CBSE Class 12 Maths: Relations and Functions Miscellaneous Exercise

Understanding the Miscellaneous Exercise

The 'Miscellaneous Exercise' in your NCERT textbook for Relations and Functions is strategically placed to offer a comprehensive review of the entire chapter. Unlike preceding exercises that often focus on individual concepts, these problems are designed to integrate your understanding across different topics. You might encounter questions that ask you to prove a function is both injective and surjective, determine if a complex relation is an equivalence relation, or find the inverse of a composite function.

To effectively tackle these problems, it's essential to recall definitions precisely and understand the conditions under which properties hold. For instance, proving a function is one-one requires demonstrating that distinct elements always map to distinct images, while proving it's onto involves showing that every element in the codomain has a pre-image. These exercises demand not just calculation, but also logical reasoning and structured proof writing. Approach each problem as a puzzle that needs a careful, step-by-step solution, ensuring every condition is met and every property is justified. This exercise is key to truly mastering Relations and Functions.

Key Concepts Revisited

Equivalence Relation

A relation R on a set A is an equivalence relation if it is reflexive (aRa), symmetric (aRb implies bRa), and transitive (aRb and bRc implies aRc).

Injective (One-One) Function

A function f: A → B is injective if distinct elements of A have distinct images in B. Mathematically, if f(x₁) = f(x₂), then x₁ = x₂ for all x₁, x₂ ∈ A.

Surjective (Onto) Function

A function f: A → B is surjective if every element in the codomain B has at least one pre-image in the domain A. This means for every y ∈ B, there exists an x ∈ A such that f(x) = y.

Bijective Function

A function is bijective if and only if it is both injective (one-one) and surjective (onto).

Composition of Functions (gof)

If f: A → B and g: B → C are two functions, then their composition, denoted as gof, is a function from A to C defined by (gof)(x) = g(f(x)).

Inverse of a Function (f⁻¹)

A function f: A → B has an inverse function f⁻¹: B → A if and only if f is bijective. If f(x) = y, then f⁻¹(y) = x.

Fully Worked Examples from NCERT

• Example 1: Proving Bijectivity and Finding Inverse Show that the function f: N → Y, defined by f(x) = 4x + 3, where Y = {y ∈ N: y = 4x + 3 for some x ∈ N}, is invertible. Also, find the inverse of f. Solution: Step 1: Prove f is One-One (Injective) Let x₁, x₂ ∈ N such that f(x₁) = f(x₂). So, 4x₁ + 3 = 4x₂ + 3 Subtract 3 from both sides: 4x₁ = 4x₂ Divide by 4: x₁ = x₂ Since f(x₁) = f(x₂) implies x₁ = x₂, the function f is one-one. Step 2: Prove f is Onto (Surjective) Let y be an arbitrary element of the codomain Y. By the definition of Y, we know that y = 4x + 3 for some x ∈ N. We need to express x in terms of y. From y = 4x + 3, we get: 4x = y - 3 x = (y - 3) / 4 Since y ∈ Y, y must be of the form 4k + 3 for some k ∈ N (e.g., if x=1, y=7; if x=2, y=11). This implies y-3 is a multiple of 4 and y-3 ≥ 4(1)+3-3 = 4 (since x>=1). Thus, (y-3)/4 is a natural number (x ∈ N). So, for every y ∈ Y, there exists an x = (y - 3) / 4 ∈ N such that f(x) = f((y - 3) / 4) = 4((y - 3) / 4) + 3 = (y - 3) + 3 = y. Therefore, f is onto. Step 3: Conclusion on Invertibility Since f is both one-one and onto, it is a bijective function. Hence, f is invertible. Step 4: Find the Inverse Function (f⁻¹) Let f(x) = y. Then y = 4x + 3. To find the inverse, we express x in terms of y: x = (y - 3) / 4 So, the inverse function f⁻¹: Y → N is defined by f⁻¹(y) = (y - 3) / 4. Final Answer: The function f is invertible, and its inverse is f⁻¹(y) = (y - 3) / 4.
• Example 2: Determining Relation Properties Let R be a relation on the set A = {1, 2, 3, ..., 14} defined by R = {(x, y): 3x - y = 0}. Determine if R is reflexive, symmetric, and transitive. Solution: Step 1: List the elements of the relation R The condition is 3x - y = 0, which means y = 3x. Since x, y ∈ A = {1, 2, ..., 14}: If x = 1, y = 3(1) = 3. So (1, 3) ∈ R. If x = 2, y = 3(2) = 6. So (2, 6) ∈ R. If x = 3, y = 3(3) = 9. So (3, 9) ∈ R. If x = 4, y = 3(4) = 12. So (4, 12) ∈ R. If x = 5, y = 3(5) = 15, which is not in A. So we stop here. Thus, R = {(1, 3), (2, 6), (3, 9), (4, 12)}. Step 2: Check for Reflexivity A relation R on set A is reflexive if (a, a) ∈ R for every a ∈ A. For this relation, we need (1, 1), (2, 2), ..., (14, 14) to be in R. However, (1, 1) ∉ R because 3(1) - 1 = 2 ≠ 0. Therefore, R is not reflexive. Step 3: Check for Symmetry A relation R is symmetric if (x, y) ∈ R implies (y, x) ∈ R. Consider (1, 3) ∈ R. For R to be symmetric, (3, 1) must be in R. But 3(3) - 1 = 9 - 1 = 8 ≠ 0. Therefore, (3, 1) ∉ R. Hence, R is not symmetric. Step 4: Check for Transitivity A relation R is transitive if (x, y) ∈ R and (y, z) ∈ R implies (x, z) ∈ R. Consider the elements of R: We have (1, 3) ∈ R. We also have (3, 9) ∈ R. For transitivity, (1, 9) must be in R. Let's check: 3(1) - 9 = 3 - 9 = -6 ≠ 0. Therefore, (1, 9) ∉ R. Hence, R is not transitive. Final Answer: The relation R is neither reflexive, nor symmetric, nor transitive.

