CCEA A2 maths - C3 20th may 2016

Reply 80

Rmcmurray21

2

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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Reply 81

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

Reply 82

Camderman106

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 not found

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

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

\frac{dy}{dx}

, but as long as you wrote the formula for the gradient you shouldn't lose marks!

Reply 84

will872

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

\frac{dy}{dx}

, 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.

Reply 85

Mr.bob

16

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?

Reply 86

will872

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.

Reply 87

Phizzle125

Anyone have any solutions to put my mind at ease? Especially Q6, 7, 8(iii) thanks

Reply 88

harrytqo

1

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

Reply 89

Mr.bob

16

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

Reply 90

Smcdonald98

2

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...

(edited 7 years ago)

Reply 91

Camderman106

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)

(edited 7 years ago)

Reply 92

username1763791

4

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

Reply 93

Camderman106

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