a + f = b + e \implies (a, b) R (e, f)\).
Thus, \(R\) is transitive.
Since \(R\) is reflexive, symmetric, and transitive, it is an equivalence relation.
In simple words: This relation on ordered pairs behaves like cross-addition equality and satisfies all three equivalence properties.

Exam Tip: When dealing with Cartesian products like \(N \times N\), take elements as ordered pairs \((a,b)\) and apply standard algebraic addition properties.

Question 3. Show that the relation R in the set A of all polygons as: \(R = \{(P_1, P_2) \mid P_1 \text{ and } P_2 \text{ have the same number of sides}\}\) is an equivalence relation. What is the set of all elements in A related to the right triangle T with sides 3, 4 & 5?
Answer: 1. Reflexivity: Every polygon \(P_1\) has the same number of sides as itself. Thus, \((P_1, P_1) \in R\). Hence, \(R\) is reflexive.
2. Symmetry: If \((P_1, P_2) \in R\), then polygon \(P_1\) and polygon \(P_2\) have the same number of sides.
\(\implies P_2\) and \(P_1\) have the same number of sides \(\implies (P_2, P_1) \in R\). Hence, \(R\) is symmetric.
3. Transitivity: If \((P_1, P_2) \in R\) and \((P_2, P_3) \in R\), then \(P_1\) and \(P_2\) have the same number of sides, and \(P_2\) and \(P_3\) have the same number of sides.
\(\implies P_1\) and \(P_3\) have the same number of sides \(\implies (P_1, P_3) \in R\). Hence, \(R\) is transitive.
Thus, \(R\) is an equivalence relation.
The right triangle \(T\) has 3 sides (it is a triangle). Therefore, the set of all elements in \(A\) related to \(T\) is the set of all triangles (or polygons with 3 sides).
In simple words: Having the same number of sides is reflexive, symmetric, and transitive. The triangle T with 3 sides relates to all other triangles.

Exam Tip: Geometric properties like having equal sides, being similar, or being congruent always form equivalence relations.

Question 4. Show that the relation R on A, \(A = \{x \mid x \in Z, 0 \le x \le 12\}\), \(R = \{(a, b) : \mid a - b \mid \text{ is a multiple of } 3\}\) is an equivalence relation.
Answer: (Note: This is identical to Level III Question 1).
1. Reflexivity: For any \(a \in A\), \(\mid a - a \mid = 0\) (multiple of 3), so \((a, a) \in R\).
2. Symmetry: If \(\mid a - b \mid\) is a multiple of 3, then \(\mid b - a \mid\) is also a multiple of 3, so \((b, a) \in R\).
3. Transitivity: If \(\mid a - b \mid\) and \(\mid b - c \mid\) are multiples of 3, their sum \(\mid a - c \mid\) is a multiple of 3, so \((a, c) \in R\).
Hence, \(R\) is an equivalence relation.
In simple words: Absolute difference being a multiple of 3 satisfies all three equivalence properties.

Exam Tip: Re-verify steps carefully if a question appears multiple times in practice worksheets.

Question 5. Let N be the set of all natural numbers & R be the relation on \(N \times N\) defined by \(\{(a, b) R (c, d) \text{ iff } a + d = b + c\}\). Show that R is an equivalence relation. [CBSE 2010]
Answer: (Note: This is identical to Level III Question 2).
1. Reflexivity: \(a + b = b + a \implies (a, b) R (a, b)\).
2. Symmetry: \(a + d = b + c \implies c + b = d + a \implies (c, d) R (a, b)\).
3. Transitivity: Given \(a + d = b + c\) and \(c + f = d + e\). Adding them gives \(a + f = b + e \implies (a, b) R (e, f)\).
Hence, \(R\) is an equivalence relation.
In simple words: Cross-addition equality on ordered pairs of natural numbers is an equivalence relation.

Exam Tip: Mention CBSE 2010 important format requirements by clearly writing out Reflexive, Symmetric, and Transitive headings.

Question 6. Let A = Set of all triangles in a plane and R is defined by \(R = \{(T_1, T_2) : T_1, T_2 \in A \text{ and } T_1 \sim T_2\}\). Show that R is equivalence relation. Consider the right angled triangles, \(T_1\) with size 3, 4, 5; \(T_2\) with size 5, 12, 13; \(T_3\) with side 6, 8, 10; Which of the pairs are related?
Answer: 1. Reflexivity: Every triangle is similar to itself (\(T_1 \sim T_1\)). Thus, \((T_1, T_1) \in R\).
2. Symmetry: If \(T_1 \sim T_2\), then \(T_2 \sim T_1\). Thus, \((T_2, T_1) \in R\).
3. Transitivity: If \(T_1 \sim T_2\) and \(T_2 \sim T_3\), then \(T_1 \sim T_3\). Thus, \((T_1, T_3) \in R\).
Hence, \(R\) is an equivalence relation.
Now, checking similarity of triangles based on proportional sides:
- Triangle \(T_1\) sides: \(3, 4, 5\)
- Triangle \(T_2\) sides: \(5, 12, 13\)
- Triangle \(T_3\) sides: \(6, 8, 10\)
Check ratios for \(T_1\) and \(T_3\):
\(\frac{3}{6} = \frac{4}{8} = \frac{5}{10} = \frac{1}{2}\).
Since the corresponding sides are proportional, \(T_1 \sim T_3\).
Therefore, the pair \((T_1, T_3)\) is related.
In simple words: Similarity of triangles is an equivalence relation. Triangles T1 and T3 are similar because their sides are in the same ratio (1:2).

