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

Did anyone else think the sum/integral one was really badly put. Like I'm sure it wanted the integral but technically the limit is equal to zero.

last part Q14

in the end, dy/dx = y(lnx+1), y is the initial equation so you equal y and lnx+1 to 0. In the end, the answer was e^-1

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

Gg(0)=10 no?

this paper was harder than the first one, but I did better due to actually having proper rest before the exam

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) [5 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. [2 mark]
c) Interpret the value of 0.86 [1 mark]
d) 369 soaps were required, to make a profit [3 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 [4 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 [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 [4 marks]
b) Show that there is a change of sign [2 marks]
c) x3 = 1.673, 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 (thanks @ThE MaRkSmAn):

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

I agree. I got one hundred and fourty or something like that

I agree. I got one hundred and fourty or something like that

Yep, and it might've cost me the A* as well - I ended up getting 0 for the question as I was thinking algebraically rather than logically.

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) [5 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. [2 mark]
c) Interpret the value of 0.86 [1 mark]
d) 369 soaps were required, to make a profit [3 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 [4 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 [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 [4 marks]
b) Show that there is a change of sign [2 marks]
c) x3 = 1.673, 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 (thanks @ThE MaRkSmAn):

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

g(0) was 5, which is greater than 2, so 4(5) - 7 = 1