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Exam (elaborations)

markscheme for as pure set a paper 1

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It is designed according to a level 2015 specification. it is an As exam paper and all pure mathematics topics are covered










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Uploaded on
June 8, 2021
Number of pages
7
Written in
2020/2021
Type
Exam (elaborations)
Contains
Questions & answers

Subjects

  • pure mathematics

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



Q Answer Mark Comments
1
1 Gradient =  B1
2
1
y 7    x  4 M1 Substituting “their” value for the gradient
2
into the equation y  y1  m( x  x1 )
1
y   x9 A1 Equation correct, in this or equivalent form
2

2 Stretch; B1
Parallel to y-axis; B1 Accept ‘vertical stretch’
Scale factor 2 B1


2 x   23 
3 x 1
3 a B1
x  3(3x  1) M1 Attempt at an equation in x using index
laws or taking logs on both sides
3
x A1
8
 x  x2  1
b log  1  or log x  2log x  log x M1 Apply either
 2  2
 x   ab 
log a  log b  log c  log   or
 c 
log a  k log a for each term
k

 5 5
log  x 2  or log x M1 Use laws of indices or
  2 collect up like terms (this scores both M1
& A1 marks)
5
Apply log a  k log a and express as
k
log x A1
2
single logarithm in log x




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

, 4 Gradient PR × Gradient QR  1 M1 Use of m1  m2  1 for perpendicular lines
4 3 k A1
  1  k  7
8 2
PQ is the diameter of the circle, since
angle subtended at circumference is 90°
Centre of circle is at
midpoint of PQ: (2, 3)
B1 Correct deduction; leading to correct
r   5  2    7  3
2 2 2
method to find radius

r 5
Area of circle:  r 2  25 M1 Correctly calculating the radius
A1 Accept 78.5

dy
5  2x  3 M1 Attempt at differentiating equation of C
dx
A1 Gradient of tangent at x  2
At x  2 , m = 2(2) – 3 =1
B1 Evaluate y at x  2
M1 Use of equation of straight line
y  0 at x  2
A1
y  x2

6 1  cos x   3cos x  2  0
2 M1 Use of identity
cos2 x  3cos x  3  0
M1 Find solutions for cos 𝑥
3  32  4(1)(3) 3  21
cos x  
2 2 A1
cos x  0.7913 or –3.7913 (reject)
M1 Subtract first solution from 360° or other
method to find second solution
x = 37.7° A1 Both correct
or 322.3°
7 a 18 + 8C1(1)7(2x)1 + 8C2(1)6(2x)2 +
8 M1 Uses binomial theorem to expand bracket
C3(1)5(2x)3

1 + 16x + 112x2 + 448x3 A1 1 + 16x
A1 Completely correct

b 1 + 16(0.01) + 112(0.01)2 + 448(0.01)3 B1 Selects x  0.01
M1 Substitute chosen value of x into a
1.171648
A1




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