B1 = 2 + = 2 y x x y – oe 2x2 + 5x + 1 = their (2 + x) = 2(y 2)2 + 5(y 2) + 1 M1 y – – 2y2 8y + 8 + 5y 10 + 1 = 0 y – – – M1dep 2y2 4y 1 = 0 – – 2x2 + 4x 1 = 0 – −4 ± 42 − (4 × 2 −−4 ± (−4)2 − (4 × 2 ×−1) ×−1) 21 M1 2 × 2 2 × 2 or 24 or 4 ± 4 x = and x = 0.2(…) or x y = 2.2(…) and y = or y –2.2(…) –0.2(…) = and = or A1 = 2.2(…) and = 0.2(…) or = y x x y –2.2(…) –0.2(…) = 0.2(…) and = 2.2(…) and = y x –0.2(…) –2.2(…) x = and y = and y = 2.2 and x = 0.2 and y –2.2 –0.2 A1 x = 0.2 and y = 2.2 = and x = –0.2 –2.2 Q Additional Guidance Mark 21 BEWARE, roots of 2x2 + 5x + 1 = 0 are 0.22 and 2.28 – – Correctly substituting their values from their quadratic scores M1, e.g. 2x2 + 5x + 1 = 0 scores M0M0M1A0A0 −5± 52 −(4×2×1) 2×2 All four solutions are required to score full marks
21 y = 2 + x B1 x = y – 2 2x2 + 5x + 1 = their (2 + x) M1 oe y = 2(y – 2)2 + 5(y – 2) + 1 2y2 – 8y + 8 + 5y – y – 10 + 1 = 0 2x2 + 4x – 1 = 0 M1dep 2y2 – 4y – 1 = 0 M1 −4 ± 42 − (4 × 2 ×−1) −−4 ± (−4)2 − (4 × 2 ×−1) 2 × 2 or 2 × 2 or 4 ± 4 24 x = –2.2(…) and x = 0.2(…) or x = –2.2(…) and y = –0.2(…) or x = 0.2(…) and y = 2.2(…) A1 y = 2.2(…) and y = –0.2(…) or y = 2.2(…) and x = 0.2(…) or y = –0.2(…) and x = –2.2(…) x = –2.2 and y = –0.2 and x = 0.2 and y = 2.2 A1 y = 2.2 and x = 0.2 and y = –0.2 and x = –2.2 Q Additional Guidance Mark 21 BEWARE, roots of 2x2 + 5x + 1 = 0 are – 0.22 and – 2.28 Correctly substituting their values from their quadratic scores M1, e.g. 2x2 + 5x + 1 = 0 −5± 52 −(4×2×1) scores M0M0M1A0A0 2×2 All four solutions are required to score full marks
November-2014-Paper-2-Mark-Scheme1.docx