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Original post by Lay-Z
I'm normally good at trig aswell, how do you do that one with 4x^3-11x-3=0

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You used the identity (1-2sin^2) for cos2A. Then multiplied out the bracket and then simplified to give you -4sin^3+11sin+3=0 Then take it to the other side. to give 4sin^3-11sin-3=0 and then sub sin=x to give 4x^3+11x+3=0
Reply 501
Original post by LewisRichardson
You use your first part by splitting the f(x) into 3 factors equalling 0, then replace x with sinO.


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Werent it two factors (2x+3) then (x^2+ax+c)?
Reply 502
I think I dropped at least 20 marks. Hardest paper I've ever done.
Reply 503
Original post by LewisRichardson
You use your first part by splitting the f(x) into 3 factors equalling 0, then replace x with sinO.


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I'm not so sure.. If i remember correctly you actually had to use the double angle theory for sin2A and then expand that out. Then substitute cos^2A for 1-sin^a etc...
It was a longggg paper! I didn't finish it.. Did anyone else??


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Yeah then use quadratic to give the other 2 and only one of the numbers gave solutions
Original post by ItsLewis
I'm not so sure.. If i remember correctly you actually had to use the double angle theory for sin2A and then expand that out. Then substitute cos^2A for 1-sin^a etc...


Shows how well I did haha


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Reply 507
Original post by georgeherby
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Yeah then use quadratic to give the other 2 and only one of the numbers gave solutions


Yup what i did.
I can't even join in on the 'what did you get/this is what I got' discussions because I feel I did so badly on it. Loads of questions I worked through and all was hunky dory until the final line and I'd think, "Hmmm, this doesn't look right...",

I guess I'll throw some answers out just with the off-chance I was correct...

For the cartesian equation of the parametric curve... xy+4y-4x = 0 ???? (I managed to work through to get the correct equation, but the '=0' can't be right. Maybe I did a sign error somewhere. If so, then it might be =32... or =16?) Anyway I got =0...

For the 2/15(k=root5)... k=5, but it was really odd to work out and I'll be honest, my calculator can answer in surds so I kind of jumped from one line to randomly having the answer.

For the vector question, part c, I also got each co-ordinate as being something/17.
I didn't even look at part d.

For the final question....... First part, I did integration by parts to get something like tcos(tpi/4) + 16/pi^2 sin(tpi.4)
And for the next part I used that to do the various integrations. Only problem is my '+c' came out as 256-16/pi^2, which is different to what other people got.
And then when I plugged it all back into the original equation for when t=45... I forgot to add my +c into it!!! So my final answer was negative. I then realised I'd done it wrong but only had about 15 seconds left, so instead just wrote a note saying, "argh, forgot the +c, boohoo." Haha.

To conclude:

Hardest C4 test I've ever done, by a very long shot. And I did about ten of them over the last few days.

I predict a B or C for me. Lucky for me I've done alright in everything else so far, so I'm still on track for getting an A. Woohoo. Must admit though, I was sort of getting my hopes up for an A*, but after that ordeal, no way!
Reply 509
Original post by Qwob
That was not at all fun.

I got full UMS on the 4 most recent past papers, and seeing as the standard grades of papers should be equal, I hope the grade boundaries are low. Maybe 63-65 for full UMS? 63 was full UMS for Jan 2013 remember.

Vectors: anyone else get D (5, 1, 2)? or something similar

I couldn't find the coordinates of E though. I got some massive quadratic involving mu, where mu was like 2.3821 and -2.5832. I kind of gave up on that one. Hope I get some method marks though for working out the area of the smaller triangle ( (sqrt1802)/2) anyone?

Differential question was deceivingly tricky. It looked almost too simple. I'm not sure. As it said write down the differential equation I assumed we did not have to do any working out for that part.

dh/dt = a.cos(kt) right?

then max. when a.cos(kt) = 1.3

when kt=0, a = 1.3

then I just guessed k = 12? As it occurs once every 12 hours.

The last question I got something like 364~ cm, give or take 5cm.

Also question 2d I found difficult, only realised after the exam you could draw triangles to find sin(beta) and cos(beta). Hoping here I get 2 marks for method though.

Overall I think I've dropped around 8-10 marks. By far my lowest mark out of all the past papers I did.


Yeah you got the last question right... i forgot to divide by 16 before square rooting it, such an idiot! but i got K as pi/24, not sure if that's right though!.
Reply 510
Yeah, the Cartesian equation one was definitely = 32
Reply 511
Original post by ItsLewis
*your


Correct my grammar? I'll caress your grandma!
Original post by Lay-Z
I'm normally good at trig aswell, how do you do that one with 4x^3-11x-3=0

Posted from TSR Mobile


I don't know if this is right but this is what I did. (I bet it's wrong)
In the previous question you had already factorised 4x^3-11x-3=0 to get one linear equation multiplied by a quadratic equation, which both equal 0.
I used to quadratic formula to solve the quadratic equation.

I only got 2 values of theta though :s-smilie:
Reply 513
Cardesian Q

K was 32?

And point Q was (-50,0)??
To be fair this paper was a piece of **** and we all agree on that. Did anyone else run out of space for the first question :/
Original post by MSI_10
Cardesian Q

K was 32?

And point Q was (-50,0)??


I got that :wink:
Reply 516
Original post by georgeherby
i think you had 256 but you had to subtract 16 as the sin when to zero but the cos was 16 so C=240


it was 4tsin((pi/4)t) if t = 0 then it is 4(0) lots of the expression which is = 0 therefore you dont subtract 16 i got the c as 248 may have mad a mistake though
Reply 517
stuart, I also 256-16/pi^2 for c. Hoping this is the exact form of c, as that =254.4...

Can anyone remember the differential equation?

Was it dx/dt = (tcos(pi/4 t)/2x

?

Trying to redo my answer, but not sure if I've remembered the equation correctly.
Original post by georgeherby
You used the identity (1-2sin^2) for cos2A. Then multiplied out the bracket and then simplified to give you -4sin^3+11sin+3=0 Then take it to the other side. to give 4sin^3-11sin-3=0 and then sub sin=x to give 4x^3+11x+3=0


Hey could you tell me if the last sub part of vectors Q was 4 or 6 marks?
Original post by Mr Clingfilm
To be fair this paper was a piece of **** and we all agree on that. Did anyone else run out of space for the first question :/


I did and had to wait for ages for them to give me some :/ What did everyone get as the answer to the last question on the first page??? Out of everyone I've spoken to it seems like I was the only one who did that question...

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