Problem 134 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Given \(A=\\{-1,0,2,5,6,11\\}\) and \(B=\\{-2,-1,0,18,28,108\\}\) and \(f(x)=x^{2}-x-2\). Find \(f(A)\). Whether \(f(A)=B\) or not.

Step-by-Step Solution

Answer

\(f(A) = \{-2, 0, 0, 18, 28, 108\}\) and \(f(A) \neq B\).

Step 1: Calculate f(A)

To find the image of A under the function f, we need to calculate the value of f(x) for each x in A, i.e., for x = -1, 0, 2, 5, 6, and 11. \(f(-1) = (-1)^2 - (-1) - 2 = 1 + 1 - 2 = 0\) \(f(0) = (0)^2 - (0) - 2 = -2\) \(f(2) = (2)^2 - (2) - 2 = 4 - 2 - 2 = 0\) \(f(5) = (5)^2 - (5) - 2 = 25 - 5 - 2 = 18\) \(f(6) = (6)^2 - (6) - 2 = 36 - 6 - 2 = 28\) \(f(11) = (11)^2 - (11) - 2 = 121 - 11 - 2 = 108\) So, the values obtained are 0, -2, 0, 18, 28, 108, therefore, \(f(A) = \{-2, 0, 0, 18, 28, 108\}\).

Step 2: Compare f(A) with B

Now we'll compare the set \(f(A) = \{-2, 0, 0, 18, 28, 108\}\) with the given set B, \(B = \{-2, -1, 0, 18, 28, 108\}\). We can see that the two sets are different because -1 is present in B but not in f(A). Also, the value 0 appears twice in f(A) but only once in B. Thus, the sets f(A) and B are not equal, so \(f(A) \neq B\).

Key Concepts

Understanding Set Theory

Set theory is a fundamental part of mathematics that deals with the study of sets, which are collections of objects. In the context of functions and mapping, sets play a crucial role as they define the domain and codomain of a function.

Considering our previous exercise, we had two sets, A and B, which consisted of specific numbers. A function is a rule that assigns each element of one set, known as the domain, to another set, known as the codomain, typically with some relationship between the elements.

For example, in the function f(x), every x value in set A is paired with a corresponding value in set B through the rule defined by the function, in this case, f(x) = x^2 - x - 2. We can see how set theory underpins the concept of mapping elements from one set to another via a function. Understanding the basic principles of set theory is essential for grasping more complex mathematical concepts.

The Image of a Function

The image of a function refers to the set of all output values it can produce. It is obtained by applying the function to every element of the domain.

In our exercise with function f(x) = x^2 - x - 2 and domain A, we computed the image f(A) by evaluating f(x) for every element in the set A. The resulting set of values constitutes the image of A under the function, f.

Evaluating the Image

After evaluating f(x) for all elements in set A, we found the image of A, denoted as f(A), which turned out to be {-2, 0, 0, 18, 28, 108}. In contrast to a set in normal circumstances, where repeated elements are not acknowledged, in the image of a function, such repetition underscores the function's behavior at different inputs that yield the same output. However, when comparing the image of a function to another set for equality, we would consider the set notation conventions, where repeated elements do not count multiple times.

Mastering Function Evaluation

Function evaluation is the process of determining the output of a function for a given input. To evaluate a function, you substitute the input value into the function's rule.

The recent problem involved function evaluation for a polynomial function where we worked through the process step by step. For instance, when we evaluated f at -1, we calculated f(-1) to be 0 by substituting -1 into the function's formula.

Tackling Polynomial Functions

With polynomial functions like f(x) = x^2 - x - 2, it’s crucial to follow the correct order of operations: first, square the input (x^2), then subtract the product of the input and 1 (x), and finally subtract 2. By carefully computing these steps for each element of set A, the correct image f(A) is derived.

The function evaluation process is integral to finding the image of a function and highlights the importance of accurately applying the function's rule to each element of the domain to establish the connection between the domain and the image.