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AQA FP2 9th June post exam discussion

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slips

the p+13 was the worst bit definitely got that wrong...on the last question think it said to leave it in terms of root 2?
also on question one i got the wq/29 or whatever it was, but used my calculator to do tanh(ln2.5). was another way?

I did tanhx = sinhx/coshx = (e^2x-1)/(e^2x+1) with a bit of manipulation then sub'd the x in.

It was worth 6/7 marks that part so I thought using you're calculator would be too easy!

Reply 21

Martial44

kota

Could anyone remember the marks on each questions??

For instance, 1a - 2 marks etc
1b - 7 marks

Q2 a) 3 marks, b) 5 marks
Q6 a) 2 marks, b) 6 marks

Reply 22

slips

Adam92

I did tanhx = sinhx/coshx = (e^2x-1)/(e^2x+1) with a bit of manipulation then sub'd the x in.

It was worth 6/7 marks that part so I thought using you're calculator would be too easy!

yeah it was worth 7, but i came back to it at the end and didn't have a lot of time so just went with the calculator...i'll def get some marks for all the working before i think

Reply 23

Tomato_Soup1992

Adam92

I did tanhx = sinhx/coshx = (e^2x-1)/(e^2x+1) with a bit of manipulation then sub'd the x in.

It was worth 6/7 marks that part so I thought using you're calculator would be too easy!

I just subbed 5/2 into [(e^x)-(e^-x)]/[(e^x)+(e^-x)]

Reply 24

Adam92

Tomato_Soup1992

I just subbed 5/2 into [(e^x)-(e^-x)]/[(e^x)+(e^-x)]

It's all good.

Reply 25

Martial44

Did anybody else get a = 128 and Q5 being 2 / root 3 ?

Reply 26

SoManyFaces

I got a=8 and for q5, arc length=pi/6

Reply 27

SoManyFaces

Also, for the last question.. How is everyone getting e^17ipi/12 ?
I just tried it again and got 77ipi/12 :|

Reply 28

Martial44

SoManyFaces

I got a=8 and for q5, arc length=pi/6

At least someone got a = 8 :p:

. I was convinced I was wrong not long ago but remembering back I think it is the correct answer, did you have any doubts about a = 8?

Reply 29

2^Oscar

For question 4. b (ii) You needed to use the information in part (i) about substituting alpha into the equation resulting in 0. So if you do the same for beta and gamma you get 0 too. So all you had to do was add together f(alpha), f(beta) and f(gamma) and set it equal to 0, then the answer just dropped out and you could use the sum of the cube of the roots and what not to find a value for the sum of the squares of the roots in terms of p.

Reply 30

slips

i got arc length = pi/6, and checked it on my calculator, so pretty sure that's right

Reply 31

Adam92

SoManyFaces

Also, for the last question.. How is everyone getting e^17ipi/12 ?
I just tried it again and got 77ipi/12 :|

(1 + root3i)^8 = 2^8e^i8pi/3

(1 - i)^5 = root2^5e^-i5pi/4

Timesing the e's together gives you e^(i8pi/3 - i5pi/4) which equals e^i17pi/12

Reply 32

Adam92

slips

i got arc length = pi/6, and checked it on my calculator, so pretty sure that's right

& same, pretty sure it's right.

Reply 33

SoManyFaces

Adam92

(1 + root3i)^8 = 2^8e^i8pi/3

(1 - i)^5 = root2^5e^-i5pi/4

Timesing the e's together gives you e^(i8pi/3 - i5pi/4) which equals e^i17pi/12

Arghhh.. I got the argument of 1-i as 3pi/4. Do you know how many marks the last q was worth?

Reply 34

Adam92

SoManyFaces

Arghhh.. I got the argument of 1-i as 3pi/4. Do you know how many marks the last q was worth?

Part a) 2/3, b) 3, c) 4

Reply 35

ALICEMOON

Original post by &#946

At least someone got a = 8 :p:

. I was convinced I was wrong not long ago but remembering back I think it is the correct answer, did you have any doubts about a = 8?

What question had finding an a??? ahhh

Reply 36

Martial44

ALICEMOON

What question had finding an a??? ahhh

You didnt have to find a, it was question 7 which said to put in the form a root 2 e . I put 128, I made an error in earlier posts. For some reason im adamant it was 128. anyways :p:

Reply 37

ALICEMOON

Original post by &#946

You didnt have to find a, it was question 7 which said to put in the form a root 2 e . I put 128, I made an error in earlier posts. For some reason im adamant it was 128. anyways :p: