# Mark Scheme. Mathematics General Certificate of Education examination June series. MPC1 Pure Core 1

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1 Version 1.0: 0706 abc General Certificate of Education Mathematics 660 MPC1 Pure Core 1 Mark Scheme 006 examination June series Mark schemes are prepared by the Principal Examiner and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation meeting attended by all examiners and is the scheme which was used by them in this examination. The standardisation meeting ensures that the mark scheme covers the candidates responses to questions and that every examiner understands and applies it in the same correct way. As preparation for the standardisation meeting each examiner analyses a number of candidates scripts: alternative answers not already covered by the mark scheme are discussed at the meeting and legislated for. If, after this meeting, examiners encounter unusual answers which have not been discussed at the meeting they are required to refer these to the Principal Examiner. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of candidates reactions to a particular paper. Assumptions about future mark schemes on the basis of one year s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper. Copyright 006 AQA and its licensors. All rights reserved.

3 AQA GCE Mark Scheme, 006 June series MPC1 MPC1 1(a)(i) Gradient AB = = 6 = = 1.5 Must be y on top and subtr n of cords Any correct equivalent y 7 = m( x 1) or y 1 = m( x 5) Verifying points or y = x+ c leading to x +y = 17 AG (or grad & 1 point verified) (b) Attempt to eliminate x or y : 7x = etc x = 6 y = 1 Solving x y = 8; x+ y= 17 C is point (6, 1 ) (c) Grad of perp = 1 / their gradient AB = Or mm 1 = 1 used or stated ft their gradient AB (a) y 7 = ( x 1) or y x = 19 ( x + ) + CSO Any correct form of equation p = q = (b) ( x + ) = or their No real square root of y ( x + p) = q Or discriminant = 6 76 Disc < 0 so no real roots (all correct figs) (c) 19 Minimum (, ) graph x ft their p and q (or correct) Parabola (vertex roughly as shown) Crossing at y = 19 marked or (0, 19) stated (d) Translation (and no additional transf n) E1 Not shift, move, transformation, etc One component correct eg units up through All correct if not vector must say units in negative x- direction, to left etc (a) dy 10x dx = kx condone extra term Correct derivative unsimplified (b) When x = 1, gradient = 10 FT their gradient when x = 1 (c) Tangent is y 5= 10( x 1) or y + 10x = 15 etc When x = ( d y dx dy = 160 ( or < 0 ) dx < 0 hence) y is decreasing E1 Total 7 Attempt at y & tangent (not normal) CSO Any correct form Value of their d y when x = dx ft Increasing if their d y dx > 0

4 MPC1 AQA GCE Mark Scheme, 006 June series MPC1 (cont) (a) Multiplied out At least terms with 5term ( 5) ( 5) = 5 ( = 0) Answer = (b) Either 75 = 5 or 7 = 9 Or multiplying top and bottom by Expression = 5 or = CSO Total 6 5(a)(i) dy x 0x 8 dx 5 81 One term correct Another term correct All correct ( no + c etc) or 5 9 or 5 Their d y = 0 for stationary point dx ( x )(x 1) = 0 x = 1 or x = m1 Or realising condition for stationary pt Attempt to solve using formula/ factorise Award, for verification that dy x = = 0 then may earn m1 later dx (b)(i) x 10x + 1 x ( + c) ( 0) = 56 1 One term correct unsimplified Another term correct unsimplified All correct unsimplified (condone missing + c) Attempt to sub limit into their (b)(i) AG Integration, limit sub n all correct (iii) Area of triangle = 1 1 Shaded Area = 56 1 triangle area = Total 15 Correct unsimplified 1 1 Or equivalent such as 99

5 AQA GCE Mark Scheme, 006 June series MPC1 MPC1 (cont) 6(a) p() = p() = 0 x is a factor Finding p() and not long division Shown = 0 plus a statement (b) xx ( x+ ) or ( x )( x x) attempt Or p(1) = 0 x 1 is a factor attempt p( x) = x( x 1)( x ) Condone x + 0 or x 0 as factor (c)(i) p() = (Remainder is) Must use p() and not long division Attempt to multiply out and compare Or long division ( terms of quotient) 7(a)(i) coefficients a = b = 1 r = SC for r = if M0 scored ( x ) centre has x-coordinate = and y-coordinate = 0 x x... 1 Withhold final for long division unless written as ( x )( x x 1) Attempt to complete square for x implied if value correct or Centre (,0) RHS = 18 Radius = 18 Radius = Withhold if circle equation RHS incorrect Square root of RHS of equation (if > 0) (b) Perpendicular bisects chord so need to use Length of d = (radius) d = so perpendicular distance = d 18 (c)(i) x + ( k x) x 1 = 0 ( k x) = k kx+ x x + k kx x 1= 0 ( x + k kx x 7= 0) x ( k + 1) x+ k 7 = 0 AG (be convinced about algebra and = 0) ( k + 1) (k 7) b ac in terms of k (either term k 8k = 0 or k correct) k 8= 0 b ac = 0 correct quadratic equation in k ( k )( k + ) = 0 m1 Attempt to factorise, solve equation k =, k = SC, for, (if M0 scored) (iii) Line is a tangent to the circle E1 1 Line touches circle at one point etc Total 17 TOTAL 75 5

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