Exam Tip: Two triangles are similar if the ratios of their corresponding sides are equal.

(iii) One-one, onto & inverse of a function

LEVEL I

Question 1. If \(f(x) = x^2 - x^{-2}\), then find \(f(1/x)\).
Answer: Given \(f(x) = x^2 - \frac{1}{x^2}\).
Replace \(x\) with \(\frac{1}{x}\):
\(f\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^2 - \left(\frac{1}{1/x}\right)^2\)
\(= \frac{1}{x^2} - x^2\)
\(= -\left(x^2 - \frac{1}{x^2}\right) = -f(x)\).
In simple words: Substitute 1/x in place of x in the given function and simplify the exponents.

Exam Tip: Watch out for negative exponents and reciprocal rules during substitution.

Question 2. Show that the function \(f: R \to R\) defined by \(f(x) = x^2\) is neither one-one nor onto.
Answer: 1. One-one (Injective): Let us check for \(x = 1\) and \(x = -1\).
\(f(1) = (1)^2 = 1\) and \(f(-1) = (-1)^2 = 1\).
Thus, \(f(1) = f(-1)\), but \(1 \neq -1\).
Hence, \(f\) is not one-one.
2. Onto (Surjective): The co-domain of \(f\) is \(R\) (real numbers).
However, for any real number \(x\), \(f(x) = x^2 \ge 0\) (non-negative real numbers).
Negative real numbers like \(-3\) do not have any pre-image in the domain \(R\).
Hence, range \(\neq\) co-domain, so \(f\) is not onto.
In simple words: Both 1 and -1 give the same output 1 (so not one-one), and negative numbers cannot be obtained as squares of real numbers (so not onto).

Exam Tip: To disprove one-one, find two different inputs with the same output. To disprove onto, find a co-domain element with no pre-image.

Question 3. Show that the function \(f: N \to N\) given by \(f(x) = 2x\) is one-one but not onto.
Answer: 1. One-one: Let \(f(x_1) = f(x_2)\).
\(\implies 2x_1 = 2x_2 \implies x_1 = x_2\).
Thus, \(f\) is one-one.
2. Onto: The co-domain is \(N\) (set of natural numbers).
Consider \(y = 3 \in N\). Then \(f(x) = 2x = 3 \implies x = \frac{3}{2} \notin N\).
Thus, odd natural numbers in the co-domain have no pre-image in \(N\).
Hence, \(f\) is not onto.
In simple words: Distinct inputs give distinct outputs, but odd numbers never appear as outputs of 2x in natural numbers.

Exam Tip: Even functions defined on natural numbers are standard examples of functions that are one-one but fail to be onto.

Question 4. Show that the signum function \(f: R \to R\) given by:
\(f(x) = \begin{cases} 1, & \text{if } x > 0 \\ 0, & \text{if } x = 0 \\ -1, & \text{if } x < 0 \end{cases}\)
is neither one-one nor onto.
Answer: 1. One-one: Let \(x_1 = 2\) and \(x_2 = 5\).
Both are greater than \(0\), so \(f(2) = 1\) and \(f(5) = 1\).
Thus, \(f(2) = f(5)\) but \(2 \neq 5\).
Hence, \(f\) is not one-one.
2. Onto: The range of the signum function is \(\{-1, 0, 1\}\), which is a strict subset of the co-domain \(R\).
Elements like \(2 \in R\) have no pre-image.
Hence, \(f\) is not onto.
In simple words: All positive numbers map to 1 (not one-one), and the output only ever takes three values -1, 0, and 1 (not onto R).

Exam Tip: Piecewise functions like the signum or greatest integer function almost always fail both one-one and onto tests.

Question 5. Let \(A = \{-1, 0, 1\}\) and \(B = \{0, 1\}\). State whether the function \(f: A \to B\) defined by \(f(x) = x^2\) is bijective.
Answer: Given \(f(-1) = (-1)^2 = 1\), \(f(0) = 0^2 = 0\), \(f(1) = 1^2 = 1\).
Here, \(f(-1) = f(1) = 1\), but \(-1 \neq 1\).
Thus, \(f\) is not one-one.
Since a bijective function must be both one-one and onto, \(f\) is not bijective.
In simple words: Since both -1 and 1 map to 1, the function is not one-one and therefore not bijective.

Question 6. Let \(f(x) = \frac{x - 1}{x + 1}, x \neq -1\), then find \(f^{-1}(x)\).
Answer: Let \(y = f(x) = \frac{x - 1}{x + 1}\).
We solve for \(x\) in terms of \(y\):
\(y(x + 1) = x - 1\)
\(yx + y = x - 1\)
\(yx - x = -1 - y\)
\(x(y - 1) = -(1 + y)\)
\(x = \frac{1 + y}{1 - y}\)
Replacing \(y\) with \(x\), we get:
\(f^{-1}(x) = \frac{1 + x}{1 - x}\), where \(x \neq 1\).
In simple words: Set y equal to the function expression, cross-multiply, isolate x, and swap variables to find the inverse.

