[mathslover1]

(Original post by SamHLFC)
I got my complex number z to be z= -3 -2i. So I got that wrong:/ Also forgot to put my p values back into the equation to find the x value. Think I got the matrices, except the M^2004 question. What did everyone get for the Newton Raphson method?

0.1934

[bratwast]

(Original post by ryanboi)
Mate you can use either your method or the 360degree + alpha one. Both gives the right answer, just different way of showing it. You won't get mark off for using different general solutions.

yeah I thought not, cheers bro

[SamHLFC]

(Original post by mathslover1)
0.1934

Good, at least I got that right. I got -31 for the gradient question, asymptotes y=0, x=1 and another x asymptote. Got interval 1.175<x<2. Think I dropped 4 on the discriminant question. Think I dropped 3 on complex numbers and 3 on the sum & product somehow

[kirstyy93]

(Original post by croasdaj)
The general solution question, you had to do sin-1 of (cos20) to get 70. Then just work from there like normal.
You get x=-540n
x=-540n -60
our teacher went through this so im pretty sure its right

Ahh that's what I got, but I stupidly added 2npi instead of 360 degrees so I ended up with x=-3npi and x=-60-3npi

How many marks do you reckon I would lose? Did your teacher give you a rough idea of marks etc?

[Nebula]

Here are my solutions. The questions are probably in the wrong order. If people can correct my question order/ typos/ incorrect answers it would be appreciated.

x = (105-270n-(105x(-1)^n))

This is correct, just messy (IMO). I don't understand why anyone would think this is somehow better than just using two equations.

[kirstyy93]

(Original post by Nebula)
Here are my solutions. The questions are probably in the wrong order. If people can correct my question order/ typos/ incorrect answers it would be appreciated.

This is correct, just messy (IMO). I don't understand why anyone would think this is somehow better than just using two equations.

I got pretty much the same apart from the matrices question, I reckon I've dropped 4 marks on that.

Also, on the trig question I stupidly added 2npi nested of 360 degrees, I still got the same answer only it wasn't completely in degrees :/ any ideas how many marks I will lose?

[Miyata]

(Original post by Nebula)
Here are my solutions. The questions are probably in the wrong order. If people can correct my question order/ typos/ incorrect answers it would be appreciated.

This is correct, just messy (IMO). I don't understand why anyone would think this is somehow better than just using two equations.

I just did on wolfram, that trig question definitely had alpha as 70 :// (too bad for me). I also got the solving inequality wrong, if the method is correct in your work. Was it done by the teacher? I got something like:
x/(x+1)(x-2) + 0.5 *((x+1)(x-2)/(x+1)(x-2)) =< 0
simplifying to:
x/(x+1)(x-2) + ((x+1)(x-2)/2(x+1)(x-2)) =< 0
x/(x+1)(x-2) + 1/(x+1)(x-2) =< 0
x+1/(x+1)(x-2) =< 0
so Critical values: x= -1* (asymptote), x = 2
Drew a sign diagram, were values below -1 gives negative, values above 2 give positive, and values below 2 and above -1 are negative, so for the solution I wrote
x < -1, and -1< x =< 2
Is that right or wrong? If wrong how many marks would I get?

And for trig question, I used correct method, and everything, just had used wrong alpha value so I got -540n with different angles, out of how 6 marks how many would I get? because when looking at mark schemes, they always a lot of marks like 4 marks even if its wrong solution, and then 2 marks for the final correct solution.

[smallfish008]

I was surprised when the paper didn't ask any question on summation or linear form. But the paper was quite nice.

[bratwast]

I'm guessing the mark scheme is going to be tight. 65/75 for an A maybe

[Miyata]

(Original post by bratwast)
I'm guessing the mark scheme is going to be tight. 65/75 for an A maybe

I hope they won't be tight, better for everyone, getting higher UMS.

[bratwast]

yeah but they have to change the ums so that a certain percentage of students in the country get an A (stupid I know).

