Mathematics > A/As Level Mark Scheme > GCE Mathematics A H240/01: Pure Mathematics Advanced GCE Mark Scheme for Autumn 2021. 100% Approved (All)

GCE Mathematics A H240/01: Pure Mathematics Advanced GCE Mark Scheme for Autumn 2021. 100% Approved pass rate

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Question Answer Marks AO Guidance 1 16 – 4(k + 3) M1* 1.1 Attempt discriminant Allow b2 + 4ac for M1, but nothing else –4k – 12 + 16 > 0 A1 2.3 Obtain correct inequality Not necessarily expand... ed 4k – 4 < 0 M1dep * 1.1 Attempt to solve their inequality or equation for k k < 1 A1 1.1 Obtain k < 1 [4] OR (completing the square or differentiating) M1* – attempt to complete the square, or differentiate, and link minimum point to 0 A1 – obtain (k + 3) – 4 < 0 M1d* – solve their inequality or equation A1 – obtain k < 1 OR (using perfect square) M1* – link k + 3 to 4 A1 – obtain k + 3 < 4 M1d* – solve their inequality or equation A1 – obtain k < 1 2 (a) (C =) 4000 + 4m B1 3.3 Correct equation / expression for A Or 40 + 0.04m (C =) 6(m – 100) B1 3.3 Correct equation / expression for B Or 0.06(m – 100) [2] B1B0 if units inconsistent in two equations SC B1 for both 44 + 0.04m and 0.06m (or 4400 + 4m and 6m) – from using m = 0 at 100 minutes (b) 4000 + 4m = 6(m – 100) 2m = 4600 M1 1.1 Attempt to solve simultaneously, from two linear equations in m At least one equation must have constant term Could be implied by final answer of 38hrs 20 mins m = 2300 A1 3.4 Obtain 2300 (minutes) isw once 2300 seen [2]H240/01 Mark Scheme October 2021 7 Question Answer Marks AO Guidance 3 (a) x ky z = 2 30 4 3 = × × k k = 2.5 M1 3.1a Attempt to find value for k From x ky z = 2 or x k z y = 2 only Using sum, not product, is M0 but watch for + z being used for positive square root 2 x y z = 2.5 A1 1.1 Correct equation Ignore modulus sign if used around z Allow BOD if initial equation stated explicitly, k found correctly but then final equation not seen or seen as now incorrect [2] (b) x = × × 2.5 9 5 M1 1.1 Attempt to find x, from equation in terms of y, z and numerical k Could be from using direct proportion and not their equation from (a) x = 112.5 A1 1.1 Obtain 112.5 Or any exact equiv [2] 4 DR (a) f(0.5) = 0.25 – 0.75 – 5.5 + 6 = 0 B1 2.1 Attempt f(0.5) and show equal to 0 Must be using factor theorem so B0 for alternative methods B0 for just f(0.5) = 0 Condone 2(0.5)3 – 3(0.5)2 – 11(0.5 [Show More]

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