ACSBR 2023 4E5N Add Math Prelim P2 - MS_updated

Marking Scheme Secondary 4 Express / 5 Normal (Academic) Additional Mathematics Paper 2 Preliminary Examination 2023 Anglo-Chinese School (Barker Road) 1 (a) $1000 (b) = (c) 3 months or August $4000 2 (a) Solving (1), (2) (b) Sub. (1) into (2): For the line to intersect the curve at two distinct points,

Marking Scheme Secondary 4 Express / 5 Normal (Academic) Additional Mathematics Paper 2 Preliminary Examination 2023 Anglo-Chinese School (Barker Road) 3 (a) Alternative Method (b) ,

Marking Scheme Secondary 4 Express / 5 Normal (Academic) Additional Mathematics Paper 2 Preliminary Examination 2023 Anglo-Chinese School (Barker Road) 4 (a) Using similar triangles (b) Using chain rule When h = 8, 5 (a) or (reject) (shown) Since

Marking Scheme Secondary 4 Express / 5 Normal (Academic) Additional Mathematics Paper 2 Preliminary Examination 2023 Anglo-Chinese School (Barker Road) (b) Midpoint of PR = Gradient of QS = Equation of QS : Sub (1) into Subst. into (1) : (c) Area = =. or 28.6 (3sf) 6 (a) Given that AE = EB and CF = FB, E is the midpoint of AB and F is midpoint FB. Hence by mid-point theorem, AC and EF are parallel (b) angle AED = angle DCE (tangent chord theorem) angle DCE = angle CEF

Marking Scheme Secondary 4 Express / 5 Normal (Academic) Additional Mathematics Paper 2 Preliminary Examination 2023 Anglo-Chinese School (Barker Road) (alternate angles, AC is parallel to EF) Hence angle AED = angle CEF (c) In triangle AED and triangle CEF Let angle CFE = Angle CDE = (angles in opposite segment) Angle ADE = (adjacent angles on a a straight line) Angle CFE = Angle ADE or Angle CFE = Angle ADE (exterior angle of cyclic quadrilateral ) or angle DAE = angle FEB (corresponding angles, AC parallel EF) Angle FEB = angle ECF (tangent chord theorem) Angle DAE = angle ECF From part (a) angle AED = angle CEF Triangle AED is similar to triangle CE