If not enough people get over the boundary and too many are under they will lower/heighten the boundary accordingly. Seems like nearly everyone found it easy... could be a record breaking low for C2 and a record breaking high for FP1 by the sounds of it. AQA have got messy haha

[James181]

Can anyone give me some reassurance on this?

For the general solution I got x = 540n, x = 540n - 60

I realise I probably should have put -540n, but I thought that since n is any integer the sign wouldn't matter? Do you think I will still get credit for this solution?

Cheers

[Nebula]

(Original post by Miyata)
x/(x+1)(x-2) + 0.5 *((x+1)(x-2)/(x+1)(x-2)) =< 0
simplifying to:
x/(x+1)(x-2) + ((x+1)(x-2)/2(x+1)(x-2)) =< 0
x/(x+1)(x-2) + 1/(x+1)(x-2) =< 0

I have no idea how you did this step. Next steps should be:
2x/2(x+1)(x-2) + (x^2-x-2)/2(x+1)(x-2) ≤ 0
(x^2+x-2)/2(x+1)(x-2)≤ 0
(x+2)(x-1)/(x+1)(x-2)≤ 0
x=+-1 and +-2 are critical values ... -2≤x<-1 1≤x<2.

x+1/(x+1)(x-2) =< 0
so Critical values: x= -1* (asymptote), x = 2
Drew a sign diagram, were values below -1 gives negative, values above 2 give positive, and values below 2 and above -1 are negative, so for the solution I wrote
x < -1, and -1< x =< 2
Is that right or wrong? If wrong how many marks would I get?

I think this is wrong. The question was only 2 or 3 marks (I think) but I don't know what you would get (max 1 or 2 because it's wrong).

As for the trigonometry, not really sure how many marks maybe 4 or 5?

For the inequality Q, i put down -2≤x≤-1 and 1≤x≤2 so i had wrong inequality signs, would that still get me 2 out of 3, do you think?

You would lose 1 mark for this on past mark schemes so would probably only lose one here (If the mistake is repeated it only loses one mark).

[kasiasunshine]

For the inequality Q, i put down -2≤x≤-1 and 1≤x≤2 so i had wrong inequality signs, would that still get me 2 out of 3, do you think?

[Kyyp]

How many marks lost if you did the cyclical part of the general solution in radians and got the correct answer?
Also, 1 method mark for (m^4)^503 = m^2012?

[Marsters]

(Original post by Nebula)
I have no idea how you did this step. Next steps should be:
2x/2(x+1)(x-2) + (x^2-x-2)/2(x+1)(x-2) ≤ 0
(x^2+x-2)/2(x+1)(x-2)≤ 0
(x+2)(x-1)/(x+1)(x-2)≤ 0
x=+-1 and +-2 are critical values ... -2≤x<-1 1≤x<2.

Surely x could not have been equal to -2 or 1 as there were asymptotes at those points.

[mathslover1]

(Original post by kasiasunshine)
For the inequality Q, i put down -2≤x≤-1 and 1≤x≤2 so i had wrong inequality signs, would that still get me 2 out of 3, do you think?

Yeah me too, i got the wrong signs..
Yeah i was acctually thinking waht the 3rd mark might be for, thought it would be the signs?

[mathslover1]

(Original post by Nebula)
Here are my solutions. The questions are probably in the wrong order. If people can correct my question order/ typos/ incorrect answers it would be appreciated.

This is correct, just messy (IMO). I don't understand why anyone would think this is somehow better than just using two equations.

lost a mark acctually really happy (maybe a few more for B1 marks)

[croasdaj]

(Original post by kirstyy93)
Ahh that's what I got, but I stupidly added 2npi instead of 360 degrees so I ended up with x=-3npi and x=-60-3npi

How many marks do you reckon I would lose? Did your teacher give you a rough idea of marks etc?

I've actually seen on a couple of mark schemes were they allow the use of 2npi, but our teacher said probably 2 marks.
You would get marks for working out 2 solutions, re-arranging to get x=...

[Nebula]

(Original post by Marsters)
Surely x could not have been equal to -2 or 1 as there were asymptotes at those points.

no, asymptotes were at x=-1 and x=2.