I can't remember exactly but I know that it didn't simplify 😊(Original post by Smcdonald98)
Can anyone remember what they got for 8a ii, i couldn't get mine to simplify nicely
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CCEA A2 maths  C3 20th may 2016
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 81
 21052016 11:12

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 21052016 11:14
The paper was tough enough but this is what I got for each question, anyone else get the same?
1.45,153
2.1x+3/2(x^2)5/2(x^3)
3. Cant draw the graphs obviously but I'm fairly sure they were right
b. 11/(x^2)2/(x)+2/(x+1)
4.8
5.dy/dx=2x+1+ln(x) and I used the point (1,1) hence dy/dx=3
ii. dy/dx=0 at stationary point, change in sign between x=0.2 and x=0.3 therefore passes through 0 hence stationary point is between those values
iii. 0.232 (but I used it twice instead of once so the answer should have been 0.230. Any idea how many marks I'd lose for this? I've got all the working out for one iteration done above it)
6. 1700 years
7.I tried to prove it was always positive by saying when: x>1 : x^2+2> x therefore positive
x<1: x^2+2> x therefore positive
1<x<1: x^2x< 2 therefore positive hence for any value of x it must be positive. Not sure how many marks that will get me though
ii. dy/dx=2e^(2x)(x^2+x2): 2e^(2x) is always positive and (x^2+x2) is always negative, +ve multiplied by a ve = ve which is less then 0 hence dy/dx<0
8. 4cosx+ 2x/(1x^2)
ii. (2cotxtan3x+3sec^(2)3x)/(sin^2(x)(tan3x)^2)
b. ended up integrating (sec^(2)2x) to get (0.5tan2x) which after subbing in the values gave root 3 and multiplied by the 4/3 outside gave (4(3)^0.5)/3 
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 83
 21052016 11:39
(Original post by will872)
The paper was tough enough but this is what I got for each question, anyone else get the same?
1.45,153
2.1x+3/2(x^2)5/2(x^3)
3. Cant draw the graphs obviously but I'm fairly sure they were right
b. 11/(x^2)2/(x)+2/(x+1)
4.8
5.dy/dx=2x+1+ln(x) and I used the point (1,1) hence dy/dx=3
ii. dy/dx=0 at stationary point, change in sign between x=0.2 and x=0.3 therefore passes through 0 hence stationary point is between those values
iii. 0.232 (but I used it twice instead of once so the answer should have been 0.230. Any idea how many marks I'd lose for this? I've got all the working out for one iteration done above it)
6. 1700 years
7.I tried to prove it was always positive by saying when: x>1 : x^2+2> x therefore positive
x<1: x^2+2> x therefore positive
1<x<1: x^2x< 2 therefore positive hence for any value of x it must be positive. Not sure how many marks that will get me though
ii. dy/dx=2e^(2x)(x^2+x2): 2e^(2x) is always positive and (x^2+x2) is always negative, +ve multiplied by a ve = ve which is less then 0 hence dy/dx<0
8. 4cosx+ 2x/(1x^2)
ii. (2cotxtan3x+3sec^(2)3x)/(sin^2(x)(tan3x)^2)
b. ended up integrating (sec^(2)2x) to get (0.5tan2x) which after subbing in the values gave root 3 and multiplied by the 4/3 outside gave (4(3)^0.5)/3
These are the answers I got for some of the questions and I agree with you mostly 
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 84
 21052016 11:52
(Original post by will872)
5.dy/dx=2x+1+ln(x) and I used the point (1,1) hence dy/dx=3 
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 85
 21052016 11:54
(Original post by Foutre en L'air)
For this question you didn't actually need to sub in a random point and find it's gradient, it was simply asking for a general form of , but as long as you wrote the formula for the gradient you shouldn't lose marks! 
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 86
 21052016 13:17
(Original post by will872)
The paper was tough enough but this is what I got for each question, anyone else get the same?
1.45,153
2.1x+3/2(x^2)5/2(x^3)
3. Cant draw the graphs obviously but I'm fairly sure they were right
b. 11/(x^2)2/(x)+2/(x+1)
4.8
5.dy/dx=2x+1+ln(x) and I used the point (1,1) hence dy/dx=3
ii. dy/dx=0 at stationary point, change in sign between x=0.2 and x=0.3 therefore passes through 0 hence stationary point is between those values
iii. 0.232 (but I used it twice instead of once so the answer should have been 0.230. Any idea how many marks I'd lose for this? I've got all the working out for one iteration done above it)
6. 1700 years
7.I tried to prove it was always positive by saying when: x>1 : x^2+2> x therefore positive
x<1: x^2+2> x therefore positive
1<x<1: x^2x< 2 therefore positive hence for any value of x it must be positive. Not sure how many marks that will get me though
ii. dy/dx=2e^(2x)(x^2+x2): 2e^(2x) is always positive and (x^2+x2) is always negative, +ve multiplied by a ve = ve which is less then 0 hence dy/dx<0
8. 4cosx+ 2x/(1x^2)
ii. (2cotxtan3x+3sec^(2)3x)/(sin^2(x)(tan3x)^2)
b. ended up integrating (sec^(2)2x) to get (0.5tan2x) which after subbing in the values gave root 3 and multiplied by the 4/3 outside gave (4(3)^0.5)/3 
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 87
 21052016 13:27
(Original post by Mr.bob)
I did the same as you for 7. You havnet been talking to your maths teacher by any chance to find out if that was a feasible method for working it out? 
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 88
 21052016 14:29
Anyone have any solutions to put my mind at ease? Especially Q6, 7, 8(iii) thanks

