Pearson Edexcel Level 3 Advanced Subsidiary GCE in Further Mathematics (8FM0) First teaching from September 2017 First certifi cation from 2018

Pearson Edexcel Level 3 Advanced Subsidiary GCE in Further Mathematics (8FM0) First teaching from September 2017 First certifi cation from 2018Paper 1: Core Pure Mathematics Mark Scheme Question Scheme Marks AOs 1(a) 5 5   1 15             M1 1.1b A1 1.1b  5  25   5  15  2  4  5  0      4 (4)2  4(1)(5) 2(1) or (  2)2  4  5  0    ... M1 3.1a    2  i A1 1.1b Hence the roots of f(z)  0 are 2  i, 2  i and 3 A1 2.2a (5) (b) p    "(2  i)"  "(2  i)"  "3"  p  ... M1 3.1a  p  7 cso A1 1.1b (2) 1(b) alternative f(z)  (z  3)(z 2  4z  5)  p  ... M1 3.1a  p  7 cso A1 1.1b (2) (7 marks) Notes: (a) M1: Multiplies the three given roots together and sets the result equal to 15 or 15 A1: Obtains a correct equation in  M1: Forms a quadratic equation in  and attempts to solve this equation by either completing the square or using the quadratic formula to give   .... A1:   2  i A1: Deduces the roots are 2  i, 2  i and 3 (b) M1: Applies the process of finding  of their three roots found in part (a) to give p  ... A1: p  7 by correct solution only (b) Alternative M1: Applies the process expanding (z  "3")(z  (their sum)z  their product) in order to find p  ... A1: p  7 by correct solution only 25 Pearson Edexcel Level 3 Advanced Subsidiary GCE in Further Mathematics – Sample Assessment Materials – Issue 1 – August 2017 © Pearson Education Limited 2017 Question Scheme Marks AOs 2(a) r  3 1 2            5 3 4            3 1 2           M1 1.1b 3 2 10 xy z   A1 2.5 (2) (b) 3 1 2            1 5 3            8 B1 1.1b (3)2  (1)2  (2)2 . (1)2  (5)2  (3)2 cos  " 3  5  6" M1 1.1b 8 8 90 arccos or sin 14. 35 14. 35             M1 2.1   21.2 (1 dp)  * cso A1* 1.1b (4) (c) 3(7 ) (3 5 ) 2( 2 3 ) 10 ...            M1 3.1a    1 2 A1 1.1b M1 1.1b X (7.5, 5.5, 3.5)  A1ft 1.1b (4) (10 marks) Notes: (a) M1: Attempts to apply the formula r.n a.n  A1: Correct Cartesian notation. e.g. 3 2 10 or 3 2 10 x  y z xy z       Note: Do not allow final answer given as rij k (3 2 ) 10,    o.e. (b) B1: OA n  8   M1: An attempt to apply the correct dot product formula between n and d M1: Depends on previous M mark. Applies the dot product formula to find the angle between  and l A1*: 21.2 cso 26 Pearson Edexcel Level 3 Advanced Subsidiary GCE in Further Mathematics – Sample Assessment Materials – Issue 1 – August 2017 © Pearson Education Limited 2017 Question 2 notes continued: (c) M1: Substitutes l into  and solves the resulting equation to give   .... A1:    1 2 o.e. M1: Depends on previous M mark. Substitutes their  into l and finds at least one of the coordinates A1ft: (7.5, 5.5, 3.5) but follow through on their value of   27 Pearson Edexcel Level 3 Advanced Subsidiary GCE in Further Mathematics – Sample Assessment Materials – Issue 1 – August 2017 © Pearson Education Limited 2017 Question Scheme Marks AOs 3 x = value of savings account, y = value of property bond account, z = value of share dealing account x  y  z  5000 x  400  y 0.015x  0.035y  0.025z  79 or 1.015x  1.035y  0.975z  5079 M1 3.1b A1 1.1b Let A  1 11 1 1 0 0.015 0.035 0.025           or 1 11 1 1 0 1.015 1.035 0.975           e.g. 1 11 1 1 0 0.015 0.035 0.025           x y z            5000 400 79           M1 3.1a A1 1.1b x y z            1 11 1 1 0 0.015 0.035 0.025           1 5000 400 79            ... ... ...           M1 1.1b x y z            1800 2200 1000           A1 1.1b Tyler invested £1800 in the savings account, £2200 in the property bond account and £1000 in the share dealing account A1ft 3.2a (7 marks) Notes: M1: Attempts to set up 3 equations with 3 unknowns A1: At least 2 equations are correct with the appropriate variables defined M1: Sets up a matrix equation of the form, e.g. ... ... ... ... ... ... ... ... ... ... ... ... x y z                 , where “…” are numerical values A1: Correct matrix equation (or equivalent) M1: Depends on previous M mark. Applies (their A) 1 5000 their "400" their "79"           and obtains at least one value of x, y or z A1: Correct answer A1ft: Correct follow through answer in context 28 Pearson Edexcel Level 3 Advanced Subsidiary GCE in Further Mathematics – Sample Assessment Materi