18 2 M1 T = 0.0096 = T = or or Method to find e.g. u, M1 cao A1 19 (-5, -7) M1 Method to start to complete the square, e.g. (x + 5)2 M1 (x + 5)2 7 – A1 cao (dep on method seen) 20 M1 For = 54°, or 66° (may be on diagram) ACD ADC= A1 For = 60° from correct working CAD C2 C2 for all correct reasons stated (C1) (C1 for one appropriate reason linked to a circle theorem used) Alternate segment theorem. Opposite angles of a cyclic quadrilateral add up to 180. Angles in a triangle add up to 180 Angles on a straight line add up to 180 21 M1 gf(x) = or f(4) = 48 A1 oe
18 2 M1 M1 A1 T = or 0.0096 = or T = Method to find u, e.g. cao 19 (-5, -7) M1 M1 A1 Method to start to complete the square, e.g. (x + 5)2 (x + 5)2 – 7 cao (dep on method seen) 20 M1 A1 C2 (C1) For ACD = 54°, or ADC= 66° (may be on diagram) For CAD = 60° from correct working C2 for all correct reasons stated (C1 for one appropriate reason linked to a circle theorem used) Alternate segment theorem. Opposite angles of a cyclic quadrilateral add up to 180. Angles in a triangle add up to 180 Angles on a straight line add up to 180 21 M1 A1 gf(x) = or f(4) = 48 oe
GCSE%201MA1%20Mark%20Scheme%202H%20Mock%20Set%202.docx