Question

Let \(\mathbb{R}\) denote the set of real numbers, \(\mathbb{Z}\) the set of integers and W the set of Non-negative integers, and define a map
(i) \(f_{1}: \mathbb{Z} \rightarrow \mathbb{Z}\) by \(f_{1}(x)=x^{3}\). Then \(f_{1}\) is one-one and not onto.
(ii) \(f_{2}: \mathbb{R} \rightarrow \mathbb{R}\) by \(f_{2}(x)=x^{3}\). Then \(f_{2}\) is one-one and onto.
(iii) \(f_{3}: \mathbb{Z} \rightarrow\) W by \(f_{3}(x)=|x|\). Then \(f_{3}\) is not one-one but is onto.
(iv) \(f_{4}: \mathbb{Z} \rightarrow \mathbb{Z}\) by \(f_{4}(x)=x^{2}\). Then \(f_{4}\) is neither one-one nor onto.
Then which of the following options is correct?

• A. (i), (iii), (iv) are true but (ii) is false
• B. All are true
• C. (ii), (iii) and (iv) are true but (i) is false
• D. (i), (ii) and (iv) are true but (iii) is false

Hint:

Check each statement separately for injectivity (distinct inputs give distinct outputs) and surjectivity (every element of the codomain is hit).

Answer 1

The Correct Option is B

(i) \(x^3\) is strictly increasing so it is one-one on \(\mathbb{Z}\). It is not onto \(\mathbb{Z}\) because, e.g., 2 is not a cube of any integer. True.
(ii) \(x^3\) is one-one and every real number has a real cube root, so it is onto \(\mathbb{R}\). True.
(iii) \(|{-1}|=|1|=1\), so not one-one. Every non-negative integer n equals \(|n|\), so it is onto W. True.
(iv) \(f_4(-1)=f_4(1)=1\), so not one-one. Negative integers and non-squares such as 2 are never attained, so not onto. True.
All four statements are true.
Answer: (2) All are true