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The function f: R → R defined by f(x) = x³ + 5 is

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One-one but not onto

Onto but not one-one

Both one-one and onto

Neither one-one nor onto

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About This Question

Subject: mathematics
Chapter: sets, relations and functions
Topic: functions
Difficulty: Medium
Year: 2025

Solution

Correct Answer:

Both one-one and onto

Test for One-one (Injection): $f (x_{1}) = f (x_{2}) ⟹ x_{1}^{3} + 5 = x_{2}^{3} + 5 ⟹ x_{1}^{3} = x_{2}^{3}$ . Since cube roots of real numbers are unique, $x_{1} = x_{2}$ . Thus, $f$ is one-one.
Test for Onto (Surjection): We need to see if every $y \in R$ has a pre-image.
- $y = x^{3} + 5 ⟹ x = 3 y - 5$ .
- Since the cube root of any real number is always real, $x \in R$ for all $y$ . Thus, $f$ is onto.
Conclusion: $f$ is a bijection.

This medium difficulty mathematics question is from the chapter sets, relations and functions, covering the topic of functions. It appeared in the 2025 exam.

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