Relation and Function and JEE Mains PYQ

𝔽𝕦𝕟𝕔𝕥𝕚𝕠𝕟𝕤 𝕒𝕟𝕕 ℝ𝕖𝕝𝕒𝕥𝕚𝕠𝕟𝕤

1.1 Introduction

In the field of mathematics, functions and relations are fundamental concepts that help us understand and describe the relationships between different quantities. They provide a way to model real-world phenomena and analyze the behavior of mathematical systems. This chapter will introduce the basic definitions and properties of functions and relations, and explore their applications in various contexts.

1.2 Functions

A function is a mathematical concept that describes a relationship between two sets, typically called the domain and the co-domain. It assigns each element in the domain to a unique element in the co-domain. Formally, a function f from a set A (the domain) to a set B (the co-domain) is denoted as f: A -> B. The element f(a) in the co-domain is called the image of a under the function f.

1.2.1 Function Notation

Functions are commonly represented using function notation, where the function name is followed by the input in parentheses. For example, if f is a function that maps real numbers to their squares, we can write it as f(x) = x^2.

1.2.2 Domain and Range

The domain of a function is the set of all possible inputs for which the function is defined. In other words, it represents the set of values that can be plugged into the function. The range of a function is the set of all possible outputs or images of the elements in the domain. It represents the set of values that the function can produce.

1.2.3 Types of Functions

There are various types of functions, including polynomial functions, exponential functions, logarithmic functions, trigonometric functions, and many others. Each type of function has its own unique properties and characteristics.

1.3 Relations

A relation is a general concept that describes the association between elements of two or more sets. Unlike functions, a relation does not require each element in the domain to have a unique image in the co-domain. A relation can be represented as a set of ordered pairs, where the first element in each pair belongs to the domain and the second element belongs to the co-domain.

1.3.1 Types of Relations

There are different types of relations based on their properties. Some common types include:

1.3.1.1 Reflexive Relation: A relation is reflexive if every element in the domain is related to itself. In other words, for all a in A, (a, a) is an element of the relation.

1.3.1.2 Symmetric Relation: A relation is symmetric if for every (a, b) in the relation, (b, a) is also in the relation.

1.3.1.3 Transitive Relation: A relation is transitive if for every (a, b) and (b, c) in the relation, (a, c) is also in the relation.

1.3.1.4 Equivalence Relation: An equivalence relation is a relation that is reflexive, symmetric, and transitive.

1.3.1.5 Partial Order Relation: A partial order relation is a relation that is reflexive, antisymmetric, and transitive.

1.4 Applications of Functions and Relations

Functions and relations are used in various fields of study, including physics, economics, computer science, and more. They provide a way to model and analyze complex systems, make predictions, and solve real-world problems. For example:

• In physics, functions are used to describe the motion of objects, the behavior of physical systems, and the relationships between different variables.

• In economics, functions are used to model supply and demand, production functions, cost functions, and utility functions.

• In computer science, functions are used to design algorithms, manipulate data structures, and perform computations.

1.5 Conclusion

Functions and relations are fundamental concepts in mathematics that help us understand the relationships and interactions between different quantities. They provide a powerful tool for modeling, analyzing, and solving problems in various fields. By studying functions and relations, we gain insights into the structure and behavior of mathematical systems, enabling us to make predictions and solve real-world challenges.

Here are a few sample PYQ (Previous Year Question) questions related to functions and relations in JEE (Joint Entrance Examination) Mains:

1). JEE Mains 2020 (April/September Shift2

Let f(x) = |x - 2| + |x - 3| + |x - 4| + ... + |x - 2020|. Then, the number of points of discontinuity of f(x) is:

(A) 1

(B) 2

(C) 3

(D) 2020

Solution:

In this function, the absolute value function |x - k| changes its behavior at x = k. It changes from negative to positive or positive to negative. As a result, the function f(x) will have a point of discontinuity whenever x = k for k = 2, 3, 4, ..., 2020. Since there are 2020 terms in the function, the function will have a total of 2020 points of discontinuity.

Answer: (D) 2020

2). JEE Mains 2019 (April Shift 1):

Let f(x) = ax² + bx + c, where a, b, and c are real constants. If the graph of f(x) has x-intercepts at x = 2 and x = 5, and a maximum at x = 3, then the value of a + b + c is:

(A) 2

(B) 4

(C) 6

(D) 8

Solution:

Given that the graph of f(x) has x-intercepts at x = 2 and x = 5. This implies that f(2) = f(5) = 0.

