Problem 6 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

If \(f:[0, \infty) \rightarrow[0, \infty)\) and \(f(x)=\frac{x}{1+x}\), then (a) one-one and onto (b) one-one but not onto (c) onto but not one-one (d) neither one-one nor onto

Step-by-Step Solution

Answer

(b) one-one but not onto.

Step 1: Check if the function is one-one (Injective)

To determine if the function is injective, we need to check if for any two values \(x_1\) and \(x_2\) within the domain \([0, \infty)\), \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\). \[f(x) = \frac{x}{1+x}\] \[f(x_1) = \frac{x_1}{1+x_1}\] \[f(x_2) = \frac{x_2}{1+x_2}\] Set them equal: \[\frac{x_1}{1+x_1} = \frac{x_2}{1+x_2}\] Cross-multiply to get: \[x_1(1 + x_2) = x_2(1 + x_1)\] Expand both sides: \[x_1 + x_1x_2 = x_2 + x_2x_1\] Simplify: \[x_1 = x_2\] Thus, \(f(x)\) is one-one.

Step 2: Determine if the function is onto (Surjective)

To check if the function is onto, we must see if for every \(y\) in the codomain \([0, \infty)\), there exists an \(x\) in the domain such that \(f(x) = y\). \[y = \frac{x}{1+x}\] Solve for \(x\): \[y(1 + x) = x\] \[y + yx = x\] \[yx - x = -y\] \[x(y - 1) = -y\] \[x = \frac{-y}{y - 1}\] The only issue occurs when \(y = 1\). At \(x \to \infty\), \(f(x) \to 1\), but 1 is not in the range of the function because \(x = \frac{-1}{0}\) is undefined. Hence, the function does not cover all values in the codomain \([0, \infty)\), so \(f(x)\) is not onto.

Key Concepts

Function Injectivity

To determine if a function is injective, we look for situations where two different inputs might result in the same output.
In simpler terms, a function is injective (or one-to-one) if each element of the domain maps to a unique element in the codomain.
To check injectivity, consider two inputs from the domain:

• Suppose you have two different numbers, \(x_1\) and \(x_2\), such that \(f(x_1) = f(x_2)\).
• If it turns out that \(x_1\) must equal \(x_2\) for that situation to hold true, the function is injective.

In the example of the function \(f(x) = \frac{x}{1+x}\), setting \(f(x_1) = f(x_2)\) leads to an equation that simplifies down to \(x_1 = x_2\).
This simplified equation confirms that no two different inputs produce the same output, making the function injective.
Understanding injectivity helps us avoid redundancy by ensuring distinct inputs are required for each output.

Function Surjectivity

A function is surjective (or onto) if every member of its codomain is an output for some input from its domain.
In practical terms, this means that no element in the codomain is left out.
This is checked by ensuring that for every possible output value \(y\), there is an input \(x\) such that \(f(x) = y\).

• If you can find an output that cannot be obtained from any input, the function is not surjective.

In the exercise's function \(f(x) = \frac{x}{1+x}\), solving for \(x\) in terms of \(y\) results in \(x = \frac{-y}{y-1}\).
However, when \(y = 1\), the equation breaks down since it involves division by zero, meaning there is no \(x\) that produces \(f(x) = 1\).
This shows the function cannot produce every possible value in the codomain, making it non-surjective.
Recognizing surjectivity involves ensuring that every codomain element is accessible.

Domain and Codomain of Functions

The domain and codomain of a function are fundamental parts of its definition:

• The domain is the set of all possible inputs for the function.
  For instance, in the exercise \(f:[0, \infty) \rightarrow [0, \infty)\), the domain is \([0, \infty)\).
• The codomain is the set of potential outputs the function can produce, which is also \([0, \infty)\) in our case.

The domain and codomain do more than just specify ranges; they define the behavior and applicability of the function.
Understanding them allows us to apply the function to the correct set of numbers and interpret its output meaningfully.
When matching outputs with inputs as well as understanding injectivity and surjectivity, these sets are crucial.
A function's behavior, in the context of these sets, informs whether it is truly functional—or well-defined—across all intended values.