Exam Traps & Important Points to Remember

When tackling miscellaneous exercises and board exam questions on Relations and Functions, students often fall into specific traps. Here are some critical points to keep in mind:

1. Domain and Codomain are Crucial: Always pay close attention to the specified domain and codomain of a function. A function that is one-one on one domain might not be on another, or it might be onto a different codomain. For example, f(x) = x² is not one-one on R, but it is one-one on N.
2. Proving 'One-One' vs. 'Not One-One': To prove a function is one-one, you must use the general proof: assume f(x₁) = f(x₂) and show x₁ = x₂. To prove it is not one-one, a single counterexample (x₁ ≠ x₂ but f(x₁) = f(x₂)) is sufficient.
3. Proving 'Onto' vs. 'Not Onto': To prove a function is onto, take an arbitrary 'y' from the codomain and show that an 'x' exists in the domain such that f(x) = y. This often involves solving for x in terms of y. To prove it's not onto, find an element 'y' in the codomain for which no 'x' in the domain exists such that f(x) = y.
4. Inverse Function Condition: An inverse function f⁻¹ exists if and only if the original function f is bijective (both one-one and onto). Do not attempt to find an inverse if the function isn't bijective.
5. Composition Order: Remember that (gof)(x) is generally not equal to (fog)(x). The order of composition matters. (gof)(x) means applying f first, then g to the result.