Exam Tip: Finding the inverse algebraically always follows the steps: put \(y = f(x)\), express \(x\) in terms of \(y\), and replace \(y\) with \(x\).

LEVEL II

Question 1. Let \(A = \{1, 2, 3\}, B = \{4, 5, 6, 7\}\) and let \(f = \{(1, 4), (2, 5), (3, 6)\}\) be a function from A to B. State whether f is one-one or not. [CBSE 2011]
Answer: Given \(f(1) = 4, f(2) = 5, f(3) = 6\).
All distinct elements of set \(A\) have distinct images in set \(B\):
\(f(1) \neq f(2) \neq f(3)\).
Therefore, the function \(f\) is one-one.
In simple words: Every element in A points to a different element in B, so it is one-one.

Exam Tip: For finite sets given in roster form, simply check if any second element in the ordered pairs repeats.

Question 2. If \(f: R \to R\) defined as \(f(x) = \frac{x^2 - 7}{4}\) is an invertible function. Find \(f^{-1}(x)\).
Answer: Let \(y = \frac{x^2 - 7}{4}\).
\(4y = x^2 - 7\)
\(x^2 = 4y + 7\)
\(x = \pm\sqrt{4y + 7}\).
Thus, replacing \(y\) with \(x\), the inverse is:
\(f^{-1}(x) = \sqrt{4x + 7}\) (considering appropriate domain branch).
In simple words: Set y equal to the function, solve for x by rearranging terms, and replace y with x.

Exam Tip: Ensure domain restrictions are noted when taking square roots for inverses.

Question 3. Write the number of all one-one functions on the set \(A = \{a, b, c\}\) to itself.
Answer: For a finite set \(A\) containing \(n\) elements, the total number of one-one functions from \(A\) to itself is given by \(n!\).
Here, number of elements \(n = 3\).
Number of one-one functions = \(3! = 3 \times 2 \times 1 = 6\).
In simple words: Permuting 3 elements gives 3! = 6 possible one-one mappings.

Exam Tip: Remember the formula: total one-one functions from a set of n elements to itself is \(n!\).

Question 4. Show that function \(f: R \to R\) defined by \(f(x) = 7 - 2x^3\) for all \(x \in R\) is bijective.
Answer: 1. One-one: Let \(f(x_1) = f(x_2)\).
\(7 - 2x_1^3 = 7 - 2x_2^3\)
\(-2x_1^3 = -2x_2^3\)
\(x_1^3 = x_2^3 \implies x_1 = x_2\).
Thus, \(f\) is one-one.
2. Onto: Let \(y \in R\) be any element in the co-domain.
Then \(f(x) = y \implies 7 - 2x^3 = y\)
\(2x^3 = 7 - y \implies x^3 = \frac{7 - y}{2} \implies x = \left(\frac{7 - y}{2}\right)^{1/3}\).
Since \(y \in R\), \(x\) is a real number. Thus, every \(y\) has a pre-image \(x\) in \(R\).
Hence, \(f\) is onto.
Since \(f\) is both one-one and onto, it is bijective.
In simple words: Proving algebraic equality of inputs leads to equal outputs (one-one), and solving for x for any real y proves every output is covered (onto).

Exam Tip: Polynomial functions of odd degree with domain R are typically bijective.

Question 5. If \(f: R \to R\) is defined by \(f(x) = \frac{x^3 + 5}{2}\). Find \(f^{-1}\).
Answer: Let \(y = \frac{x^3 + 5}{2}\).
\(2y = x^3 + 5\)
\(x^3 = 2y - 5\)
\(x = (2y - 5)^{1/3}\)
Replacing \(y\) with \(x\), we get:
\(f^{-1}(x) = (2x - 5)^{1/3}\).
In simple words: Cross-multiply by 2, subtract 5, take the cube root, and swap variables.

Exam Tip: Always state the final inverse clearly with x as the independent variable.

LEVEL III

Question 1. Show that the function \(f: R \to R\) defined by \(f(x) = \frac{x^2 - 1}{3}, x \in R\) is one-one & onto function. Also find the \(f^{-1}\).
Answer: 1. One-one: Let \(f(x_1) = f(x_2)\).
\(\frac{x_1^2 - 1}{3} = \frac{x_2^2 - 1}{3}\)
\(x_1^2 - 1 = x_2^2 - 1 \implies x_1^2 = x_2^2\).
Since \(x_1, x_2 \in R\), \(x_1^2 = x_2^2\) does not necessarily mean \(x_1 = x_2\) (e.g., \(1\) and \(-1\)). Thus, the function as defined on full \(R\) is actually not one-one (Note: There is a typo in the original question sheet where domain should be restricted to non-negative reals or positive reals, but proceeding with standard algebraic check: \(f(1) = 0\) and \(f(-1) = 0\), so it's not one-one on \(R\)). Assuming standard textbook intended domain \(R^+ \cup \{0\}\):
If \(x \ge 0\), then \(x_1 = x_2\), making it one-one.
2. Onto: Let \(y = \frac{x^2 - 1}{3} \implies 3y = z \dots\), \(x = \sqrt{3y + 1}\).
For \(y \ge -1/3\), every element has a pre-image.
Inverse: \(f^{-1}(x) = \sqrt{3x + 1}\).
In simple words: When restricted to positive real numbers, the function becomes invertible with inverse root(3x + 1).

Exam Tip: Watch out for domain restrictions on quadratic functions to ensure they are bijective.

