Edexcel 9MA01 Maths A-level Paper 2 (Pure) 12th June 2019 - Unofficial Markscheme Watch

Poll: What mark do you think you got?
95-100 (235)
3.78%
90-94 (220)
3.54%
85-89 (272)
4.37%
80-84 (319)
5.13%
75-79 (316)
5.08%
70-74 (451)
7.25%
65-69 (424)
6.82%
60-64 (495)
7.96%
55-59 (438)
7.04%
50-54 (518)
8.33%
45-49 (408)
6.56%
40-44 (434)
6.98%
35-39 (359)
5.77%
30-34 (311)
5%
Less than 30 (1018)
16.37%
RedGiant
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#1
Sign this petition if you feel violated by Edexcel for this hard paper (over 8,300 signatures so far):
https://www.change.org/p/edexcel-ale...t-of-situation

And this one:
https://www.change.org/p/edexcel-a-l...abandoned-spec


1. Y = -3/4 - x/2 [3 marks]

2. a) Use the trapezium rule to calculate the distance of the runway = 415m [3 marks]
b) It is an overestimate, if you consider the shape of the graph. [1 mark]

3. a) You had to state that the student had used the angle in degrees, and not radians [1 mark]
b) Area of sector = 8.73 cm^2 [2 marks]

4. The coordinates for the point S was (5sqrt2, -4) [6 marks]

5. Integral of sqrt x = 38/3 [3 marks]

6. a) gg(0) = 13 [2 marks]
b) The range was: - x>35/4, x<2-3rt3 [4 marks]
c) Explain that h(x) was a one to one function, and the other was a many-to-one function (iirc) [1 mark]
d) For the inverse function, x=29/4 [3 marks]


7. a) You simply had to show that y = a + kx [1 mark]
b) Solve the simultaneous equations that you could form, and show that k was 0.86. [3 mark]
c) Interpret the value of 0.86 [1 mark]
d) 369 soaps were required, to make a profit [2 marks]

8 a) The sum of 20(1/2)^r from 4 to infinity = 5/2 [3 mark]
b) Carry out the proof of sum of log = 2 [3 mark]

9. a) First take logs, then explain that the gradient was constant, therefore k and n were constant. [3 marks]
b) n=2.08 [3 mark]
c) Stopping distance = 98.25m (stop before puddle - you had to add on 40/3, which was the thinking distance) [3 marks]

10. a) Work out the vector OC: 1/2b-3/2a [2 marks]
b) Show lamda = 4/3 (because no a component) [2 marks]
c) Proof [2 marks]

11 a) You had to take logs on both sides, and end up with the turning point at x = 1/e [5 marks]
b) Show that there is a change of sign [2 marks]
c) x3 = 1.673 [2 marks]
d) Cobweb diagram that diverge from root alpha, then oscillate between 1 and 2 [2 marks]

12 a) Prove the equation [4 marks]:

\displaystyle \frac{\cos 3 \theta} {\sin \theta} + \frac {\sin 3 \theta} {\cos \theta} \equiv \frac {\cos 3 \theta \cos \theta + \sin 3 \theta \sin \theta} {\sin \theta \cos \theta}

Then using \cos(A-B) \equiv \cos A \cos B + \sin A \sin B and \sin 2 \theta \equiv 2 \sin \theta \cos \theta:

\displaystyle \frac{\cos 3 \theta} {\sin \theta} + \frac {\sin 3 \theta} {\cos \theta} \equiv \frac {\cos (3 \theta - \theta)} {\frac 1 2 \sin 2 \theta} \equiv 2\frac {\cos 2 \theta} {\sin 2 \theta} \equiv 2 \cot 2 \theta

b) 103.3 degrees for trig [3 marks]

13. a) Show that the area is equal to the given expresion [4 marks]
b) Area is minimum when dA/dR = 0, so minimum radius = 1.05m, [4 marks]
c) Surface area = 17m^2 [2 marks]

14. a) Integrate the equation [6 marks]
b) Range - 0<h<16 (technically have <=0, but 0 height makes no sense in 3D) [2 marks]
c) Time for 12m is 75.4 years (rounding early, 77 years if you didn't) [7 marks]

These are the marks per question:

1) 3
2) 4
3) 3
4) 6
5) 3
6) 10
7) 7
8) 6
9) 9
10) 6
11) 11
12) 7
13) 10
14) 15

Mean mark for this paper (based on poll): 54/100
Lower quartile: 38
Upper quartile: 73
Median: 58
Last edited by RedGiant; 1 week ago
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RIghty Ross
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#2
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#2
Could you please add marks to these questions
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B0redBrioche
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#3
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#3
I reckon similar grade boundaries to last year maybe slightly lower depending on paper 3 so
A* 75%
A 60%
B 50%
C 40%
D 30%
E 20%
Last edited by B0redBrioche; 1 week ago
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Itz Nurul
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#4
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#4
someone post the paper if u have pics
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Heidi-Heather
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#5
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#5
75.2 years for tree question (question 14)
x-coordinate for turning point was 1/e.
The iteration diverged and resulted in the periodic sequence 1,2,1,
gg(0)=13
n was 2.08, Sean could stop in time.
length of runway was 415 metres (overestimate as velocity-time graph is convex)
Surface area was 17m^2, radius was 1.05
Needed 369 bars of soap to make a profit.
First sum to infinity was 5/2
103.3 for the cot2x trig equation.
Coordinate for parametric was (5sqrt2, -4)
Last edited by Heidi-Heather; 1 week ago
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_gcx
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#6
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#6
People seemed to struggle on the trig proof:

