C1 reduces to 2(2n−3) or 4n 6 – C1 for conclusion e.g. 2(2n−3) always even, 4n 6 is always even both are even numbers, they – are multiples of 2. 17 for or or oroe seen on diagram or in a calculation P1 P1 for × or × or oe for × or × or or or oe for × + × for 1 ( × + × ) or 1 ("" + "") P1 – ‒ or "" + ""oe or 1 "" oe ‒ A1 oe SC B1 for B2 for 18 P1 evidence of understanding that is perpendicular to or states the gradient of as 0.5 AC DB, DB = + 21 y −2x oe P1 shows a process to find the gradient of a perp. line e.g. use of or states = + c or states the gradient of AC as y −2x −2 P1 (dep on P2) for sub. of = 5, = 11 into = + where is their found gradient for AC. x y y mx c m A1 oe 1MA1/1H Question Working Answer Mark Notes 19 for step to solve the problem e.g. = a + c or = a + c or demonstrates the location of and D − P1 on the diagram X P1 for a correct vector statement using eg = + or = or = c or = 2.5c oe − for a correct equation or ratio using eg equating = 3c a = (−a + c) + c k –
C1 reduces to 2(2n−3) or 4n – 6 C1 for conclusion e.g. 2(2n−3) always even, 4n – 6 is always even both are even numbers, they are multiples of 2. 17 P1 for or or oroe seen on diagram or in a calculation P1 for × or × or oe for × or × or or or oe P1 for × + × or "" + ""oe for 1 – ( × + × ) or 1 ‒ ("" + "") or 1 ‒ "" oe A1 oe SC B1 for B2 for 18 y = −2x + 21 P1 evidence of understanding that AC is perpendicular to DB, or states the gradient of DB as 0.5 oe P1 shows a process to find the gradient of a perp. line e.g. use of or states y = −2x + c or states the gradient of AC as −2 P1 (dep on P2) for sub. of x = 5, y = 11 into y = mx + c where m is their found gradient for AC. A1 oe 1MA1/1H Question Working Answer Mark Notes 19 P1 for step to solve the problem e.g. = − a + c or = a + c or demonstrates the location of D and X on the diagram P1 for a correct vector statement using eg = + or = − or = c or = 2.5c oe for a correct equation or ratio using k eg equating = 3c – a = (−a + c) + c
1ma1_1h_-_may_2017_mark_scheme.docx