Question 2. Consider a function \(f: R^+ \to [-5, \infty)\) defined \(f(x) = 9x^2 + 6x - 5\). Show that f is invertible & \(f^{-1}(y) = \frac{\sqrt{y + 6} - 1}{3}\), where \(R^+ = (0, \infty)\).
Answer: Let \(y = 9x^2 + 6x - 5\).
Complete the square for the quadratic expression in terms of \(x\):
\(y = (3x)^2 + 2(3x)(1) + 1^2 - 1^2 - 5\)
\(y = (3x + 1)^2 - 6\)
\(y + 6 = (3x + 1)^2\)
Since \(x \in R^+\), \(3x + 1 > 0\), so taking the positive square root:
\(\sqrt{y + 6} = 3x + 1\)
\(3x = \sqrt{y + 6} - 1\)
\(x = \frac{\sqrt{y + 6} - 1}{3}\)
Replacing \(y\) with \(x\), we get the inverse:
\(f^{-1}(x) = \frac{\sqrt{x + 6} - 1}{3}\).
In simple words: Complete the square for the quadratic expression, solve for x, and swap variables to get the inverse.

Exam Tip: Completing the square is the standard technique for inverting quadratic functions.

Question 3. Consider a function \(f: R \to R\) given by \(f(x) = 4x + 3\). Show that f is invertible & \(f^{-1}: R \to R\) with \(f^{-1}(y) = \frac{y - 3}{4}\).
Answer: Let \(y = 4x + 3\).
Solving for \(x\):
\(4x = y - 3\)
\(x = \frac{y - 3}{4}\)
Replacing \(y\) with \(x\), we obtain:
\(f^{-1}(x) = \frac{x - 3}{4}\).
Thus, \(f\) is invertible with the given inverse function. [Note: The worksheet blank is filled by \(\frac{y - 3}{4}\)].
In simple words: Linear functions are easily inverted by isolating x and swapping variables.

Question 4. Show that \(f: R \to R\) defined by \(f(x) = x^3 + 4\) is one-one, onto. Show that \(f^{-1}(x) = (x - 4)^{1/3}\).
Answer: 1. One-one: Let \(f(x_1) = f(x_2)\).
\(x_1^3 + 4 = x_2^3 + 4 \implies x_1^3 = x_2^3 \implies x_1 = x_2\).
Thus, \(f\) is one-one.
2. Onto: Let \(y \in R\).
\(f(x) = y \implies x^3 + 4 = y \implies x^3 = y - 4 \implies x = (y - 4)^{1/3}\).
For every real \(y\), \(x\) is a real number. Thus, every element has a pre-image.
Hence, \(f\) is onto.
Inverse: Replacing \(y\) with \(x\), \(f^{-1}(x) = (x - 4)^{1/3}\).
In simple words: Cubic functions are strictly increasing, making them one-one, and solving for x shows they cover all real outputs, making them onto.

Exam Tip: Cubic functions are standard examples of bijective functions from R to R.

Question 5. Let \(A = R - \{3\}\) and \(B = R - \{1\}\). Consider the function \(f: A \to B\) defined by \(f(x) = \frac{x - 2}{x - 3}\). Show that f is one-one onto and hence find \(f^{-1}\). [CBSE 2012]
Answer: 1. One-one: Let \(f(x_1) = f(x_2)\).
\(\frac{x_1 - 2}{x_1 - 3} = \frac{x_2 - 2}{x_2 - 3}\)
Cross-multiplying:
\((x_1 - 2)(x_2 - 3) = (x_2 - 2)(x_1 - 3)\)
\(x_1 x_2 - 3x_1 - 2x_2 + 6 = x_1 x_2 - 3x_2 - 2x_1 + 6\)
Cancel \(x_1 x_2\) and \(6\):
\(-3x_1 - 2x_2 = -3x_2 - 2x_1\)
\(-x_1 = -x_2 \implies x_1 = x_2\).
Thus, \(f\) is one-one.
2. Onto: Let \(y \in B\), where \(y \neq 1\).
\(y = \frac{x - 2}{x - 3}\)
\(y(x - 3) = x - 2\)
\(yx - 3y = x - 2\)
\(yx - x = 3y - 2\)
\(x(y - 1) = 3y - 2\)
\(x = \frac{3y - 2}{y - 1}\).
Since \(y \neq 1\), \(x\) is a well-defined real number, and since \(x \neq 3\) (if \(\frac{3y - 2}{y - 1} = 3 \implies 3y - 2 = 3y - 3 \implies -2 = -3\) (impossible)), \(x \in A\).
Thus, every element in co-domain has a pre-image. Hence, \(f\) is onto.
Inverse: \(f^{-1}(x) = \frac{3x - 2}{x - 1}\).
In simple words: Cross-multiply to prove one-one, solve for x in terms of y to prove onto, and swap variables to get the inverse.

Exam Tip: Rational functions of degree 1 over 1 require careful cross-multiplication for both one-one and onto proofs.

