> GCE Further Mathematics B (MEI) Y435/01: Extra pure Advanced GCE Mark Scheme for Autumn 2021

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Mark Scheme October 2021 3 Question Answer Marks AO Guidance 1 (a) DR z = f(2, y) = 8 + 4y – 2y2 M1 1.1 Deriving correct equation of graph of section. = 10 – 2(y – 1)2⇒ max at (1, 10) or... (2, 1, 10) A1 1.1 Finding TP by completing the square, use of “–b/2a”, differentiation or mid-point between roots. Working must be shown. B1 1.1 ∩-shaped parabola which crosses horizontal axis twice. Condone incorrect variable names on axes (eg x-y for y-z). Crossing z-axis at 8, y-axis at 1 5 ± and showing (1,10) as a max A1 1.1 Coordinates of intercepts and max must be shown on graph or apparent in working. Allow decimal values (awrt –1.2 and 3.2) for the y-intercepts. z intercept must be shown as positive and max in 1st quadrant. However, scale is unimportant except that the negative y-intercept must be closer to O than the positive one. [4]Y435/01 Mark Scheme October 2021 4 (b) 2 z 3 2 x xy x ∂ = + ∂ B1 1.1 z x y 2 4 y ∂ = − ∂ B1 1.1 0 2 3 2 0 2 3 either 0 or or 3 2 zx x xy x x y y x ∂ = ∂ ⇒ + = ⇒ = = − = − M1 1.1 Setting a partial derivative to 0 and deriving condition(s) on x and/or y. Or 2 2 0 1 or 4 or 2 4 zy y x x y x y ∂ = ∂ ⇒ = = = ± 2 4 2 0, 4 9 3 9 z x y y y y y ∂ = = − ⇒ = ⇒ = ∂ M1 1.1 Substituting condition into other partial derivative equation to derive a non-zero value for x or y. 2 3 0, 2 6 0 6... z y x y x x x ∂ = = − ∂ ⇒ + = ⇒ = − 2 2 3 1 0, 4 1 3 0 6 2 z y x x x x x ∂ = = ∂ + = ⇒ = − or 32 0, 2 12 4 0 9 z x y x y y y ∂ = = ± ∂ ± = ⇒ = x = –6 A1 1.1 y = 9 z = –54 so (–6, 9, –54) A1 1.1 From correct working only. ... or 0 0 so (0, 0, 0) x y = ⇒ = A1 1.1 Derived from both z 0 x ∂ = ∂ and z 0 y ∂ = ∂ (could be by observation). If an extra SP is presented then A1 can be awarded for either SP correct and then A0. [7]Y435/01 Mark Scheme October 2021 5 2 (a) From Lagrange’s Theorem the order of any subgroup of G must be a factor of 8 and 6 is not a factor of 8 B1 2.4 Or “order of any subgroup of G (or an group of order 8) must be 1, 2 or 4 (or 8)” or “order of any subgroup must be a factor of the order of the group and 6 is not a factor of 8”. If referenced, Lagrange’s Theorem does not have to be quoted provided that it is applied. So B1 for eg “6 is not a factor of 8 so by Lagrange’s Theorem there can be no subgroup of G of order 6” but B0 for eg “By Lagrange’s Theorem there can be no subgroup of G of order 6”. [1] 2 (b) g2 (or g6) B1 2.2a May see eg gg or g◦g used here and/or throughout. Allow any multiplicative notation and any symbol for a binary operation. g6 (or g2) and no other B1 2.2a [2] 2 (c) e↔0 B1 2.2a Only needs to be seen once. g↔1, g2↔2, g3↔3, g4↔4, g5↔5, g6↔6, g7↔7 B1 2.2a Any one. g↔3, g2↔6, g3↔1, g5↔7, g6↔2, g7↔5 B1 2.2a Any other. g4↔4 does need not be seen again g↔5, g2↔2, g3↔7, g5↔1, g6↔6, g7↔3 and g↔7, g2↔6, g3↔5, g5↔3, g6↔2, g7↔1 B1 2.2a Other two. Ignore repeats. g4↔4 does need not be seen again Alternative method: e↔0 B1 Only needs to be seen once Either g↔1 or g↔3 or g↔5 or g↔7 M1 Giving all 4 possible isomorphism options for any generator of G (ie g, g3, g5 or g7) g↔1, g2↔2, g3↔3, g4↔4, g5↔5, g6↔6, g7↔7 A1 Completing the specification of any one isomorphism g↔3, g2↔6, g3↔1, g5↔7, g6↔2, g7↔5 and g↔5, g2↔2, g3↔7, g5↔1, g6↔6, g7↔3 and g↔7, g2↔6, g3↔5, g5↔3, g6↔2, g7↔1 A1 Other three. Ignore repeats. g4↔4 does need not be seen agai [Show More]

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