Inverse trigonometric functions — Uniqueness then solve 3pi/4

Question

(i) Explain why has exactly one real solution for each . Hence,

(ii) Solve .

This practice question was inspired by Question 14(c) in the 2024 NSW HSC Mathematics Extension 1 examination, © NESA. It is an original TestMum variation that practises the same mathematical idea, not an official NESA question.

Set up the solution: use the restricted domain and range that make the inverse function single-valued.(i) Sum of two strictly increasing continuous functions with range is bijective onto .
Simplify the previous line carefully, keeping exact values where possible.(ii) or ; reject the negative root by sign.
Finish the calculation, then check that the result meets the question’s conditions.Using part (i), uniqueness guarantees a single candidate once the quadratic is solved and invalid roots are rejected.
State the final answer clearly in the form the question requested.Answer: (i) continuous strictly increasing with range ; (ii) .

Common mistake: changing so many features that the new example no longer tests the same mathematical idea.

Verification: Dependent: uniqueness then solve; check rejected root.

Any mark labels are a TestMum study aid, not an official NESA marking allocation.