Question 6. Show that \(f: N \to N\) defined by \(f(x) = \begin{cases} x + 1, & \text{if } x \text{ is odd} \\ x - 1, & \text{if } x \text{ is even} \end{cases}\) is both one-one onto. [CBSE 2012]
Answer: 1. One-one: Let \(x_1, x_2 \in N\).
Case 1: If both \(x_1, x_2\) are even.
\(f(x_1) = f(x_2) \implies x_1 - 1 = x_2 - 1 \implies x_1 = x_2\).
Case 2: If both \(x_1, x_2\) are odd.
\(f(x_1) = f(x_2) \implies x_1 + 1 = x_2 + 1 \implies x_1 = x_2\).
Case 3: If one is odd and one is even (say \(x_1\) odd, \(x_2\) even).
\(f(x_1) = x_1 + 1\) (even) and \(f(x_2) = x_2 - 1\) (odd). An even number cannot equal an odd number. Thus, images cannot be equal for distinct parity inputs.
Hence, \(f\) is one-one.
2. Onto: Let \(y \in N\).
If \(y\) is odd, then \(y + 1\) is even, and \(f(y + 1) = (y + 1) - 1 = y\).
If \(y\) is even, then \(y - 1\) is odd, and \(f(y - 1) = (y - 1) + 1 = y\).
Thus, every natural number \(y\) has a pre-image in \(N\).
Hence, \(f\) is onto.
In simple words: Checking all parity combinations proves distinct inputs give distinct outputs, and every natural number has a corresponding pre-image.

Exam Tip: For piecewise functions based on even and odd numbers, test cases by separating the domain into even and odd integers.

(iv) Composition of functions

LEVEL I

Question 1. If \(f(x) = e^{2x}\) and \(g(x) = \log\sqrt{x}, x > 0\), find
(a) \((f + g)(x)\)
(b) \((f \cdot g)(x)\)
(c) \((f \circ g)(x)\)
(d) \((g \circ f)(x)\).
Answer: Given \(f(x) = e^{2x}\) and \(g(x) = \log\sqrt{x} = \frac{1}{2}\log x\).
(a) \((f + g)(x) = f(x) + g(x) = e^{2x} + \frac{1}{2}\log x\)
(b) \((f \cdot g)(x) = f(x) \cdot g(x) = e^{2x} \cdot \frac{1}{2}\log x = \frac{1}{2}e^{2x}\log x\)
(c) \((f \circ g)(x) = f(g(x)) = f(\log\sqrt{x}) = e^{2\log\sqrt{x}} = e^{\log(\sqrt{x})^2} = e^{\log x} = x\)
(d) \((g \circ f)(x) = g(f(x)) = g(e^{2x}) = \log\sqrt{e^{2x}} = \log(e^x) = x \log e = x\).
In simple words: Combine functions using standard arithmetic operations or substitute one function inside the other for composite parts.

Exam Tip: Use logarithm properties like \(\log(a^b) = b\log a\) and \(e^{\log x} = x\) to simplify composite exponential-logarithmic functions.

Question 2. If \(f(x) = \frac{x - 1}{x + 1}\), then show that
(a) \(f\left(\frac{1}{x}\right) = -f(x)\)
(b) \(f\left(-\frac{1}{x}\right) = \frac{1}{f(x)}\).
Answer: Given \(f(x) = \frac{x - 1}{x + 1}\).
(a) Substitute \(\frac{1}{x}\) for \(x\):
\(f\left(\frac{1}{x}\right) = \frac{\frac{1}{x} - 1}{\frac{1}{x} + 1} = \frac{\frac{1 - x}{x}}{\frac{1 + x}{x}} = \frac{1 - x}{1 + x} = -\frac{x - 1}{x + 1} = -f(x)\).
(b) Substitute \(-\frac{1}{x}\) for \(x\):
\(f\left(-\frac{1}{x}\right) = \frac{-\frac{1}{x} - 1}{-\frac{1}{x} + 1} = \frac{\frac{-1 - x}{x}}{\frac{-1 + x}{x}} = \frac{-(1 + x)}{x - 1} = -\frac{x + 1}{x - 1} = \frac{1}{\frac{x - 1}{x + 1}} = \frac{1}{f(x)}\).
In simple words: Substitute the given fractional inputs into the function and simplify algebraic fractions.

Exam Tip: Simplify complex fractions by finding common denominators in both numerator and denominator.

LEVEL II

Question 1. Let \(f, g: R \to R\) be defined by \(f(x) = \mid x \mid\) & \(g(x) = [x]\) where \([x]\) denotes the greatest integer function. Find \((f \circ g)\left(\frac{5}{2}\right)\) & \((g \circ f)(-\sqrt{3})\).
Answer: 1. Finding \((f \circ g)\left(\frac{5}{2}\right)\):
\((f \circ g)\left(\frac{5}{2}\right) = f\left(g\left(\frac{5}{2}\right)\right)\)
Since \(\frac{5}{2} = 2.5\), greatest integer \(g(2.5) = [2.5] = 2\).
Now, \(f(2) = \mid 2 \mid = 2\).
2. Finding \((g \circ f)(-\sqrt{3})\):
\((g \circ f)(-\sqrt{3}) = g(f(-\sqrt{3}))\)
Since \(-\sqrt{3} \approx -1.732\), \(f(-\sqrt{3}) = \mid -\sqrt{3} \mid = \sqrt{3} \approx 1.732\).
Now, \(g(\sqrt{3}) = [\sqrt{3}] = [1.732] = 1\).
In simple words: Evaluate inner function first, apply its output to the outer function using greatest integer and absolute value rules.

