Markscheme for AS pure mathematics paper 1 set B

2 hours AS level Mathematics Edexcel
100 marks Paper 1 (Set B)



Q Answer Mark Comments
1 a f  2   8  36  52  24  0 M1 Substitution (seen or implied) to indicate
use of factor theorem
Therefore 2 is a factor of f(x) A1 Must be evaluated
OR
M1 A1 Long division to show  x  2  is a factor




x 2  7 x  12
b  x  2  x3  9 x 2  26 x  24 M1 Long division or inspection (or calculator
method)
f  x    x  2   x 2 7 x  12 


f  x    x  2  x  3 x  4 
A1 Correct factors

2 ai a2 B1
ii b  16 B1
6
b log3  2 x   log3   M1 Use rules of log laws to combine RHS
 x
6
2x  M1 Forming an equation in x
x
x 3 A1 Note: x   3 is not correct since x
cannot be negative

3 A = 12 B1
12𝑒 −2𝑘 = 6.6
𝑒 −2𝑘 = 0.55 M1 Use logs to solve for k
−2𝑘 = ln(0.55)
1 A1
𝑘 = − 2 ln(0.55) (awrt 0.3)


1 M1 Substitute t = 5 into equation with values of
𝐶= 12𝑒 2 ln(0.55)×5 A and k found


= 2.7 μg/ml A1




© Oxford University Press 2017
Acknowledgements: www.oxfordsecondary.co.uk/acknowledgements AS level Mathematics Paper 1 (Set B)

, 4 𝑑𝑦
= 6𝑥 2 + 6𝑥 + 𝑘 M1 Attempt at differentiating equation of
𝑑𝑥 curve
A1 Correct expression obtained for
derivative
At x  1 , 𝑚𝑡𝑎𝑛𝑔𝑒𝑛𝑡 = 𝑘 M1 Use of derivative to find gradient of
At x  1 , 𝑚𝑛𝑜𝑟𝑚𝑎𝑙 = − 𝑘
1 tangent at x  1
M1 Use of mnormal  mtangent  1 for  lines

At x = 0, y = -1 A1
𝑥
𝑦 =− −1
𝑘
B1 Evaluate y at x  1
A1 Or equivalent form of this straight line

5 f   x   3x 2  4 x  3 B1 Derivative completely correct

Discriminant  16  4  3  3  20  0 M1 Evaluate b2  4ac and show it is less
than zero

A1
Hence f   x   0 has no solutions, and so
there are no turning points on the graph. A1 A completely correct argument which is
clear and easy to follow

6 a 1 + 7(–x2) + 21(–x2) M1 Must have coefficients
2 4
1 – 7x + 21x A1 All three terms correct
7
b C5 × (1)2 × (–1)5 M1 Including either 7C5 or 21
 21 A1

7 a 1
 tan 2 
cos 2 
 sin  
2


1
 sin 
 M1 Use of  tan 
cos   cos  
2
cos 
1  sin 2 

cos 2 
cos 2  M1 Use of sin 2   1  cos2 
 A1 A completely correct argument which
cos 2 
is clear and easy to follow
1




© Oxford University Press 2017
Acknowledgements: www.oxfordsecondary.co.uk/acknowledgements AS level Mathematics Paper 1 (Set B)