(c) = 13.7 and = 19.05 Q(20) Q(10) difference = 19.05 13.7 (M1) – 13.7 19.05) (allow – = 5.35 average = = 0.535 per minute (sign must be positive) (A1)(ft) Note: The wording may have confused some candidates. (M1)(A1)(C2) allow = 16.375 (16.4) (C2) and for answer 16.3 (which is Q(15)) award just (C1). If the interval is divided into multiple parts (eg 10) then (M1) the value averaged over these, award no marks. (d) 0.003t2 0.625t + 25 = 0 (A1) – (C2) Note: A sketch showing the first root can earn the (M1). Energy runs out after 54.0 minutes (accept 54 but no mark for 154) Note: 53.99 receives (A0)(AP), 53 is wrong, for 53 54 award (A1) – [6] 3. (a) = or = 5x or 25 = + 5k y x(5 x) y x2 c – – (M1) = 0, = 5 c k (A1)(A1)
(c) Q(20) = 13.7 and Q(10) = 19.05 difference = 19.05 – 13.7 (M1) (allow 13.7 – 19.05) = 5.35 average = = 0.535 per minute (sign must be positive) (A1)(ft) Note: The wording may have confused some candidates. allow = 16.375 (16.4) (M1)(A1)(C2) (C2) and for answer 16.3 (which is Q(15)) award just (C1). If the interval is divided into multiple parts (eg 10) then the value averaged over these, award no marks. (M1) (d) 0.003t2 – 0.625t + 25 = 0 (A1) (C2) Note: A sketch showing the first root can earn the (M1). Energy runs out after 54.0 minutes (accept 54 but no mark for 154) Note: 53.99 receives (A0)(AP), 53 is wrong, for 53 – 54 award (A1) [6] 3. (a) y = x(5 – x) or y = 5x – x2 or 25 = c + 5k (M1) c = 0, k = 5 (A1)(A1)
ms_1_review_package_for_quadratic_functions2.docx