Exam Tip: Greatest integer function [x] rounds down to the nearest integer less than or equal to x.

Question 2. Let \(f(x) = \frac{x - 1}{x + 1}\). Then find \(f(f(x))\).
Answer: Given \(f(x) = \frac{x - 1}{x + 1}\).
\(f(f(x)) = f\left(\frac{x - 1}{x + 1}\right)\)
Substitute \(\frac{x - 1}{x + 1}\) in place of \(x\) in \(f(x)\):
\(= \frac{\frac{x - 1}{x + 1} - 1}{\frac{x - 1}{x + 1} + 1}\)
Multiply numerator and denominator by \(x + 1\):
\(= \frac{(x - 1) - (x + 1)}{(x - 1) + (x + 1)}\)
\(= \frac{x - 1 - x - 1}{x - 1 + x + 1}\)
\(= \frac{-2}{2x} = -\frac{1}{x}\).
In simple words: Substitute the function into itself and simplify the resulting complex fraction.

Exam Tip: Clearing denominators by multiplying numerator and denominator by the LCD saves time.

Question 3. If \(y = f(x) = \frac{3x + 4}{5x - 3}\), then find \((f \circ f)(x)\) i.e. \(f(y)\).
Answer: Given \(f(x) = \frac{3x + 4}{5x - 3}\).
We need to find \(f(f(x))\):
\(f(f(x)) = \frac{3\left(\frac{3x + 4}{5x - 3}\right) + 4}{5\left(\frac{3x + 4}{5x - 3}\right) - 3}\)
Multiply numerator and denominator by \(5x - 3\):
\(= \frac{3(3x + 4) + 4(5x - 3)}{5(3x + 4) - 3(5x - 3)}\)
Expand terms:
Numerator = \(9x + 12 + 20x - 12 = 29x\)
Denominator = \(15x + 20 - 15x + 9 = 29\)
\(= \frac{29x}{29} = x\).
In simple words: Substituting the fractional function into itself yields x, showing that the function is its own inverse.

Exam Tip: When \(f(f(x)) = x\), it proves that the function is an involution and is its own inverse.

Question 4. Let \(f: R \to R\) be defined as \(f(x) = 10x + 7\). Find the function \(g: R \to R\) such that \(g \circ f(x) = f \circ g(x) = I_R\). [CBSE 2011]
Answer: Given \(f(x) = 10x + 7\).
We need \(g \circ f(x) = x \implies g(f(x)) = x\).
\(g(10x + 7) = x\)
Let \(10x + 7 = y \implies 10x = y - 7 \implies x = \frac{y - 7}{10}\).
Substituting back into \(g\):
\(g(y) = \frac{y - 7}{10}\)
Thus, \(g(x) = \frac{x - 7}{10}\).
In simple words: The function g is simply the inverse of f, found by solving for x in terms of y.

Question 5. If \(f: R \to R\) be defined as \(f(x) = (3 - x^3)^{1/3}\), then find \(f \circ f(x)\). [CBSE 2010]
Answer: Given \(f(x) = (3 - x^3)^{1/3}\).
\(f \circ f(x) = f(f(x)) = f\left((3 - x^3)^{1/3}\right)\)
Substitute into \(f(x)\):
\(= \left(3 - \left((3 - x^3)^{1/3}\right)^3\right)^{1/3}\)
\(= (3 - (3 - x^3))^{1/3}\)
\(= (3 - 3 + x^3)^{1/3}\)
\(= (x^3)^{1/3} = x\).
In simple words: Substitute the function inside itself and simplify the cube and cube root powers to get x.

Exam Tip: Watch out for exponent cancellation when cubing a cube root term.

Question 6. Let \(f: R \to R\) & \(g: R \to R\) be defined as \(f(x) = x^2, g(x) = 2x - 3\). Find \(fog(x)\).
Answer: Given \(f(x) = x^2\) and \(g(x) = 2x - 3\).
\(fog(x) = f(g(x)) = f(2x - 3)\)
Substitute \(2x - 3\) into \(f(x)\):
\(= (2x - 3)^2\)
\(= 4x^2 - 12x + 9\).
In simple words: Substitute g(x) into f(x) and expand using algebraic identity.

Exam Tip: Composition order matters: fog(x) means apply g first, then f.

(v) Binary Operations

LEVEL I

Question 1. Let * be the binary operation on N given by \(a * b = \text{LCM of } a \text{ & } b\). Find \(3 * 5\).
Answer: Given \(a * b = \text{LCM}(a, b)\).
\(3 * 5 = \text{LCM}(3, 5)\)
Since 3 and 5 are prime numbers, their LCM is their product: \(3 \times 5 = 15\).
In simple words: Find the least common multiple of 3 and 5, which is 15.

Exam Tip: LCM of coprime numbers is simply their product.

Question 2. Let * be the binary on N given by \(a * b = \text{HCF of } \{a, b\}, a, b \in N\). Find \(20 * 16\).
Answer: Given \(a * b = \text{HCF}(a, b)\).
\(20 * 16 = \text{HCF}(20, 16)\)
Prime factorizations:
\(20 = 2^2 \times 5\)
\(16 = 2^4\)
HCF = highest common factor = \(2^2 = 4\).
In simple words: Find the greatest common divisor of 20 and 16, which is 4.

Exam Tip: HCF is the product of the smallest powers of common prime factors.

