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 a2 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)
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 a2 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)