Practice Questions with Solutions

• Q: Let A = R - {2/3} and B = R - {4/3}. If f: A → B is defined by f(x) = (4x + 3)/(3x - 2), show that f is bijective. A: Step 1: Prove f is One-One. Assume f(x₁) = f(x₂). (4x₁ + 3)/(3x₁ - 2) = (4x₂ + 3)/(3x₂ - 2) (4x₁ + 3)(3x₂ - 2) = (4x₂ + 3)(3x₁ - 2) 12x₁x₂ - 8x₁ + 9x₂ - 6 = 12x₁x₂ - 8x₂ + 9x₁ - 6 -8x₁ + 9x₂ = -8x₂ + 9x₁ 17x₂ = 17x₁ x₁ = x₂. Thus, f is one-one. Step 2: Prove f is Onto. Let y ∈ B. We need to find x ∈ A such that f(x) = y. y = (4x + 3)/(3x - 2) y(3x - 2) = 4x + 3 3xy - 2y = 4x + 3 3xy - 4x = 2y + 3 x(3y - 4) = 2y + 3 x = (2y + 3)/(3y - 4) Since y ∈ B = R - {4/3}, 3y - 4 ≠ 0, so x is well-defined. Also, we must ensure x ≠ 2/3. If x = 2/3, then (2y+3)/(3y-4) = 2/3 => 3(2y+3) = 2(3y-4) => 6y+9 = 6y-8 => 9 = -8, which is false. So x is never 2/3. Thus, for every y ∈ B, there exists an x ∈ A. Therefore, f is onto. Final answer: Since f is both one-one and onto, it is bijective.
• Q: Let R be a relation on the set Z of all integers defined by R = {(x, y): x - y is an integer}. Determine if R is an equivalence relation. A: Step 1: Check for Reflexivity. For any integer x ∈ Z, x - x = 0. Since 0 is an integer, (x, x) ∈ R for all x ∈ Z. Thus, R is reflexive. Step 2: Check for Symmetry. Let (x, y) ∈ R. This means x - y is an integer. Let x - y = k, where k ∈ Z. Then y - x = -(x - y) = -k. Since k is an integer, -k is also an integer. So, (y, x) ∈ R. Thus, R is symmetric. Step 3: Check for Transitivity. Let (x, y) ∈ R and (y, z) ∈ R. This means x - y = k₁ (an integer) and y - z = k₂ (an integer). Consider x - z. We can write x - z = (x - y) + (y - z) = k₁ + k₂. Since k₁ and k₂ are integers, their sum k₁ + k₂ is also an integer. So, (x, z) ∈ R. Thus, R is transitive. Final answer: Since R is reflexive, symmetric, and transitive, R is an equivalence relation.
• Q: Consider f: {1, 2, 3} → {a, b, c} and g: {a, b, c} → {apple, ball, cat} defined as f(1) = a, f(2) = b, f(3) = c and g(a) = apple, g(b) = ball, g(c) = cat. Find gof and verify if it is invertible. A: Step 1: Find gof. (gof)(1) = g(f(1)) = g(a) = apple. (gof)(2) = g(f(2)) = g(b) = ball. (gof)(3) = g(f(3)) = g(c) = cat. So, gof: {1, 2, 3} → {apple, ball, cat} is defined by {(1, apple), (2, ball), (3, cat)}. Step 2: Verify if gof is one-one. The elements 1, 2, 3 in the domain {1, 2, 3} have distinct images (apple, ball, cat) in the codomain {apple, ball, cat}. Thus, gof is one-one. Step 3: Verify if gof is onto. The codomain of gof is {apple, ball, cat}. Each element in the codomain (apple, ball, cat) has a pre-image (1, 2, 3 respectively) in the domain. Thus, gof is onto. Step 4: Conclusion on invertibility. Since gof is both one-one and onto, it is bijective. Therefore, gof is invertible. Final answer: gof = {(1, apple), (2, ball), (3, cat)}, and gof is invertible because it is bijective.
• Q: Let A = {-1, 0, 1, 2}, B = {-4, -2, 0, 2} and f, g: A → B be functions defined by f(x) = x² - x and g(x) = 2|x - 1/2| - 1. Are f and g equal? Justify your answer. A: Step 1: Evaluate f(x) for each x in A. f(-1) = (-1)² - (-1) = 1 + 1 = 2. f(0) = (0)² - (0) = 0. f(1) = (1)² - (1) = 1 - 1 = 0. f(2) = (2)² - (2) = 4 - 2 = 2. So, f = {(-1, 2), (0, 0), (1, 0), (2, 2)}. Step 2: Evaluate g(x) for each x in A. g(-1) = 2|-1 - 1/2| - 1 = 2|-3/2| - 1 = 2(3/2) - 1 = 3 - 1 = 2. g(0) = 2|0 - 1/2| - 1 = 2|-1/2| - 1 = 2(1/2) - 1 = 1 - 1 = 0. g(1) = 2|1 - 1/2| - 1 = 2|1/2| - 1 = 2(1/2) - 1 = 1 - 1 = 0. g(2) = 2|2 - 1/2| - 1 = 2|3/2| - 1 = 2(3/2) - 1 = 3 - 1 = 2. So, g = {(-1, 2), (0, 0), (1, 0), (2, 2)}. Step 3: Compare f and g. Since f(x) = g(x) for all x in the common domain A, and their codomains are also the same (B), the functions f and g are equal. Final answer: Yes, f and g are equal because f(x) = g(x) for all x ∈ A and they share the same domain and codomain.

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