13 (a) f(x) = + 5x 4 shown 2 M1 method to establish at least one root in 0 < < 1, x3 x – e.g. f(0) = and f(1) = 2 –4 C1 as there is a sign change there must be at least one root in 0 < < x 1 (as f is continuous) (b) + 5x 4 = 0 shown 2 C1 at least one correct step in rearrangement x3 – + 5x = 4 C1 fully correct chain of reasoning x3 + 5) = 4 x(x2 (c) 0.709(21985) 3 B1 = 0.8 oe or M1 = oe A1 0.709(21985) or oe 14 8.3 3 M1 0.5 = or 0.5 = (1 )8 r8 – M1 can be implied by seen 0.917(…) A1 8.29 8.3 –
13 (a) f(x) = x3 + 5x – 4 shown 2 M1 method to establish at least one root in 0 < x < 1, e.g. f(0) = –4 and f(1) = 2 C1 as there is a sign change there must be at least one root in 0 < x < 1 (as f is continuous) (b) x3 + 5x – 4 = 0 x3 + 5x = 4 x(x2 + 5) = 4 shown 2 C1 at least one correct step in rearrangement C1 fully correct chain of reasoning (c) 0.709(21985) or 3 B1 = 0.8 oe M1 = oe A1 0.709(21985) or oe 14 8.3 3 M1 0.5 = r8 or 0.5 = (1 – )8 M1 can be implied by 0.917(…) seen A1 8.29 – 8.3
GCSE_Mathematics_3H_MS.docx