Question

If \(R^*\) be the set of all non-zero real numbers, then the mapping \(f: R^* \to R^*\) defined by \(f(x) = \frac{1}{x}\) is

• A. one-one and onto
• B. many-one and onto
• C. one-one, but not onto
• D. neither one-one nor onto

Hint:

The properties of a function heavily depend on its specified domain and codomain. For example, if the function was \(f: R \to R\), it would not even be a function because \(f(0)\) is undefined. If it were \(f: R^* \to R\), it would be one-one but not onto, because you could never get \(y=0\). Always pay close attention to the domain and codomain.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The question asks to classify the function \(f(x) = 1/x\) based on whether it is one-one (injective) and onto (surjective). The domain and codomain are both R*, the set of all non-zero real numbers.
One-one (Injective): A function is one-one if every distinct element in the domain maps to a distinct element in the codomain. Formally, \(f(x_1) = f(x_2) \implies x_1 = x_2\).
Onto (Surjective): A function is onto if every element in the codomain is the image of at least one element from the domain. Formally, for every \(y \in \text{Codomain}\), there exists an \(x \in \text{Domain}\) such that \(f(x) = y\).
Step 2: Key Formula or Approach:
1. Test for one-one: Assume \(f(x_1) = f(x_2)\) for \(x_1, x_2 \in R^*\). If this implies \(x_1 = x_2\), the function is one-one.
2. Test for onto: Let \(y\) be an arbitrary element in the codomain \(R^*\). Try to find an \(x\) in the domain \(R^*\) such that \(f(x) = y\). If such an \(x\) always exists, the function is onto.
Step 3: Detailed Explanation:
The function is \(f(x) = 1/x\), with \(f: R^* \to R^*\).
1. Checking for One-one (Injectivity):
Let \(x_1, x_2 \in R^*\). Assume \(f(x_1) = f(x_2)\).
\[ \frac{1}{x_1} = \frac{1}{x_2} \] By taking the reciprocal of both sides (which is valid since \(x_1, x_2 \neq 0\)), we get: \[ x_1 = x_2 \] Since \(f(x_1) = f(x_2)\) implies \(x_1 = x_2\), the function is one-one.
2. Checking for Onto (Surjectivity):
Let \(y\) be an arbitrary element in the codomain \(R^*\). We need to find if there is an \(x\) in the domain \(R^*\) such that \(f(x) = y\).
\[ f(x) = y \] \[ \frac{1}{x} = y \] Solving for \(x\), we get: \[ x = \frac{1}{y} \] Since \(y \in R^*\), it means \(y\) is a non-zero real number. Therefore, \(x = 1/y\) will also be a non-zero real number. This means \(x\) is in the domain \(R^*\).
So, for any \(y\) in the codomain, we can find a pre-image \(x = 1/y\) in the domain.
Thus, the function is onto.
Step 4: Final Answer:
Since the function is both one-one and onto, it is a bijective function. The correct option is one-one and onto.