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# AQA A Level Maths MFP2 Further Pure 2 Unofficial Mark Scheme 22 June 2018 watch

1. I'll start it below
2. Q1

(a) Show that , where . [2 marks]

(b) Hence use the method of differences to find . [3 marks]

By method of differences,

So .

Q2

can be expressed as with r &amp;amp;gt; 0 and .

(a) Show that . [2 marks]

(a) (ii) Find . [1 mark]

Q3 Prove by induction that if and [6 marks]

Q4

(a) Express in terms of . [3 marks]

(b)Find in terms of . [4 marks]

Q5

(a) On the Argand diagram, sketch the locus of points satisfying . [3 marks]

On the Argand diagram, a perpendicular bisector of the line connecting the points on the Argand diagram of (2, 0) and (0, -4)

(b) The complex number is such that and is at its maximum value. [4 marks]

Using the Argand diagram, a more geometrical or algebraic approach can be used

Q6

(a) Sketch the graph of . [2 marks]

This can be obtained off a graphical calculator

(b) Show that . [3 marks]

, since by the sketch so

(c) Show that has one stationary point and find its y-coordinate. [5 marks]

At stationary points, dy/dx = 0, , but and so

Q7

,

(a) Show that the surface area is [4 marks]

dx/dt = 2 sin 2t, dy/dt = 4 cos t

(b) Show that the surface area is [5 marks]

Q8

(a) (i) [1 mark]

5/2

(a) (ii) [1 mark]

-3/2

(a) (iii) [2 marks]

(b) (i) , show that [4 marks]

(b) (ii) [4 marks]

so and

So

Q9

(a) Use de Moivre's theorem to show that . [5 marks]

de Moire's theorem for :

Applying the binomial expansion to de Moivre's theorem, and using the identity gives

(b) (i) , and , find the possible values of . [2 marks]

Using the de Moivre's theorem result, , where

where ,

(b) (ii) Deduce the value of and show that . [4 marks]

, i.e.

, , and , so we can ignore this value

is decreasing for and so , and so

3. (Original post by Integer123)
Q1

(a) Show that , where . [2 marks]

(b) Hence use the method of differences to find . [3 marks]

By method of differences,

So .

Q8

(a) (i) [1 mark]

5/2

(a) (ii) [1 mark]

-3/2

(a) (iii) [2 marks]

(b) (i) , show that [4 marks]

(b) (ii) [4 marks]

so and

So
Fam where you applied? And got offers ?
4. I flopped so bad ffs
5. Ahahah question 8bii is so easy when you spot it but of course I couldn't work it out in the exam
6. You are honestly a lifesaver with these thank you !!
7. i thought that was quite a damn good exam. Pretty sure I got everything apart from the last part of the cubic question. I was doing it correctly but at the end for some reason my brain went oh the roots are 1/sqr root alphabeta and gamma ffs haha. Anyway hopefully it can still be 100 ums depending on boundaries.
8. Can I have your brain pls
(Original post by Anonymouspsych)
i thought that was quite a damn good exam. Pretty sure I got everything apart from the last part of the cubic question. I was doing it correctly but at the end for some reason my brain went oh the roots are 1/sqr root alphabeta and gamma ffs haha. Anyway hopefully it can still be 100 ums depending on boundaries.
9. (Original post by Integer123)
Q1

(a) Show that , where . [2 marks]

(b) Hence use the method of differences to find . [3 marks]

By method of differences,

So .

Q2

can be expressed as with r > 0 and .

(a) Show that . [2 marks]

(a) (ii) Find . [1 mark]

Q3 Prove by induction that if and [6 marks]

Q4

(a) Express in terms of . [3 marks]

(b) Find in terms of . [4 marks]

Q5

(a) On the Argand diagram, sketch the locus of points satisfying . [3 marks]

On the Argand diagram, a perpendicular bisector of the line connecting the points on the Argand diagram of (2, 0) and (0, -4)

(b) The complex number is such that and is at its maximum value. [4 marks]

Using the Argand diagram, a more geometrical or algebraic approach can be used

Q8

(a) (i) [1 mark]

5/2

(a) (ii) [1 mark]

-3/2

(a) (iii) [2 marks]

(b) (i) , show that [4 marks]

(b) (ii) [4 marks]

so and

So

Q9

(a) Use de Moivre's theorem to show that . [5 marks]

de Moire's theorem for :

Applying the binomial expansion to de Moivre's theorem, and using the identity gives

(b) (i) , and , find the possible values of . [2 marks]

Using the de Moivre's theorem result, , where

where ,

(b) (ii) Deduce the value of and show that . [4 marks]

, i.e.

