EDEXCEL CORE MATHEMATICS C1 (6663)SPECIMEN PAPER MARK SCHEME1Question numberSchemeMarks1.a= 7, d= 2B1S20= ́20 ́(2 ́7 + 19 ́2) = 520M1 A1(3 marks)2.M1 A1 A1 B1(4 marks)3.(a)Ö80 = 4Ö5B1(1)(b)(4 -Ö5)= 16 -8Ö5 + 5 = 21 -8Ö5M1 A1 A1 (3)(4 marks)4.Gradient of AB= M1 A1Gradient of l= M1y-4 = (x-3)2x -5y+ 14 = 0M1 A1(5)(5 marks)5.(a)yO xPosition, ShapeB1(0, 2), (2, 0)B1 B1(3)(b)yO xPosition, ShapeB1(0, 1), B2 (1, 0) (3)(6 marks)21ôõó++=Ö+Cxxxxx232225d)35(2÷øöçèæ-=---2573)6(45252,2,21÷øöçèæ÷øöçèæ0,23www.mymathscloud.com
EDEXCEL CORE MATHEMATICS C1 (6663)SPECIMEN PAPER MARK SCHEME2Question numberSchemeMarks6.(a)5 -2x= 2x2-3x-162x2-x-21 = 0M1 A1(2x-7)(x+ 3) = 0x= -3, x= M1 A1y= 11, y= -2M1 A1ft(6)(b)Using critical valuesx= -3,x= M1x< -3,x> M1 A1ft(3)(9 marks)7.(a)a+ (n-1)d= 250+ (10 ́50) = £750M1 A1(2)(b)n= ́20 ́(500 + 19 ́50), = £14500M1 A1, A1 (3)(c)B: ́20 ́(2A+ 19 ́60) [= 10(2A+ 1140)], = “14500”B1, M1Solve for A: A= 155M1 A1(4)(9 marks)8.(a) a= 5,(x+ 5)2 -25 + 36b= 11B1, M1 A1 (3)(b)b2-4ac= 100 -144,< 0, therefore no real rootsM1 A1(2)(c)Equal roots if b2-4ac= 04k= 100k=25M1 A1(2)(d)yO xShape, positionB1 B1(-5, 0) (0, 25)B1 B1ft(4)(11 marks)27272721[]dna)1(2-+2121www.mymathscloud.com
EDEXCEL CORE MATHEMATICS C1 (6663)SPECIMEN PAPER MARK SCHEME3Question numberSchemeMarks9.(a)f(x)=x3-4x2 + 6x+ CM1 A15 = 27 -36 + 18 + CC= -4M1 A1(4)(b)x= 2:y= 8 -16 + 12 -4 = 0M1 A1(2)(c)f ¢(3) = 27 -24 + 6 = 9, Parallel therefore equal gradientB1, M13x2-8x+ 6 = 93x2-8x-3 = 0M1(3x+ 1)(x-3) = 0Q: x= M1 A1(5)(11 marks)10.(a)= 3x2-5 -2x-2M1 A2(1,0)At both Aand B, = 3 ́1 -5 -(= -4)M1 A1(5)(b)Gradient of normal =M1 A1fty-(-2) = (x-1)4y= x-9M1 A1(4)(c)Normal at Ameets y-axis where x= 0:y= B1Similarly for normal at B:4y= x+ 9y= M1 A1Length of PQ= A1(4)(13 marks)31-xyddxydd12414149-49294949=+www.mymathscloud.com