Marks Scheme Notes 2 f(0.5) = -0.4375 (-) Either any one of f(0.5) = awrt -0.4 or f(1) = (a) M1 f(1) = 1 1 Sign change (positive, negative) f(0.5) = awrt -0.4 and f(1) = 1, sign (and is continuous) therefore (a root) A1 change and conclusion is between and (2) f(0.75) = 0.06640625() Attempt f(0.75) M1 (b) f(0.75) = awrt 0.07 and f(0.625) = f(0.625) = -0.222412109375() A1 awrt -0.2 or or A1 or equivalent in words. In (b) there is no credit for linear interpolation and a (3) correct answer with no working scores no marks. (c) Correct derivative (May be implied B1 later by e.g. 4(0.75)3 + 1) Attempt Newton-Raphson M1 Correct first application a correct – numerical expression e.g. or awrt A1 0.725 (may be implied) Awrt 0.724 A1 cao A1 A final answer of 0.724 with evidence of NR applied twice with no (5) incorrect work should score 5/5 (10 marks) Marks Scheme Notes 3(a) Focus B1 + = 0 or = - M1 x “4” x “4” Directrix + 4 = 0 or = - 4 A1 x x
Scheme Notes Marks 2 (a) f(0.5) = -0.4375 (-) f(1) = 1 Either any one of f(0.5) = awrt -0.4 or f(1) = 1 M1 Sign change (positive, negative) (and is continuous) therefore (a root) is between and f(0.5) = awrt -0.4 and f(1) = 1, sign change and conclusion A1 (2) (b) f(0.75) = 0.06640625() Attempt f(0.75) M1 f(0.625) = -0.222412109375() f(0.75) = awrt 0.07 and f(0.625) = awrt -0.2 A1 or or or equivalent in words. A1 In (b) there is no credit for linear interpolation and a correct answer with no working scores no marks. (3) (c) Correct derivative (May be implied later by e.g. 4(0.75)3 + 1) B1 Attempt Newton-Raphson M1 Correct first application – a correct numerical expression e.g. or awrt 0.725 (may be implied) A1 Awrt 0.724 A1 cao A1 A final answer of 0.724 with evidence of NR applied twice with no incorrect work should score 5/5 (5) (10 marks) Scheme Notes Marks 3(a) Focus B1 Directrix x + “4” = 0 or x = - “4” M1 x + 4 = 0 or x = - 4 A1
fp1_all-past-papers.docx