OCR GCE Further Mathematics A Y531/01: Pure Core Advanced Subsidiary GCE Mark Scheme for Autumn 2021

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Mark Scheme October 2021 3 Question Answer Marks AO Guidance 1 (a) 8 – 2λ = –6 – 3µ and –11 + 5λ = 11 + µ B1 1.1a Forming 2 correct equations in λ and µ. Could be –2 + 3λ = 8 –... µ Any two correct equations 8 – 2λ = –6 – 3µ –33 + 15λ = 33 + 3µ => –25 + 13λ = 27 M1 1.1 Attempt to solve (eg scaling one equation and adding or rewriting to a standard form for solution BC). Must reach an equation (possibly incorrect) with only one unknown. –2λ + 3µ = –14 5λ – µ = 22 3λ + µ = 10 λ = 4, µ = –2 A1 1.1 Both – 2 + 3×4 = 10 and 8 – –2 = 10 so they do intersect A1 2.4 Checking for consistency in 3rd equation and conclusion. Equation must be correct and both sides must be evaluated Allow eg 8 2 4 6 3 2 0 0 − × = − − × − = Might see λ = 4 substituted into last equation and then µ being found with this. ie – 2 + 3×4 = 8 – –2 alone is not sufficient for A1, need to see both sides becoming 10 x: 8 – 2×4 = 0 & –6 – 3×–2 = 0 y: –11 + 5×4 = 9 & 11 + –2 = 9 [4] (b) (0, 9, 10) B1 1.1 Allow as vector [1]Y531/01 Mark Scheme October 2021 4 Question Answer Marks AO Guidance 2 u = x + 1 B1 3.1a (u – 1)3 = u 3 – 3u 2 + 3u – 1 used in solution M1 1.1 Attempt to expand using binomial. 4 terms. Follow through on their u = x + 1 2x3 + 3x2 – 2x + 5 = 0 => 2(u 3 – 3u 2 + 3u – 1) + 3(u 2 – 2u + 1) – 2(u – 1) + 5 = 0 M1 1.1 Substituting into equation. Allow if no “= 0” here. Must have an attempt at expanding (u – 1)3 and (u – 1)2 Follow through on their u = x + 1 2u 3 – 3u 2 – 2u + 8 = 0 A1 2.5 Must be an equation For correct equation found using sums and products of roots allow SC2 (Method required was dictated in question) Only allocate marks using main scheme, or SC method [4] Question Answer Marks AO Guidance 3 3 + 5i is a root B1 1.2 Need to see statement that 3+5i is a root. May happen at end of question Attempt to expand (x – (3 + 5i))(x – (3 – 5i)) M1 1.1 Attempt to use the conjugate pair to derive a real quadratic May see (3 + 5i)(3 – 5i) = 9 + 25 = 34 and (3 + 5i) + (3 – 5i) = 6 instead of expansion = x2 – 6x + 34 so this must be a factor A1 2.2a x4 – 7x3 – 2x2 + 218x – 1428 = (x2 – 6x + 34)(x2 + ...x – 42) or (x2 – 6x + 34)(x2 – x + ...) M1 1.1 Attempt to factorise or divide resulting in x2 and one other term NB: This question required detailed reasoning (x2 – 6x + 34)(x2 – x – 42) A1 1.1 (x2 – x – 42) = (x – 7)(x + 6) => roots –6, 7 (and 3 + 5i) A1 1.1 3 + 5i may be mentioned as a root earlier in the solution [6]Y531/01 Mark Scheme October 2021 5 Question Answer Marks AO Guidance 4 (a) (i) Line drawn, perpendicular to line segment joining (0, -1) and (2,0) M1 1.1 Line needs to have negative gradient with |gradient| >1 and to intersect the y axis at a positive value If “shading out” is used then there needs to be an indication that the required region is below the line, such as “R” placed below line or “This region” written in etc. Region below line indicated as being the required region. A1 1.1 Exact perpendicularity not needed, but should be approximately perpendicular. [2] (a) (ii) m = –1/(½) = –2 M1 1.1 4x + 2y – 3 = 0 A1 1.1 Explicitly stated Note must be in required form ax+by+c=0 [Show More]

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