Comprehensive Revision Guide on Relations and Functions for Class 12
Welcome back to StudyBeacon's Revision Series! Today, we dive into one of the most essential chapters of Class 12 Mathematics—Relations and Functions. This topic is the backbone for various concepts in calculus and algebra. In this guide, we’ll explore the various types of relations and functions, break down their properties, and tackle some challenging questions to sharpen your skills for exams like JEE Advanced. Let's get started!
Understanding Relations and Functions
To put it simply, a relation is a rule that connects elements from one set to another. A function is a special type of relation where each input has a unique output. Let's break this down further.
1. Types of Relations
Relations help us understand how two sets of information are interconnected. Here are the various types:
- Empty Relation: If none of the elements of set A are related to any elements of set A, it’s called an empty relation.
- Universal Relation: If every element of set A is related to every element of set A, it’s a universal relation.
- Reflexive Relation: A relation R on a set A is reflexive if every element is related to itself. For example, if A = {1, 2, 3}, then R = {(1, 1), (2, 2), (3, 3)} is reflexive.
- Symmetric Relation: A relation R is symmetric if (a, b) ∈ R implies (b, a) ∈ R. For instance, if R = {(1, 2), (2, 1)}, it is symmetric.
- Transitive Relation: A relation R is transitive if (a, b) ∈ R and (b, c) ∈ R imply (a, c) ∈ R. For example, R = {(1, 2), (2, 3), (1, 3)} is transitive.
- Equivalence Relation: A relation that is reflexive, symmetric, and transitive is called an equivalence relation. These are particularly useful in geometry and algebra.
Example:
Consider a set A = {1, 2, 3}. Define a relation R on A as "x is divisible by y." Check if R is reflexive, symmetric, or transitive.
Solution:
- Reflexive: No, because 2 is not divisible by 2.
- Symmetric: No, because if (2,1) is true, (1,2) may not be.
- Transitive: Yes, if (2,1) is true and (3,2), (3,1) holds.
2. Types of Functions
A function is a relation where each input has a unique output. Understanding different types of functions helps in many competitive exams.
- One-One Function (Injective): Each element of set A maps to a unique element of set B. No two elements in A have the same image in B.
- Onto Function (Surjective): Every element of B has a pre-image in A. All elements of B are covered.
- Bijective Function: A function that is both one-one and onto. For example, f(x) = x + 2 where x is a real number.
- Constant Function: All elements of A map to the same element in B. For example, f(x) = 5.
Example:
Let f: R → R be defined by f(x) = 2x + 3. Is this function one-one, onto, or both?
Solution:
- One-One: Yes, since every distinct x maps to a distinct y.
- Onto: Yes, as every real number y can be written as y = 2x + 3.
3. Composite and Inverse Functions
Composite Functions: If f: A → B and g: B → C are two functions, then the composite function (g ∘ f): A → C is defined by (g ∘ f)(x) = g(f(x)). Understanding composite functions is vital as they frequently appear in calculus and coordinate geometry.
Inverse Functions: A function f: A → B is said to be invertible if there exists a function g: B → A such that (g ∘ f)(x) = x for all x ∈ A and (f ∘ g)(y) = y for all y ∈ B. The function g is the inverse of f, denoted by f⁻¹.
Example:
If f(x) = 3x + 2 and g(x) = (x - 2)/3, verify if g is the inverse of f.
Solution:
- f(g(x)) = f((x - 2)/3) = 3((x - 2)/3) + 2 = x.
- g(f(x)) = g(3x + 2) = ((3x + 2) - 2)/3 = x.
Hence, g is indeed the inverse of f.
4. Properties of Functions
Some important properties that you must know include:
- Even Function: A function f(x) is even if f(-x) = f(x) for all x in its domain.
- Odd Function: A function f(x) is odd if f(-x) = -f(x) for all x in its domain.
- Periodic Functions: A function f(x) is periodic if there exists a T > 0 such that f(x + T) = f(x) for all x. For example, sine and cosine functions are periodic with period 2π.
Practice Questions (JEE Advanced Level)
Question 1:
If f(x) = |x| and g(x) = x² - 2x, find (f ∘ g)(x) and discuss its nature.
Solution: The composite function (f ∘ g)(x) = f(g(x)) = |x² - 2x|. This function is always non-negative because it is an absolute value function. It is continuous for all x ∈ R, and it achieves its minimum at x = 1.
Question 2:
Let f: R → R be a function defined as f(x) = ex. Show that f(x) is one-one and onto when its co-domain is (0, ∞).
Solution:
- One-One: For any two real numbers x₁ and x₂, if f(x₁) = f(x₂), then ex₁ = ex₂. Taking the natural logarithm on both sides, we get x₁ = x₂. Therefore, f(x) is injective.
- Onto: For any y > 0, there exists an x = ln(y) such that f(x) = y. Thus, every element in (0, ∞) is mapped by some element in R, proving that f(x) is surjective.
Question 3:
Determine whether the following function is bijective: f: R → R defined by f(x) = 3x + 7.
Solution:
- One-One: If f(x₁) = f(x₂), then 3x₁ + 7 = 3x₂ + 7. Simplifying this gives x₁ = x₂, proving that the function is injective.
- Onto: For any y ∈ R, there exists an x = (y - 7)/3 such that f(x) = y. Therefore, the function is surjective.
Since f(x) is both one-one and onto, it is bijective.
Question 4:
Find the range of the function f(x) = √(x² - 4) and discuss its properties.
Solution:
The domain of the function is x ∈ R such that x² - 4 ≥ 0, which implies x ≥ 2 or x ≤ -2. The range of f(x) is [0, ∞) since the square root function produces non-negative values only. The function is neither injective nor surjective over R, but it can be made bijective on the restricted domain.
Conclusion
Relations and Functions are crucial not just for your board exams but also for competitive exams like JEE Advanced. Make sure to grasp each type of relation and function, understand their properties, and practice various problems to solidify your understanding. Try solving the above questions, and don’t hesitate to revisit the concepts if needed.
Keep practicing and stay tuned to StudyBeacon for more such in-depth revision notes and practice questions. If you found this helpful, please share it with your friends and let us know your thoughts in the comments!
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