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 89
 21052016 15:42
(Original post by Camderman106)
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 
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 90
 21052016 16:18
(Original post by harrytqo)
How were you meant to work out k?? surely it was just 0.04 
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 91
 21052016 16:27
(Original post by harrytqo)
How were you meant to work out k?? surely it was just 0.04
In 100 years, N = 0.96No
so 0.960No = No x e^kt
0.960 = e^k(100)
using logs will give you the answer to k
k = 0.000408...Last edited by Smcdonald98; 21052016 at 16:28. 
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 92
 21052016 17:22
(Original post by harrytqo)
How were you meant to work out k?? surely it was just 0.04
M=(initialM)e^kt
You cannot just assume k=0.04, you fill in for the initial conditions to get
96=100e^k(100)
So then
96/100=e^100k
Ln(0.96)=100k
(Ln(0.96)/100)=k
K turns out to be 0.00040821994......Then you substitute for the second conditions
50%=100%e^kt
1/2= e^kt
(Ln(1/2))= kt
(Ln(0.5))/k=t
Sub in k
(Ln(0.5))/0.00040821994... = 1697.97482347...
=1700years(3sf)Last edited by Camderman106; 21052016 at 17:23. 
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 93
 21052016 17:29
(Original post by Camderman106)
No it wasn't, let me explain. The initial conditions told you it lost 4% of its mass every 100 years. And the equation is going to be...
M=(initialM)e^kt
You cannot just assume k=0.04, you fill in for the initial conditions to get
96=100e^k(100)
So then
96/100=e^100k
Ln(0.96)=100k
(Ln(0.96)/100)=k
K turns out to be 0.00040821994......Then you substitute for the second conditions
50%=100%e^kt
1/2= e^kt
(Ln(1/2))= kt
(Ln(0.5))/k=t
Sub in k
(Ln(0.5))/0.00040821994... = 1697.97482347...
=1700years(3sf) 
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 94
 21052016 17:39
(Original post by Foutre en L'air)
Answer should probably be to the nearest year by the way 
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 95
 18082016 22:35
I got an A* in maths. So happy after the disaster that was this paper for me, I managed to get 80/100 in c3 but 100/100 in c4.

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 96
 19082016 15:53
Congrats! I got an A* too and hope you get in your firm choice.

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 97
 21082016 01:28
(Original post by 9clpc9)
Congrats! I got an A* too and hope you get in your firm choice. 
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 98
 21082016 18:22
(Original post by Golden hawk)
I did , hopefully you did as well
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Updated: August 21, 2016
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