CCEA A2 maths - C3 20th may 2016

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Well after looking at this again, I know I've done badly now. What was the boundary for an A* last year? Whats the lowest its ever been for an A* as well. This paper was definitely harder than any other c3 paper I've ever done

Reply 61

Golden hawk

Original post by TomFitness

I don't think anyone will be able to tell you, but I'm sure it will be lower than it has ever been purely because not many PP compare to it. All I know was that summer 2014 pp was a 1:1 ratio from % to UMS if that makes since. So 60/75 was an A etc

I know in summer 2014 an A* was 70/75 and in summer 2013 it was 68/75 for an A*. If it's the summer 2014 grade boundary this year I'm screwed. I just hope it's below 67 or below for the A* otherwise i not get into uni

(edited 7 years ago)

Reply 62

Mr.bob

Original post by Golden hawk

just emailed CCEA there now to find out the mark boundaries in terms of raw marks for the last 3 years under the FOI, so here's hoping they reply soon :smile:

Reply 63

ozmo19

Original post by Mr.bob

just emailed CCEA there now to find out the mark boundaries in terms of raw marks for the last 3 years under the FOI, so here's hoping they reply soon :smile:

What email did you use?

Reply 64

Camderman106

Original post by harrytqo

and did your newton rhapson method work? mine didnt

The problem here was what you let f(X) to be. You should have had the equation for the differential and not the initial equations so it is actually f'(X)/f''(X) which is quite sneaky. Once you did this it worked fine

Reply 65

username1763791

Original post by harrytqo

and did your newton rhapson method work? mine didnt

You had to use

\frac{dy}{dx}

f(x)

and then differentiate THAT and use it in your NR formula :smile:

Reply 66

Camderman106

Original post by harrytqo

same^ ln0.5/-0.0004 = 1730 years

That's actually not correct. The correct answer is 1700 not 1730 because k is not 0.04.
You go like this
96=100e^-100k
Ln(96/100) = -100k
K = something near 0.04 but not exactly.
Then use that value of k to find t when
1/2=e^-kt
Ln(1/2)/-k = t
T= 1697.???
=1700(3sf)

Reply 67

username1763791

Original post by hampshir3

exactly this, no way i would have thought of a method to complete it without doing a whole other subject, ccea get your finger out

I don't do physics and I thought it was OK after a little guessing and knowing that they were trying to poke at an exponential question.

Reply 68

Camderman106

I hear people completed the square in q7 but I just differentiated, set that =0 for max/min. Then showed that the minimum occured at X=1/2 and then found the minimum value to be a positive number. Do you think this will be accepted?

Reply 69

username1763791

Original post by Camderman106

I've never seen this method before?

Reply 70

username1763791

Original post by Camderman106

Yeah, thats a valid method. You could have also sketched the graph, or used the discriminant/ solved it to show there were no real roots etc. There were a few different methods and I'm sure they will all be accepted.

Reply 71

Camderman106

Original post by Smcdonald98

What did you guys do for 7 part ii? I differentiated it twice to see if d2y/dx2 was greater than 0 but not sure if thats correct

This one was good. When you differentiate I believe the answer was
dy/dx = 2e^(-2x) X (x-(x^2+2))
The value of 2e(-2x) can never be negative, so all you have to do is prove that (x-(x^2+2)) is always negative. But this multiples out to be
-x^2+x-2<0
Which can be rewritten as
0<x^2+x+2
Which was proven in part 1. Therefore it is always negative because a positive X a negative is always a negative which is your proof

Reply 72

Camderman106

Original post by Foutre en L'air

Great, my method for these questions was slightly umconventional but still technically valid so I'm hoping I get credit haha

Reply 73

Camderman106

Original post by Foutre en L'air

I've never seen this method before?

It's the same method used in every single past paper. The equation for exponential decay is
VALUE=INITIALVALUEe^-kt
V=IVe^-kt
That's the only way I knew the equation is from past paper questions. But you have to figure out k before you can answer it

Reply 74

username1763791

Original post by Camderman106

I knew the whole

M = M_{0}e^{-kt}

thing, but I've never had to find k before, so I just set it as the rate at which the mass was lost..

Reply 75

Camderman106

Original post by Foutre en L'air

I knew the whole

M = M_{0}e^{-kt}

thing, but I've never had to find k before, so I just set it as the rate at which the mass was lost..

Yeah it's kind of a wee trick. It's just because k is not that simple. You do have to calculate k unless it's given. I know a lot of people who did the same thing as you so do t worry you'll still get most of the marks

Reply 76

username1763791

Original post by Camderman106

I've never seen finding k like that in past papers or anything, it's not even in the textbook :frown:

Reply 77

Camderman106

Original post by Foutre en L'air

I've never seen finding k like that in past papers or anything, it's not even in the textbook :frown:

Yeah its a question about applying your understanding and unfortunately its very very sneaky. I only knew that I needed to calculate k because of physics. Many of the questions in this paper were like that though. Q7 had never come up like that before and Q8 combined about 4 different facts into 1 question. It was undoubtedly a hard paper