Zadanie 8b, Paper 1, maj 2025, TZ2, Matematyka AA IB HL

Put $u=\arcsin x$ and $v=\arcsin(2x)$. Taking cosine and expanding gives $\sqrt{1-x^2}\sqrt{1-4x^2}-2x^2=0$. Squaring yields 1 − 5x² + 4x⁴ = 4x⁴, so $x=\pm1/\sqrt5$. Both inverse-sine terms have the sign of x; their sum can be negative only for the negative candidate. For $x=-1/\sqrt5$, the cosine is zero and the sum belongs to [−π, 0), so it is −π/2. Thus $x=-1/\sqrt5$.

M1: use a compound-angle identity. M1: express the complementary trig values with square roots. A1: correct expanded equation. A1: correct trig value of -π/2. M1: eliminate radicals and solve the resulting algebraic equation. A1: only the negative valid root, with extraneous candidates excluded. Accept equivalent sine or angle-substitution methods.