Problem 133 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Let \(A=\\{-2,-1,0,1,2\\}\) and \(f: A \rightarrow I\) given by \(f(x)=x^{2}-2 x-3\). Find range of \(f\). Also find preimages of \(6,-3\) and 5. \\{Ans. \(f(A)=\\{-4,-3,0,5\\} ;\) no pre-image of \(6 ; 0\) and 2 are pre-images of \(-3 ;-2\) is the pre-image of 5\(\\}\)

Step-by-Step Solution

Answer

The range of f is \{-4, -3, 0, 5\}; there is no pre-image of 6; 0 and 2 are pre-images of -3; and -2 is the pre-image of 5.

Step 1: Identify function and given set

We have been given the function f(x) = x^2 - 2x - 3, with the set \(A=\{-2, -1, 0, 1, 2\}\).

Step 2: Calculate f(x) for each element of A

Now we will calculate the function's value for each element in A: 1. \(f(-2)=(-2)^{2}-2(-2)-3=4+4-3=5\) 2. \(f(-1)=(-1)^{2}-2(-1)-3=1+2-3=0\) 3. \(f(0)=0^{2}-2(0)-3=0-0-3=-3\) 4. \(f(1)=1^{2}-2(1)-3=1-2-3=-4\) 5. \(f(2)=2^{2}-2(2)-3=4-4-3=-3\)

Step 3: Find the range of f

By evaluating f(x) for each element in A, we get f(A) = \{-4, -3, 0, 5\}, which is the range of the function.

Step 4: Find preimages of 6, -3, and 5

1. Pre-image of 6: There is no pre-image of 6 because none of the f(x) values were equal to 6. 2. Pre-image of -3: f(0) = -3 and f(2) = -3. Both 0 and 2 are pre-images of -3. 3. Pre-image of 5: f(-2) = 5, so -2 is the pre-image of 5. So, we have our solution: The range of f is {-4, -3, 0, 5}; there is no pre-image of 6; 0 and 2 are pre-images of -3; and -2 is the pre-image of 5.

Key Concepts

Preimage

The preimage of a function refers to the set of all input values that map to a particular output value. In simpler terms, if you start with a specific known output, the preimage consists of all inputs that produce that output when the function is applied.
In our example, the function is given by \(f(x) = x^2 - 2x - 3\). To find the preimage for a specific value, like \(-3\) or \(5\), we need to check which values of \(x\) from the set \(A=\{-2, -1, 0, 1, 2\}\) give us these outputs when substituted into the function.

• For example, for \(f(x) = -3\), substituting \(x = 0\) and \(x = 2\) results in \(f(0) = -3\) and \(f(2) = -3\), so the preimages of \(-3\) are \(0\) and \(2\).
• For \(f(x) = 5\), substituting \(x = -2\) results in \(f(-2) = 5\), so the preimage of \(5\) is \(-2\).
• Sometimes, no preimage exists for a given output, as in the case of \(6\), because none of the inputs from set \(A\) result in an output of \(6\).

Function Evaluation

Evaluating a function means substituting the input value into the function expression to find the output. It's like asking the function what it gives us for a particular input.
To evaluate the function \(f(x) = x^2 - 2x - 3\) for each element in the set \(A = \{-2, -1, 0, 1, 2\}\), follow these simple steps:

• Substitute each element from the domain set \(A\) into the function expression.
• Calculate the result to find the output value, \(f(x)\), for each input.
• For instance, when you substitute \(x = 1\), we get \(f(1) = 1^2 - 2(1) - 3 = -4\).
• Another example, substitute \(x = -1\), we get \(f(-1) = (-1)^2 - 2(-1) - 3 = 0\).

By evaluating the function for each element of \(A\), you uncover how the function behaves over its domain.

Polynomial Functions

Polynomial functions are mathematical expressions involving a sum of powers of one or more variables multiplied by coefficients. Our function \(f(x) = x^2 - 2x - 3\) is a second-degree polynomial, because it includes \(x^2\), which defines it as quadratic.
Polynomial functions, like our example, can take the form \(ax^n + bx^{n-1} + ... + constant\), where \(a, b,...\) are coefficients, and \(n\) denotes the degree of the polynomial.

• The degree determines the shape and number of turning points the graph might have. For quadratic functions like this one, the graph is a parabola.
• Polynomial functions are continuous and differentiable, making them predictable and smooth curves.
• They play a fundamental role in calculus as their properties are often used to describe more complex functions.

Understanding polynomial functions helps you easily approach how to analyze them, like determining whether there's a maximum or minimum value, and anticipating their general shape just by looking at their equation.

Domain and Range

The domain of a function is the set of all acceptable inputs, while the range is the collection of possible outputs. For a function to fully represent a mathematical model, it’s vital to define what can go into it (domain) and what comes out (range).
In our case, \(f(x) = x^2 - 2x - 3\) has a domain derived from the set \(A = \{-2, -1, 0, 1, 2\}\). This means the function is only defined for these particular values of \(x\).

• The range is identified by evaluating the function for each element of the domain and collecting the results. From the previous steps, we found that the range \(f(A)\) is \{-4, -3, 0, 5\}.
• Knowing the range lets us understand the values that come out when we put domain values into the function.
• This concept is crucial for applications where you need to understand limitations or potential outputs, such as ensuring particular conditions aren't exceeded in engineering scenarios.