, , and , so we can ignore this value

is decreasing for and so , and so

for 4a I didn't notice it saying give it in terms of cosh x so i left it as 2+2cosh2x but in part b i converted this into 4cosh^2x during the integration. Does that mean I'll lose a mark
11. (Original post by dhchgch)
I'm sure you did better than you thought! A lot of people skipped some of the 4 markers (including me), so already everyone's lowered the grade boundaries for this year😂
12. (Original post by jess2157)
I'm sure you did better than you thought! A lot of people skipped some of the 4 markers (including me), so already everyone's lowered the grade boundaries for this year😂
do u remember what answer(s) you put for question 6 and 7? Like the last parts of them
13. Anyone got any rough ideas on grade boundaries?
14. (Original post by Anonymouspsych)
do u remember what answer(s) you put for question 6 and 7? Like the last parts of them
For question 6 my answer was weird. I tried over and over but couldn't fix it. I got that coshy = +-5/12, but then I couldn't do the inverse of either of those answers soooooooo pffftt I don't know about that.
For question 7, my final answer was π/3[64sqrt(2)-32].
15. (Original post by fos99e)
Anyone got any rough ideas on grade boundaries?
I'm guessing they'll be pretty similar to last year, if not a couple of marks lower just because some of the 4 markers were quite hard.
Last year's grade boundaries were this by the way:

58/52/45/38 for A*/A/B/C...
16. (Original post by jess2157)
For question 6 my answer was weird. I tried over and over but couldn't fix it. I got that coshy = +-5/12, but then I couldn't do the inverse of either of those answers soooooooo pffftt I don't know about that.
For question 7, my final answer was π/3[64sqrt(2)-32].
yess i think i put down the same as you for q 7 and yea for q6 you get coshy = +-5/12 but you need to reject the negative solution as coshy>=1
17. (Original post by Anonymouspsych)
yess i think i put down the same as you for q 7 and yea for q6 you get coshy = +-5/12 but you need to reject the negative solution as coshy>=1 so u just use y=arccosh(5/12) = ln2 i believe.
But I don't understand ughhhhhhhh
For question 6, I got that coshy = +-5/12 and rejected the negative (because of the graph). But when I tried to do cosh-1(5/12), I couldn't get it on my calculator or by using the formula... I don't know where I went wrong haha!
18. (Original post by jess2157)
But I don't understand ughhhhhhhh
For question 6, I got that coshy = +-5/12 and rejected the negative (because of the graph). But when I tried to do cosh-1(5/12), I couldn't get it on my calculator or by using the formula... I don't know where I went wrong haha!
Oh my bad it wouldn't be coshy=5/12 as obviously coshy can't be less than 1. I did get an answer with ln something but i can't remember lol do you remember the question xD?
19. (Original post by Anonymouspsych)
Oh my bad it wouldn't be coshy=5/12 as obviously coshy can't be less than 1. I did get an answer with ln something but i can't remember lol do you remember the question xD?
Ugh dammit XD I just tried your answer (ln2) it came out with cosh(ln2) = 5/4, if that looks familiar?
The first part said you had to prove the differential of cosh-1x, which I got. In the end, you had to find the y-coordinate of the stationary point for y = (a constant) + (a constant)x - cosh-1(3x) or something, but that's all I remember
20. (Original post by jess2157)
Ugh dammit XD I just tried your answer (ln2) it came out with cosh(ln2) = 5/4, if that looks familiar?
The first part said you had to prove the differential of cosh-1x, which I got. In the end, you had to find the y-coordinate of the stationary point for y = (a constant) + (a constant)x - cosh-1(3x) or something, but that's all I remember
yesss ah i remember oh you're gonna be annoyed when you see this. so you worked out x correctly to be 5/12 but it was cosh-1(3x) not cosh-1(x)
so you are actually evaluating cosh-1(3*5/12) =cosh-1(5/4) and since 5/4>1 this will work with the cosh-1(x) formula with ln.

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