[Wa 007]

(Original post by Arsey)
Lower than I thought

100UMS = 75
90UMS = 68
80UMS = 61
70UMS = 54
60UMS = 48
50UMS = 42
40UMS = 36

Thats not too bad to be honest, anyone else nervous about tomorrow?

[Helghast]

(Original post by Wa 007)
Thats not too bad to be honest, anyone else nervous about tomorrow?

No, I got mine like 14 hours ago.

[Mouth]

(Original post by Helghast)
No, I got mine like 14 hours ago.

How did you do?

[Machoo]

97 ums. Boom.

[Mr Tough]

100 UMS. Bang.

[Putch1]

Hi,

Does anyone have the question paper as pdf? The reason im asking is because i wanted to print it and try it myself. (And yes, i do know the pdf is in the first post, but the solutions are already written in the paper itself)

Any help will be appreciated.

[raheem94]

(Original post by Putch1)
Hi,

Does anyone have the question paper as pdf? The reason im asking is because i wanted to print it and try it myself. (And yes, i do know the pdf is in the first post, but the solutions are already written in the paper itself)

Any help will be appreciated.

The pdf file of the original paper, without answers, is attached with this post.

[Putch1]

(Original post by raheem94)
The pdf file of the original paper, without answers, is attached with this post.

Thanks raheem ^_^

[raheem94]

(Original post by Putch1)
Thanks raheem ^_^

You are welcome.

If you need anymore Jan 2012 past paper or mark scheme, feel free to contact me.

[amz1209]

i dont understand 9ii. I got up to the formation of the equations b/a=pie/10 and b/a=3pie/5-pie/a. how do i solve the two. thanks so much

[Aahmsil]

(Original post by amz1209)
i dont understand 9ii. I got up to the formation of the equations b/a=pie/10 and b/a=3pie/5-pie/a. how do i solve the two. thanks so much

Well first off, we know that sin(ax - b) = 0
So, ax - b = 0, Pie, 2Pie
Take 'b' to the other side
ax = b, Pie + b, 2Pie + b
Divide the equation by 'a'
x = b/a, (Pie + b)/a, (2Pie + b)/a

From the question: x co-ordinates = Pie/10, 3Pie/5, 11Pie/10

So, b/a = Pie/10
And b = a*Pie/10

(Pie + b)/a can be written as: Pie/a + b/a
This is equivalent to x co-ordinate 3Pie/5
So, Pie/a + b/a = 3Pie/5
Substitute Pie/10 in place of b/a
Pie/a + Pie/10 = 3Pie/5
Pie/a = 3Pie/5 - Pie/10
Pie/a = 0.5Pie
a = Pie/0.5Pie
a = 2

We know b = a*Pie/10
b = 2*Pie/10
b = Pie/5

Hence, solutions are a = 2 and b = Pie/5
Hope you got what I meant.
Cheers!

[amz1209]

(Original post by Aahmsil)
Well first off, we know that sin(ax - b) = 0
So, ax - b = 0, Pie, 2Pie
Take 'b' to the other side
ax = b, Pie + b, 2Pie + b
Divide the equation by 'a'
x = b/a, (Pie + b)/a, (2Pie + b)/a

From the question: x co-ordinates = Pie/10, 3Pie/5, 11Pie/10

So, b/a = Pie/10
And b = a*Pie/10

(Pie + b)/a can be written as: Pie/a + b/a
This is equivalent to x co-ordinate 3Pie/5
So, Pie/a + b/a = 3Pie/5
Substitute Pie/10 in place of b/a
Pie/a + Pie/10 = 3Pie/5
Pie/a = 3Pie/5 - Pie/10
Pie/a = 0.5Pie
a = Pie/0.5Pie
a = 2

We know b = a*Pie/10
b = 2*Pie/10
b = Pie/5

Hence, solutions are a = 2 and b = Pie/5
Hope you got what I meant.
Cheers!

Thanks so much, really well explained.

[Aahmsil]

(Original post by amz1209)
Thanks so much, really well explained.

