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: Verify one-to-one property.
A function \( f(x) \) is one-to-one if for any \( x_1, x_2 \) in the domain, \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \). For \( f(x) = x^3 - 1 \): If \( f(x_1) = f(x_2) \), then \( x_1^3 - 1 = x_2^3 - 1 \), which simplifies to \( x_1^3 = x_2^3 \). This further implies \( x_1 = x_2 \). Therefore, \( f(x) \) is one-to-one.
Step 2: Verify onto property.
A function \( f(x) \) is onto if every element in the codomain \( \mathbb{Z} \) is an output for some input in the domain \( \mathbb{Z} \). The outputs of \( f(x) = x^3 - 1 \) are of the form \( x^3 - 1 \). 
However, not all integers can be generated by this formula. 
For instance, there is no integer \( x \) for which \( f(x) = 0 \), because \( x^3 - 1 = 0 \) leads to \( x^3 = 1 \), and \( 1 \) is not the cube of any integer. 
Consequently, \( f(x) \) is not onto.
Step 3: Conclusion
The function \( f(x) = x^3 - 1 \) is one-to-one but not onto.