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# CCEA A2 maths - C3 20th may 2016

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1. (Original post by Smcdonald98)
Can anyone remember what they got for 8a ii, i couldn't get mine to simplify nicely
I can't remember exactly but I know that it didn't simplify 😊

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2. The paper was tough enough but this is what I got for each question, anyone else get the same?
1.45,153
2.1-x+3/2(x^2)-5/2(x^3)
3. Cant draw the graphs obviously but I'm fairly sure they were right
b. 1-1/(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^2-x< 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+x-2): 2e^(-2x) is always positive and (-x^2+x-2) is always negative, +ve multiplied by a -ve = -ve which is less then 0 hence dy/dx<0
8. 4cosx+ 2x/(1-x^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
3. (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
2.1-x+3/2(x^2)-5/2(x^3)
3. Cant draw the graphs obviously but I'm fairly sure they were right
b. 1-1/(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^2-x< 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+x-2): 2e^(-2x) is always positive and (-x^2+x-2) is always negative, +ve multiplied by a -ve = -ve which is less then 0 hence dy/dx<0
8. 4cosx+ 2x/(1-x^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
Attachment 535615535617535600

These are the answers I got for some of the questions and I agree with you mostly
Attached Images

4. (Original post by will872)
5.dy/dx=2x+1+ln(x) and I used the point (1,1) hence dy/dx=3
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!
5. (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!
Yeah I thought that might have been it but the wording was a bit ambiguous so just incase I decided sub in the values.
6. (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
2.1-x+3/2(x^2)-5/2(x^3)
3. Cant draw the graphs obviously but I'm fairly sure they were right
b. 1-1/(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^2-x< 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+x-2): 2e^(-2x) is always positive and (-x^2+x-2) is always negative, +ve multiplied by a -ve = -ve which is less then 0 hence dy/dx<0
8. 4cosx+ 2x/(1-x^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
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?
7. (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?
No not yet, I don't see why it wouldn't be accepted though since technically it is correct. Just depends how picky they are in the ms.
8. Anyone have any solutions to put my mind at ease? Especially Q6, 7, 8(iii) thanks
9. (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
How were you meant to work out k?? surely it was just 0.04
10. (Original post by harrytqo)
How were you meant to work out k?? surely it was just 0.04
I think what they expected students to do was n/2 = (0.96)^x X n where x is the number of centuries and n is the mass of radium. The n cancel out btw
11. (Original post by harrytqo)
How were you meant to work out k?? surely it was just 0.04
The general equation is N =No e^-kt (it is minus because it's exponential decay)

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...
12. (Original post by harrytqo)
How were you meant to work out k?? surely it was just 0.04
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)
13. (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)
Answer should probably be to the nearest year by the way
14. (Original post by Foutre en L'air)
Answer should probably be to the nearest year by the way
I was thinking about that which is why I gave both answers in my paper. However they almost always want the answers rounded with these questions so that's why I am saying its 1700
15. 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.
16. Congrats! I got an A* too and hope you get in your firm choice.
17. (Original post by 9clpc9)
Congrats! I got an A* too and hope you get in your firm choice.
I did , hopefully you did as well
18. (Original post by Golden hawk)
I did , hopefully you did as well
That's great, congrats!! I did too

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