My pleasure, just glad I could be of help!

[deniu3545]

(Original post by Arsey)
I will create two types of thread a pre-exam and post exam thread.

Pre-exam threads understandably need to be locked, but it seem to take ages for them to be unlocked.

Here are my model answers, written on the paper. I thought it was a reasonable paper in comparison to previous ones.

Question analysis

Q1. GP, very easy (6m)

Q2. Circles, Joke (4m)

Q3. Binomial, pretty easy (7m)

Q4. Logs, I thought this was easy, but I can see it causing problems. (6m)

Q5. Factor theorem (6m)

Q6. Trap rule, calculus. Trap rule was very easy, integration was pretty easy too (11m)

Q7. Radians. I can see this causing problems (12m)

Q8. Calculus. Not too bad but it will cause problems. (13m)

Q9. a) was pretty easy but b) was very hard (8m)


Q1-Q6 should be bankers, so 45 marks should be nailed on.

Q7 and Q8, whilst not very hard they will cause probs and are worth a third of the paper.

It was 65 for an A last Jan, I would say this paper was easier on the whole!

I am going to go for

100UMS = 75
90UMS = 69.5
80UMS = 64
70UMS = 56
60UMS = 48
50UMS = 40
40UMS = 32

Can anyone please tell where i find all the ARSEY model C2 exam papers (like this paper, for all the past papers!) like he did for S1. thanks !

[gdunne42]

(Original post by matthurry)
Can anyone please tell where i find all the ARSEY model C2 exam papers (like this paper, for all the past papers!) like he did for S1. thanks !

Use the search function

or

Click on his name, then click on the MyStats tab and view all threads started by him

[nm786]

(Original post by Arsey)
I will create two types of thread a pre-exam and post exam thread.

Pre-exam threads understandably need to be locked, but it seem to take ages for them to be unlocked.

Here are my model answers, written on the paper. I thought it was a reasonable paper in comparison to previous ones.

Question analysis

Q1. GP, very easy (6m)

Q2. Circles, Joke (4m)

Q3. Binomial, pretty easy (7m)

Q4. Logs, I thought this was easy, but I can see it causing problems. (6m)

Q5. Factor theorem (6m)

Q6. Trap rule, calculus. Trap rule was very easy, integration was pretty easy too (11m)

Q7. Radians. I can see this causing problems (12m)

Q8. Calculus. Not too bad but it will cause problems. (13m)

Q9. a) was pretty easy but b) was very hard (8m)


Q1-Q6 should be bankers, so 45 marks should be nailed on.

Q7 and Q8, whilst not very hard they will cause probs and are worth a third of the paper.

It was 65 for an A last Jan, I would say this paper was easier on the whole!

I am going to go for

100UMS = 75
90UMS = 69.5
80UMS = 64
70UMS = 56
60UMS = 48
50UMS = 40
40UMS = 32

Hi,
could you please tell me how to do question 7c and 9ii from this C2 (jan 2012) paper: http://www.thestudentroom.co.uk/atta...7&d=1334340533
i don't know how to do it
and could you also please help me on question 7c and d on this M1 (jan 2012) paper: http://www.thestudentroom.co.uk/atta...1&d=1334340533
i don't know how to do this too
please help me,
thanks mate.

[Aahmsil]

(Original post by nm786)
Hi,
could you please tell me how to do question 7c and 9ii from this C2 (jan 2012) paper: http://www.thestudentroom.co.uk/atta...7&d=1334340533
i don't know how to do it
and could you also please help me on question 7c and d on this M1 (jan 2012) paper: http://www.thestudentroom.co.uk/atta...1&d=1334340533
i don't know how to do this too
please help me,
thanks mate.

