A mapping f : A → B defined as \(f(x)=\frac{2 x+3}{3 x+5}, x \in A\) If f is to be onto, then what are A and B equal to ?
Finding Domain and Codomain for an Onto Mapping
The given function is \(f(x)=\frac{2 x+3}{3 x+5}\). We are asked to find the sets A (domain) and B (codomain) such that the mapping \(f: A \rightarrow B\) is onto (surjective).
Determining the Domain A
The domain of a function is the set of all possible input values (x) for which the function is defined. For a rational function like \(f(x)=\frac{2 x+3}{3 x+5}\), the denominator cannot be zero.
So, we must have:
\(3x + 5 \neq 0\)
\(3x \neq -5\)
\(x \neq -\frac{5}{3}\)
Therefore, the domain A is the set of all real numbers excluding \(-\frac{5}{3}\).
A = \(R \backslash \{-\frac{5}{3}\}\)
Determining the Codomain B for an Onto Function
For a function \(f: A \rightarrow B\) to be onto (surjective), the codomain B must be equal to the range of the function \(f\). The range is the set of all possible output values (y) of the function.
To find the range, we set \(y = f(x)\) and solve for x in terms of y:
\(y = \frac{2x+3}{3x+5}\)
Multiply both sides by \((3x+5)\) (assuming \(3x+5 \neq 0\)):
\(y(3x+5) = 2x+3\)
Distribute y on the left side:
\(3xy + 5y = 2x + 3\)
Gather terms with x on one side and terms without x on the other side:
\(3xy - 2x = 3 - 5y\)
Factor out x from the left side:
\(x(3y - 2) = 3 - 5y\)
Solve for x by dividing by \((3y - 2)\) (assuming \(3y-2 \neq 0\)):
\(x = \frac{3 - 5y}{3y - 2}\)
For x to be a real number in the domain A, the expression on the right side must be defined. This expression is undefined when the denominator is zero.
So, we must have:
\(3y - 2 \neq 0\)
\(3y \neq 2\)
\(y \neq \frac{2}{3}\)
This means that the value \(y = \frac{2}{3}\) is not in the range of the function. The range is therefore the set of all real numbers excluding \(\frac{2}{3}\).
For the function to be onto, the codomain B must be equal to this range.
B = \(R \backslash \{\frac{2}{3}\}\)
Conclusion
For the mapping \(f(x)=\frac{2 x+3}{3 x+5}\) to be onto, the domain A and the codomain B must be:
- A = \(R \backslash \{-\frac{5}{3}\}\)
- B = \(R \backslash \{\frac{2}{3}\}\)
Comparing with Options
Let's compare our result with the given options:
| Option | A | B | Match? |
|---|---|---|---|
| 1 | \(R \backslash\{-\frac{5}{3}\}\) | \(R \backslash\{-\frac{2}{3}\}\) | No |
| 2 | \(R\) | \(R \backslash\{-\frac{5}{3}\}\) | No |
| 3 | \(R \backslash\{-\frac{3}{2}\}\) | \(R \backslash\{0\}\) | No |
| 4 | \(R \backslash\{-\frac{5}{3}\}\) | \(R \backslash\{\frac{2}{3}\}\) | Yes |
The results match Option 4.
Revision Table: Key Concepts for Onto Functions
| Concept | Definition | How to Find |
|---|---|---|
| Domain (A) | Set of valid input values (x). | Identify values of x that make the function expression undefined (e.g., denominator = 0, square root of negative number) and exclude them from the set of real numbers (R) or the specified base set. |
| Codomain (B) | The target set where the output values (y) are expected to lie. | Specified in the function definition \(f: A \rightarrow B\). For an onto function, it must equal the range. |
| Range | Set of all actual output values (y) the function produces for the given domain. | Solve \(y = f(x)\) for x in terms of y. Identify values of y for which x is undefined in the domain. Exclude these values from the codomain (or R) to find the range. |
| Onto Function (Surjective) | A function where every element in the codomain B has at least one corresponding element in the domain A. | This means the range of the function is exactly equal to its codomain B. |
Additional Information: Types of Function Mappings
Understanding different types of function mappings is crucial in set theory and calculus. Here are the main types:
- One-to-One (Injective) Function: A function where distinct elements in the domain A are mapped to distinct elements in the codomain B. If \(f(x_1) = f(x_2)\), then \(x_1 = x_2\).
- Onto (Surjective) Function: A function where every element in the codomain B is the image of at least one element in the domain A. The range equals the codomain.
- One-to-One Correspondence (Bijective) Function: A function that is both one-to-one and onto. For every element in the codomain B, there is exactly one corresponding element in the domain A. Bijective functions have inverse functions.
- Into Function: A function that is not onto. The range is a proper subset of the codomain.
In this problem, the condition that the function is "onto" is key to determining the codomain B, as it forces B to be equal to the range of the function over the specified domain A.
Important Questions from Relations and Functions
Consider the following statements:
1. The relation f defined by \(f(x)= \begin{cases}x^3, & 0 \leq x \leq 2 \\ 4 x, & 2 \leq x \leq 8\end{cases}\) is a function.
2. The relation g defined by \(g(x)= \begin{cases}x^2, & 0 \leq x \leq 4 \\ 3 x, & 4 \leq x \leq 8\end{cases}\) is a function.
Which of the statements given above is/are correct?
A function satisfies \(f(x-y)=\frac{f(x)}{f(y)}\), where f(y) ≠ 0. If f(1) = 0.5, then what is f(2) + f(3) + f(4) + f(5) + f(6) equal to ?
If f(x) = x(4x2 - 3), then what is f(sinθ) equal to ?
Let R be a relation on the set N of natural numbers defined by ‘nRm ⟺ n is a factor of m’. Then which one of the following is correct?
f(xy) = f(x) + f(y) is true for all
