Question

Let \( \mathbb{Z} \) denote the set of integers. Then, the function \( f : \mathbb{Z} \to \mathbb{Z} \) defined as \( f(x) = x^3 - 1 \) is:

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

Hint:

To check if a function is one-one, use the condition \( f(x_1) = f(x_2) \implies x_1 = x_2 \). To check if it is onto, ensure every element in the codomain has a preimage in the domain.

Answer 1

The Correct Option is B

Step 1: Check if the function is one-one
A function \( f(x) \) is one-one if distinct inputs give distinct outputs, i.e., \( f(x_1) = f(x_2) \implies x_1 = x_2 \). For \( f(x) = x^3 - 1 \): \[ f(x_1) = f(x_2) \implies x_1^3 - 1 = x_2^3 - 1 \implies x_1^3 = x_2^3 \implies x_1 = x_2. \] Thus, \( f(x) \) is one-one. 
Step 2: Check if the function is onto
A function \( f(x) \) is onto if every element in the codomain \( \mathbb{Z} \) has a preimage in the domain \( \mathbb{Z} \). The function \( f(x) = x^3 - 1 \) outputs values of the form \( x^3 - 1 \). 
However, not all integers can be expressed in this form. 
For example, there is no \( x \in \mathbb{Z} \) such that \( f(x) = 0 \), as \( x^3 - 1 = 0 \) implies \( x^3 = 1 \), which does not hold for any integer \( x \). 
Thus, \( f(x) \) is not onto.
Step 3: Conclusion
The function \( f(x) = x^3 - 1 \) is one-one but not onto.