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

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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

2. a) Use the trapezium rule to calculate the distance of the runway = 415m

b) It is an overestimate, if you consider the shape of the graph.

3. a) You had to state that the student had used the angle in degrees, and not radians

b) Area of sector = 8.73 cm^2

4. The coordinates for the point S was (5sqrt2, -4)

5. Integral of sqrt x = 38/3

6. a) gg(0) = 13

b) The range was: - x>35/4, x<2-3rt3

c) Explain that h(x) was a one to one function, and the other was a many-to-one function (iirc)

d) For the inverse function, x=29/4

7. a) You simply had to show that y = a + kx

b) Solve the simultaneous equations that you could form, and show that k was 0.86.

c) Interpret the value of 0.86

d) 369 soaps were required, to make a profit

8 a) The sum of 20(1/2)^r from 4 to infinity = 5/2

b) Carry out the proof of sum of log = 2

9. a) First take logs, then explain that the gradient was constant, therefore k and n were constant.

b) n=2.08

c) Stopping distance = 98.25m (stop before puddle - you had to add on 40/3, which was the thinking distance)

10. a) Work out the vector OC: 1/2b-3/2a

b) Show lamda = 4/3 (because no a component)

c) Proof

11 a) You had to take logs on both sides, and end up with the turning point at x = 1/e

b) Show that there is a change of sign

c) x3 = 1.673

d) Cobweb diagram that diverge from root alpha, then oscillate between 1 and 2

12 a) Prove the equation

Then using and :

b) 103.3 degrees for trig

13. a) Show that the area is equal to the given expresion

b) Area is minimum when dA/dR = 0, so minimum radius = 1.05m,

c) Surface area = 17m^2

14. a) Integrate the equation

b) Range - 0<h<16 (technically have <=0, but 0 height makes no sense in 3D)

c) Time for 12m is 75.4 years (rounding early, 77 years if you didn't)

These are the marks per question:

Mean mark for this paper (based on poll): 54/100

Lower quartile: 38

Upper quartile: 73

Median: 58

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]:**Then using and :

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)**15Mean mark for this paper (based on poll): 54/100

Lower quartile: 38

Upper quartile: 73

Median: 58

Last edited by RedGiant; 1 year ago

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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%

A* 75%

A 60%

B 50%

C 40%

D 30%

E 20%

Last edited by B0redBrioche; 1 year ago

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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)

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 year ago

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#7

the stopping distance was greater. I don't think he accounted for the thinking distance.

(Original post by

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)

**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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#8

(Original post by

the stopping distance was greater. I don't think he accounted for the thinking distance.

**SkylineS4**)the stopping distance was greater. I don't think he accounted for the thinking distance.

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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

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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?

Taking the 4th value as a, then r= 1/2?

I got 2.5?

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#12

Same here

(Original post by

How was sum to infinity 1.05?

Taking the 4th value as a, then r= 1/2?

I got 2.5?

**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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#13

I'm confused I thought that was the speed of the car...

(Original post by

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.

**_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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#14

**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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#17

(Original post by

I'm confused I thought that was the speed of the car...

**SkylineS4**)I'm confused I thought that was the speed of the car...

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#18

**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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