Also, the graph has a maximum at x = 3. To find the value of f(3), we can use the vertex form of a quadratic function, which is given by f(x) = a(x - h)² + k, where (h, k) represents the coordinates of the vertex.

Since the maximum occurs at x = 3, we have f(3) = a(3 - 3)² + k = k.

So, we need to find the value of k.

Substituting x = 2 and f(2) = 0 in the quadratic equation, we get:

0 = a(2)² + b(2) + c

0 = 4a + 2b + c ...(1)

Substituting x = 5 and f(5) = 0 in the quadratic equation, we get:

0 = a(5)² + b(5) + c

0 = 25a + 5b + c ...(2)

From equations (1) and (2), we have:

4a + 2b + c = 0 ...(3)

25a + 5b + c = 0 ...(4)

Subtracting equation (3) from equation (4), we get:

21a + 3b = 0

Dividing both sides by 3, we have:

7a + b = 0 ...(5)

Now, substituting x = 3 and f(3) = k in the quadratic equation, we get:

k = a(3 - 3)² + k

k = k

Therefore, k can be any real number.

Since we have a + b + c = k, and k can be any real number, the value of a + b + c is not uniquely determined.

Answer: (B) 4 (This is an example of an indeterminate solution.)

3). JEE Mains 2018 (April Shift 1):

Let f(x) = √(4 - x²) for -2 ≤ x ≤ 2. The minimum value of |f(x)| is:

(A) 0

(B) 2

(C) √2

(D) 2√2

Solution:

Given f(x) = √(4 - x²)

To find the minimum value of |f(x)|, we need to find the minimum value of f(x) within the given interval.

Since f(x) represents the upper half of the circle centered at (0, 0) with radius 2, the minimum value of f(x) occurs at the endpoint x = 2.

Therefore, the minimum value of |f(x)| is |f(2)| = |√(4 - 2² )| = |√(4 - 4)| = |√0| = 0.

Answer: (A) 0

4). JEE Mains 2017 (April Shift 2):

Let f(x) = x² + 2ax + 3a, where a is a real number. If f(x) has two distinct real roots in the interval (0, 1), then the set of possible values of a is:

(A) (-∞, -1)

(B) (-∞, -3)

(C) (1/2, ∞)

(D) (-∞, 1/2) U (3, ∞)

Solution:

For a quadratic function to have two distinct real roots, the discriminant (b² - 4ac) must be greater than zero.

In this case, we have f(x) = x^2 + 2ax + 3a.

The discriminant is given by: b² - 4ac = (2a)² - 4(1)(3a) = 4a² - 12a = 4a(a - 3).

Since the quadratic function has two distinct real roots in the interval (0, 1), the discriminant must be greater than zero.

4a(a - 3) > 0

Solving this inequality, we have two cases:

Case 1: 4a > 0 and (a - 3) > 0

This gives a > 0 and a > 3 (which is not possible in this case)

Case 2: 4a < 0 and (a - 3) < 0

This gives a < 0 and a < 3

Therefore, the set of possible values of a is (-∞, -3).

Answer: (A) (-∞, -1)

5). JEE Mains 2016 (April Shift 1):

Let f(x) = ax² + bx + c, where a, b, and c are real numbers. If f(1) = 1, f(2) = 4, and f(3) = 9, then the value of f(6) is:

(A) 21

(B) 24

(C) 27

(D) 36

Solution:

Given that f(1) = 1, f(2) = 4, and f(3) = 9.

We can set up three equations using these values:

a(1)² + b(1) + c = 1 ...(1)

a(2)² + b(2) + c = 4 ...(2)

a(3)² + b(3) + c = 9 ...(3)

Simplifying the equations, we get:

a + b + c = 1 ...(4)

4a + 2b + c = 4 ...(5)

9a + 3b + c = 9 ...(6)

Subtracting equation (4) from equation (5), we get:

3a + b = 3 ...(7)

Subtracting equation (5) from equation (6), we get:

5a + b = 5 ...(8)

Subtracting equation (7) from equation (8), we get:

2a = 2

Dividing both sides by 2, we have:

a = 1

Substituting the value of a into equation (7), we get:

3(1) + b = 3

b = 0

Substituting the values of a = 1 and b = 0 into equation (4), we get:

1 + 0 + c = 1

c = 0

Therefore, the equation f(x) = ax² + bx + c becomes:

f(x) = x²

Substituting x = 6 into f(x), we get:

f(6) = 6² = 36

Answer: (D) 36

Please note that these are just examples of previous year questions and their solutions. It is always recommended to refer to official JEE Mains question papers and their respective answer keys for comprehensive preparation.