Question 3. Let * be a binary operation on the set Q of rational numbers defined as \(a * b = \frac{ab}{5}\). Write the identity of *, if any.
Answer: Let \(e \in Q\) be the identity element for *.
Then \(a * e = a\) for all \(a \in Q\).
\(\frac{ae}{5} = a\)
Since \(a \neq 0\) (for general identity element definition), dividing by \(a\):
\(\frac{e}{5} = 1 \implies e = 5\).
Thus, the identity element is \(5\).
In simple words: Set the operation formula equal to a with an identity e, and solve for e to get 5.

Exam Tip: Identity element satisfies \(a * e = e * a = a\).

Question 4. If a binary operation * on the set of integer Z, is defined by \(a * b = a + 3b^2\). Then find the value of \(2 * 4\).
Answer: Given \(a * b = a + 3b^2\).
Substitute \(a = 2\) and \(b = 4\):
\(2 * 4 = 2 + 3(4)^2\)
\(= 2 + 3(16)\)
\(= 2 + 48 = 50\).
In simple words: Substitute 2 for a and 4 for b into the given formula and calculate the result.

Exam Tip: Follow BODMAS rules: evaluate exponents before multiplication and addition.

LEVEL II

Question 1. Let \(A = N \times N\) & * be the binary operation on A defined by \((a, b) * (c, d) = (a + c, b + d)\). Show that * is (a) Commutative (b) Associative (c) Find identity for * on A, if any.
Answer: (a) Commutativity: Let \((a, b), (c, d) \in N \times N\).
\((a, b) * (c, d) = (a + c, b + d)\)
Since addition of natural numbers is commutative (\(a + c = c + a\) and \(b + d = d + b\)):
\(= (c + a, d + b) = (c, d) * (a, b)\).
Thus, * is commutative.
(b) Associativity: Let \((a, b), (c, d), (e, f) \in N \times N\).
\([(a, b) * (c, d)] * (e, f) = (a + c, b + d) * (e, f) = ((a + c) + e, (b + d) + f)\)
Using associativity of addition in N:
\(= (a + (c + e), b + (d + f)) = (a, b) * (c + e, d + f) = (a, b) * [(c, d) * (e, f)]\).
Thus, * is associative.
(c) Identity element: Let \((e_1, e_2) \in N \times N\) be the identity.
\((a, b) * (e_1, e_2) = (a, b)\)
\((a + e_1, b + e_2) = (a, b)\)
\(a + e_1 = a \implies e_1 = 0\)
\(b + e_2 = b \implies e_2 = 0\)
Since \(0 \notin N\) (natural numbers start from 1), there is no identity element in \(N \times N\). (If \(N\) included 0, identity would be \((0, 0)\)).
In simple words: Addition in components is commutative and associative, but zero is required for an identity which is not in natural numbers N.

Exam Tip: Always check whether the calculated identity element belongs to the specified set.

Question 2. Let \(A = Q \times Q\). Let * be a binary operation on A defined by \((a, b) * (c, d) = (ac, ad + b)\). Find: (i) the identity element of A (ii) the invertible element of A.
Answer: (i) Identity element: Let \((e_1, e_2) \in Q \times Q\) be the identity element.
\((a, b) * (e_1, e_2) = (a, b)\)
\((ae_1, ad_2 + b) = (a, b)\) [Wait, applying formula: \((a, b) * (e_1, e_2) = (ae_1, ae_2 + b)\)]
Comparing coordinates:
\(ae_1 = a \implies e_1 = 1\)
\(ae_2 + b = b \implies ae_2 = 0 \implies e_2 = 0\).
Thus, the identity element is \((1, 0)\).
(ii) Invertible elements: Let \((x, y)\) be the inverse of \((a, b)\).
\((a, b) * (x, y) = \text{Identity} = (1, 0)\)
\((ax, ay + b) = (1, 0)\)
\(ax = 1 \implies x = \frac{1}{a}\) (provided \(a \neq 0\))
\(ay + b = 0 \implies ay = -b \implies y = -\frac{b}{a}\).
Thus, every element \((a, b)\) with \(a \neq 0\) is invertible, and its inverse is \(\left(\frac{1}{a}, -\frac{b}{a}\right)\).
In simple words: Solve component equations with the identity (1,0) to find the identity, and repeat for the inverse pair.

Question 3. Examine which of the following is a binary operation
(i) \(a * b = \frac{a + b}{2}\); \(a, b \in N\)
(ii) \(a * b = \frac{a + b}{2}\); \(a, b \in Q\)
For binary operation check commutative & associative law.
Answer: (i) For \(a, b \in N\): Let \(a = 1, b = 2\).
\(1 * 2 = \frac{1 + 2}{2} = \frac{3}{2} \notin N\).
Since the result does not always belong to \(N\), this is not a binary operation on \(N\).
(ii) For \(a, b \in Q\): Since the average of two rational numbers is always a rational number, this is a binary operation on \(Q\).
Checking properties for \(a * b = \frac{a + b}{2}\) on \(Q\):
- Commutativity: \(a * b = \frac{a + b}{2} = \frac{b + a}{2} = b * a\). (Commutative)
- Associativity: \((a * b) * c = \frac{\frac{a + b}{2} + c}{2} = \frac{a + b + 2c}{4}\).
\(a * (b * c) = \frac{a + \frac{b + c}{2}}{2} = \frac{2a + b + c}{4}\).
Since \(\frac{a + b + 2c}{4} \neq \frac{2a + b + c}{4}\) (in general), it is not associative.
In simple words: Average of natural numbers is not always natural (not a binary op on N), but is rational (is a binary op on Q). It is commutative but not associative.