Here's how I did question 9(ii)

Well first off, we know that sin(ax - b) = 0
So, ax - b = 0, Pie, 2Pie
Take 'b' to the other side
ax = b, Pie + b, 2Pie + b
Divide the equation by 'a'
x = b/a, (Pie + b)/a, (2Pie + b)/a

From the question: x co-ordinates = Pie/10, 3Pie/5, 11Pie/10

So, b/a = Pie/10
And b = a*Pie/10

(Pie + b)/a can be written as: Pie/a + b/a
This is equivalent to x co-ordinate 3Pie/5
So, Pie/a + b/a = 3Pie/5
Substitute Pie/10 in place of b/a
Pie/a + Pie/10 = 3Pie/5
Pie/a = 3Pie/5 - Pie/10
Pie/a = 0.5Pie
a = Pie/0.5Pie
a = 2

We know b = a*Pie/10
b = 2*Pie/10
b = Pie/5

Hence, solutions are a = 2 and b = Pie/5
Hope you got what I meant.
Cheers!

[dslc]

(Original post by nm786)
Hi,
could you please tell me how to do question 7c and 9ii from this C2 (jan 2012) paper: http://www.thestudentroom.co.uk/atta...7&d=1334340533
i don't know how to do it
and could you also please help me on question 7c and d on this M1 (jan 2012) paper: http://www.thestudentroom.co.uk/atta...1&d=1334340533
i don't know how to do this too
please help me,
thanks mate.

See arsey's working here for the Jan 12 paper:
https://docs.google.com/viewer?a=v&q...gDsMJUxOGdfB3A

(7c) is not a nice Q at all (until you assign an 'x'!)

[nm786]

(Original post by Aahmsil)
Here's how I did question 9(ii)

Well first off, we know that sin(ax - b) = 0
So, ax - b = 0, Pie, 2Pie
Take 'b' to the other side
ax = b, Pie + b, 2Pie + b
Divide the equation by 'a'
x = b/a, (Pie + b)/a, (2Pie + b)/a

From the question: x co-ordinates = Pie/10, 3Pie/5, 11Pie/10

So, b/a = Pie/10
And b = a*Pie/10

(Pie + b)/a can be written as: Pie/a + b/a
This is equivalent to x co-ordinate 3Pie/5
So, Pie/a + b/a = 3Pie/5
Substitute Pie/10 in place of b/a
Pie/a + Pie/10 = 3Pie/5
Pie/a = 3Pie/5 - Pie/10
Pie/a = 0.5Pie
a = Pie/0.5Pie
a = 2

We know b = a*Pie/10
b = 2*Pie/10
b = Pie/5

Hence, solutions are a = 2 and b = Pie/5
Hope you got what I meant.
Cheers!

sorry mate, i don't get this method, the third method by arsey looks easier to understand though, but thanks anyway

(Original post by dslc)
See arsey's working here for the Jan 12 paper:
https://docs.google.com/viewer?a=v&q...gDsMJUxOGdfB3A

(7c) is not a nice Q at all (until you assign an 'x'!)

thanks, third method was quite easy to understand for 9ii

[nm786]

(Original post by Aahmsil)
Here's how I did question 9(ii)

Well first off, we know that sin(ax - b) = 0
So, ax - b = 0, Pie, 2Pie
Take 'b' to the other side
ax = b, Pie + b, 2Pie + b
Divide the equation by 'a'
x = b/a, (Pie + b)/a, (2Pie + b)/a

From the question: x co-ordinates = Pie/10, 3Pie/5, 11Pie/10

So, b/a = Pie/10
And b = a*Pie/10

(Pie + b)/a can be written as: Pie/a + b/a
This is equivalent to x co-ordinate 3Pie/5
So, Pie/a + b/a = 3Pie/5
Substitute Pie/10 in place of b/a
Pie/a + Pie/10 = 3Pie/5
Pie/a = 3Pie/5 - Pie/10
Pie/a = 0.5Pie
a = Pie/0.5Pie
a = 2

We know b = a*Pie/10
b = 2*Pie/10
b = Pie/5

Hence, solutions are a = 2 and b = Pie/5
Hope you got what I meant.
Cheers!

could you please help me on question 7c and d on this M1 (jan 2012) paper: http://www.thestudentroom.co.uk/atta...1&d=1334340533 i don't understand how to do it.