Bijective Functions
Let f from real numbers to real numbers be defined by f(x) = x^3 + x + 1; which statement about the invertibility of f is correct based on monotonicity analysis?
Select the correct option:
Solution
f is bijective, hence invertible on all reals
A function from reals to reals is bijective when it is both injective and surjective, and strict monotonicity from a positive derivative is the standard JEE Advanced criterion for injectivity. The derivative f'(x) = 3x^2 + 1 is strictly positive for every real x, since 3x^2 ≥ 0 and the added 1 keeps it above zero, so f is strictly increasing and therefore one-to-one. As a continuous cubic, f tends to negative \infty as x decreases and positive \infty as x increases, so by the intermediate value property it attains every real value, making it onto. Being both injective and surjective, f is bijective and hence invertible on all reals. Option many-one is false because the derivative never vanishes. Option into contradicts the full real range of a cubic. Option constant is absurd given the cubic term. Hence f is bijective. Plausibility check: a strictly increasing continuous function with unbounded range in both directions must hit each output exactly once, confirming a well-defined inverse exists everywhere. Monotonicity inferred from the sign of the derivative is the cleanest available injectivity test, while continuity together with an unbounded range in both directions secures surjectivity through the intermediate value theorem. Combining these two observations yields a fully rigorous, exam-ready proof of invertibility without ever constructing the inverse function explicitly, which is exactly the reasoning examiners expect here.
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About This Question
- Subject
- mathematics
- Chapter
- sets, relations and functions
- Topic
- bijective functions
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
f is bijective, hence invertible on all reals
A function from reals to reals is bijective when it is both injective and surjective, and strict monotonicity from a positive derivative is the standard JEE Advanced criterion for injectivity. The derivative f'(x) = 3x^2 + 1 is strictly positive for every real x, since 3x^2 ≥ 0 and the added 1 keeps it above zero, so f is strictly increasing and therefore one-to-one. As a continuous cubic, f tends to negative \infty as x decreases and positive \infty as x increases, so by the intermediate value property it attains every real value, making it onto. Being both injective and surjective, f is bijective and hence invertible on all reals. Option many-one is false because the derivative never vanishes. Option into contradicts the full real range of a cubic. Option constant is absurd given the cubic term. Hence f is bijective. Plausibility check: a strictly increasing continuous function with unbounded range in both directions must hit each output exactly once, confirming a well-defined inverse exists everywhere. Monotonicity inferred from the sign of the derivative is the cleanest available injectivity test, while continuity together with an unbounded range in both directions secures surjectivity through the intermediate value theorem. Combining these two observations yields a fully rigorous, exam-ready proof of invertibility without ever constructing the inverse function explicitly, which is exactly the reasoning examiners expect here.
This medium difficulty mathematics question is from the chapter sets, relations and functions, covering the topic of bijective functions. It appeared in the 2025 exam.
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