MTH1010_09_04_2

Question

MTH1010_09_04_2

BlackTom AI · Accepted Answer

We are asked to select the function which has an inverse function.
Option a: f : R → R where f(x) = sin(x). Over the entire real line, the sine function is not one-to-one because it repeats values in every 2π interval. Therefore it is not invertible as a function from R to R.
Option b: f : (0, 2π) → R where f(x) = sin(x). Although this interval is smaller, sin(x) still takes the same values at x and π − x within (0, 2π), so the function is not injective on this domain either. Hence it lacks an inverse in the sense of a bijection onto its codomain.
Option c: f : (−π/2, π/2) → R where f(x) = sin(x). On this domain, sin(x) is strictly increasing, so it is one-to-one. Its range is (−1, 1), so if we consider the natural inverse, it maps (−1, 1) back to (−π/2, π/2) via arcsin. In the usual sense of having an inverse function, a one-to-one function with an appropriate codomain (here effectively (−1, 1)) does have an inverse. However, since the stated codomain is R, the function is not onto R, which would violate the requirement for a bijection with codomain R. If we interpret the inverse in the conventional way (restricting attention to the natural range), this option is the one that yields an inverse.
Option d: f : (0, π) → R where f(x) = sin(x). On this interval, sin(x) increases from 0 to 1 then decreases back to 0, so it is not one-to-one. Therefore no inverse function exists for this option.
In summary: options a, b, and d fail the injectivity requirement (and/or the bijection requirement with the stated codomain), while option c is the one where sine is one-to-one on the given domain. Consequently, option c is the correct choice when considering the standard notion of an inverse function on the natural range.

MCD2130 - T2 - 2025 10.7 Online Quiz - MCD2130_QUIZ_NINE Circular Functions, Continuity and Differentiability

MTH1010_09_04_2

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