Zadanie 1e, Paper 3, maj 2022, TZ2, Matematyka AA IB HL

Put u=x−1>2. On the upper branch y=√(u³−4u), $y\prime\prime=\frac{12u(u^3-4u)-(3u^2-4)^2}{4(u^3-4u)^{3/2}}$. The lower branch has the opposite sign, so the zero condition is the same. The numerator simplifies to 3u⁴−24u²−16. Set v=u²:3v²−24v−16=0, so v=4±8√3/3. Only the positive value is possible, and u>2 selects u=√(4+8√3/3). Therefore x=1+√(4+8√3/3). The numerator is strictly increasing for u>2 (derivative 12u(u²−4)>0), passing from negative to positive at that root, so concavity changes.

M1:differentiate the first derivative or use implicit second differentiation. A1:correct first numerator contribution and denominator. A1:correct second contribution. M1:set the second derivative to 0. A1:3u⁴−24u²−16=0. M1:quadratic formula in u². A1:x=1+√(4+8√3/3), with invalid branches excluded. Equivalent forms are accepted. Subsequent accuracy marks remain available when the second-derivative denominator is absent or wrong but the required numerator equation and ensuing algebra are correct. For implicit differentiation, use M1 for differentiating 2yy′=3u²-4, A1 for 2(y′)^2+2yy″=6u, M1 for y″=0, A1 for substituting y′ into (y′)^2=3u, then the same final A1 M1 A1 for the quartic equation, quadratic method and valid radical. Product-rule work earns the two intermediate accuracy marks for its two correct terms. No extra mark condition for a separate concavity-change proof is imposed.