Inverse trigonometric functions — Uniqueness then solve with 4x and 6x

Question

(i) Explain why , where , has exactly one real solution.

(ii) Hence 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) Both and are strictly increasing and continuous. Their sum is strictly increasing and continuous with range , so for each there is exactly one .
Finish the calculation, then check that the result meets the question’s conditions.(ii) Take tan: . Factor: or . Reject (sum of arctangents is negative). Accept (both arctangents positive; product of arguments exceeds , so the sum equals ).
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 hence solve; check rejected root by sign.

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