Exam Tip: Always check if the output of an operation stays within the specified set to test if it's a binary operation.

LEVEL III

Question 1. Let \(A = N \times N\) & * be a binary operation on A defined by \((a, b) \times (c, d) = (ac, bd)\) for \((a, b), (c, d) \in N \times N\)
(i) Find \((2, 3) * (4, 1)\)
(ii) Find \([(2, 3) * (4, 1)] * (3, 5)\) and \((2, 3) * [(4, 1) * (3, 5)]\) & show they are equal
(iii) Show that * is commutative & associative on A.
Answer: Given \((a, b) * (c, d) = (ac, bd)\).
(i) \((2, 3) * (4, 1) = (2 \times 4, 3 \times 1) = (8, 3)\).
(ii) First expression: \([(2, 3) * (4, 1)] * (3, 5) = (8, 3) * (3, 5) = (8 \times 3, 3 \times 5) = (24, 15)\).
Second expression: \((2, 3) * [(4, 1) * (3, 5)] = (2, 3) * (4 \times 3, 1 \times 5) = (2, 3) * (12, 5) = (2 \times 12, 3 \times 5) = (24, 15)\).
Since both results are equal, associative property holds for these elements.
(iii) Commutativity: \((a, b) * (c, d) = (ac, bd) = (ca, db) = (c, d) * (a, b)\) (since multiplication in N is commutative). Thus, * is commutative.
Associativity: \([(a, b) * (c, d)] * (e, f) = (ac, bd) * (e, f) = ((ac)e, (bd)f) = (a(ce), b(df)) = (a, b) * (ce, df) = (a, b) * [(c, d) * (e, f)]\). Thus, * is associative.
In simple words: Component-wise multiplication inherits commutativity and associativity from natural number multiplication.

Exam Tip: Component-wise binary operations inherit properties directly from the underlying set operations.

Question 2.

Define a binary operation \(*\) on the set \(A=\{0,1,2,3,4,5\}\) as

\[ a*b= \begin{cases} a+b, & \text{if } a+b<6\\ a+b-6, & \text{if } a+b\geq 6 \end{cases} \]

Show that zero is the identity element for this operation and each element \(a\) of the set has an inverse given by:

\[ \boxed{ a^{-1}= \begin{cases} 0, & a=0\\ 6-a, & a\neq 0 \end{cases} } \]

Answer: 1. Identity element: Let \(e = 0\).
For any \(a \in \{0, 1, 2, 3, 4, 5\}\), since \(a + 0 = a < 6\),
\(a * 0 = a + 0 = a\) and \(0 * a = 0 + a = a\).
Thus, \(0\) is the identity element.
2. Invertibility: Let \(b = 6 - a\) be the inverse of \(a\).
For \(a = 0\), inverse is \(6 - 0 = 6 \notin \text{set} \dots\) wait, for \(a = 0\), \(0 * 0 = 0\), so 0 is its own inverse. For \(a \neq 0\), \(b = 6 - a\).
Check \(a * (6 - a)\):
\(a + (6 - a) = 6 \ge 6\).
Using the second rule: \(a + (6 - a) - 6 = 6 - 6 = 0\) (which is the identity element).
Thus, every element \(a\) is invertible, with \(6 - a\) as its inverse (and 0 is its own inverse).
In simple words: Zero acts as standard zero-sum identity, and adding (6-a) triggers the minus-6 rule to yield identity 0.

Exam Tip: Modular arithmetic style operations like addition modulo 6 follow these exact piecewise definitions.

Question 3. Consider the binary operations \(*: R \times R \to R\) and \(o: R \times R \to R\) defined as \(a * b = \mid a - b \mid\) and \(a \text{ o } b = a, a, b \in R\). Show that * is commutative but not associative, o is associative but not commutative. 
Answer: 1. For operation * (\(a * b = \mid a - b \mid\)):
- Commutativity: \(a * b = \mid a - b \mid = \mid -(b - a) \mid = \mid b - a \mid = b * a\). (Commutative)
- Associativity: Let \(a = 1, b = 2, c = 3\).
\((1 * 2) * 3 = \mid 1 - 2 \mid * 3 = 1 * 3 = \mid 1 - 3 \mid = 2\).
\(1 * (2 * 3) = 1 * \mid 2 - 3 \mid = 1 * 1 = \mid 1 - 1 \mid = 0\).\br /> Since \(2 \neq 0\), * is not associative.
2. For operation o (\(a \text{ o } b = a\)):
- Commutativity: Let \(a = 2, b = 3\).
\(2 \text{ o } 3 = 2\) and \(3 \text{ o } 2 = 3\). Since \(2 \neq 3\), o is not commutative.
- Associativity: \((a \text{ o } b) \text{ o } c = a \text{ o } c = a\).
\(a \text{ o } (b \text{ o } c) = a \text{ o } b = a\).
Since both sides are equal, o is associative.
In simple words: Absolute difference is symmetric (commutative) but fails grouping (not associative), while selection operation (taking first element) groups associatively but is order-dependent.