\displaystyle \frac{\cos 3 \theta} {\sin \theta} + \frac {\sin 3 \theta} {\cos \theta} \equiv \frac {\cos 3 \theta \cos \theta + \sin 3 \theta \sin \theta} {\sin \theta \cos \theta}

Then using \cos(A-B) \equiv \cos A \cos B + \sin A \sin B and \sin 2 \theta \equiv 2 \sin \theta \cos \theta:

\displaystyle \frac{\cos 3 \theta} {\sin \theta} + \frac {\sin 3 \theta} {\cos \theta} \equiv \frac {\cos (3 \theta - \theta)} {\frac 1 2 \sin 2 \theta} \equiv 2\frac {\cos 2 \theta} {\sin 2 \theta} \equiv 2 \cot 2 \theta
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SkylineS4
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#7
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#7
the stopping distance was greater. I don't think he accounted for the thinking distance.
(Original post by RedGiant)
Post answers below:

(credits to @radishcoffee)

1. y = -3/4 - x/2
2. runway = 415m (overestimate)
3. sector = 8.73 cm^2
4. (5sqrt2, -4)
5. integral sqrt x = 38/3
6. gg(0) = 13, range - x>35/4, x<2-3rt3, for inverse x=29/4
7. 369 soaps = profit
8. first sum of 20(1/2)^r from 4 to infinity = 5/2, proof of sum of log = 2
9. n=2.08, stopping distance = 98.25m (stop before puddle)
10. vectors - show lamda = 4/3 (because no a component)
11. turning point = 1/e, x3 = 1.673, cobweb diagram that diverge from root alpha, then oscillate between 1 and 2
12. 103.3 degrees for trig
13. minimum radius = 1.05m, surface area = 17m^2
14. range - 0<h<16 (technically have <=0, but 0 height makes no sense in 3D), time for 12m is 75.4 years (rounding early)
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_gcx
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(Original post by SkylineS4)
the stopping distance was greater. I don't think he accounted for the thinking distance.
No the time to stop after the brake was applied was 80 something m and the thinking distance was like 40/3 = 13.66... m.
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B0redBrioche
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#9
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#9
Thats sooo much easier, I still managed to do it but used the addition formula at the start too cos(2theta+theta) and sin(2theta+theta) which made it quite messy
(Original post by _gcx)
People seemed to struggle on the trig proof:

\displaystyle \frac{\cos 3 \theta} {\sin \theta} + \frac {\sin 3 \theta} {\cos \theta} \equiv \frac {\cos 3 \theta \cos \theta + \sin 3 \theta \sin \theta} {\sin \theta \cos \theta}

Then using \cos(A-B) \equiv \cos A \cos B + \sin A \sin B and \sin 2 \theta \equiv 2 \sin \theta \cos \theta:

\displaystyle \frac{\cos 3 \theta} {\sin \theta} + \frac {\sin 3 \theta} {\cos \theta} \equiv \frac {\cos (3 \theta - \theta)} {\frac 1 2 \sin 2 \theta} \equiv 2\frac {\cos 2 \theta} {\sin 2 \theta} \equiv 2 \cot 2 \theta
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lucasjourdan1
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#10
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#10
i got 420 for the jet question using trap. rule, might be wrong tho
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Htn_02
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#11
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#11
How was sum to infinity 1.05?
Taking the 4th value as a, then r= 1/2?
I got 2.5?
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Waynetrain17
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#12
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#12
Same here
(Original post by Htn_02)
How was sum to infinity 1.05?
Taking the 4th value as a, then r= 1/2?
I got 2.5?
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SkylineS4
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I'm confused I thought that was the speed of the car...
(Original post by _gcx)
No the time to stop after the brake was applied was 80 something m and the thinking distance was like 40/3 = 13.66... m.
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Heidi-Heather
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#14
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(Original post by Htn_02)
How was sum to infinity 1.05?
Taking the 4th value as a, then r= 1/2?
I got 2.5?
In my comment I got two of the values mixed up - edited now.
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Popey12346
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#15
Does anyone have the paper, to see the answers with the questions?
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sachinihimara
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#16
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#16
Loves that paper so much I've decided to do it again next year
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_gcx
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(Original post by SkylineS4)
I'm confused I thought that was the speed of the car...
Speed of the car was 50/3 and he had been travelling 0.8 seconds before braking, so the thinking distance was 50/3 * 0.8 = 40/3
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_gcx
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#18
(Original post by Htn_02)
How was sum to infinity 1.05?
Taking the 4th value as a, then r= 1/2?
I got 2.5?
It was 2.5, I checked on my calculator summing from 4 to 100.
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NPT
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#19
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#19
Post your answers
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matlad1
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#20
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#20
(5root2,-4) for parametric one
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