Zadanie 5, Paper 2, listopad 2024, Matematyka AA IB HL

The logarithm argument must be strictly positive for every x. Its leading coefficient is positive, so its discriminant must be negative: r²−12(r−2)<0, or r²−12r+24<0. The boundary roots are 6±2√3. Therefore $6-2\sqrt3<r<6+2\sqrt3$. Equality would give a zero logarithm argument at the quadratic minimum, so endpoints are excluded.

R1: strict positivity of the logarithm argument for all real x. A1: discriminant r²−12r+24. M1: require a negative discriminant, or a strictly positive quadratic minimum. A1: boundary values 6±2√3. A1: correct open interval. Also accept the corresponding inward-rounded closed interval 2.54≤r≤9.46, following the source decimal-bound tolerance; exact endpoints6±2√3 must remain excluded. Completing-square route: A1 for the minimum value r−2−r²/12 at x=r/6; M1 for imposing its positivity, followed by the same boundary and final marks. An equivalent extremum method applied to r=(3x²−2)/(x−1) is valid.