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# Edexcel A2 C3 Mathematics 12th June 2015 watch

1. (Original post by randlemcmurphy)
We know two points so we can form two equations:

1=a|6-b|-1This means a|6-b|=0 So a could equal zero and b could equal 6

We also have:

11=a|0-b|-1This means ab=1
This means that a cannot equal zero, so b equals 6 and a equals 1/6 (hope that is the right answer!)
A = 2 and i didn't get how you did it :/

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2. (Original post by lam12)
A = 2 and i didn't get how you did it :/

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I realised I wrote 1 instead of 11, post edited.
3. (Original post by mismash)
hello, i am really sorry to but in.. but im a year 12 student and i am really worried because i dont think i have done v well in my c1 and c2 maths exams.
would u guys recommend retaking those modules in y13 or simply is it going to waste time and increase work load too much...???
It really won't increase your workload too much since you need to know C1 and C2 to be able to do C3 and C4 anyway - 90% of my class are resitting them since they seem so much easier now so think they might as well improve their grade.

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4. (Original post by randlemcmurphy)
We know two points so we can form two equations:

1=a|6-b|-1This means a|6-b|=0 So a could equal zero and b could equal 6

We also have:

11=a|0-b|-1This means ab=12
This means that a cannot equal zero, so b equals 6 and a equals 1/6 (hope that is the right answer!)
If ab=12, that'd make b 2 which is the right answer! I didn't think of solving it this way (used that it had moved 6 units to the right) but might use in future! My problem was that I stupidly used (11,0) instead of (0,11)... :/ Thanks

(Original post by lam12)
The range changes because it's 3theta. You use 2pi... I don't exactly know what you're exactly asking tbh haha

For the first bit because it's a ^2 it can't be a negative and therefore the sin bit = 0

For the second part it would be 3 theta - 1.107 = 2pi as the range increases as its 3theta and now we can go up to 3pi but it's not = to 3pi and therefore we use just 2pi

I don't think I explained this well but I tried, I may be wrong but it's how it was explained to me :-)

Attachment 412775

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Ahh thankyou! My problem was that I didn't see the 3theta, assuming it was theta so ended up witht eh wrong answer! Thanks for drawing it out and everything though ))

(Original post by studentwiz)
coordinates (6,-1)

a l6-bl -1 = -1
a l6-bl =0
divide by a
6-b=0
b=6
Hey, I had found b but couldn't find a, this is another way to approach it that I didn't think of though so thanks
5. useful material for the upcoming exams

http://www.thestudentroom.co.uk/show....php?t=3353445

http://www.thestudentroom.co.uk/show....php?t=3361867

http://www.thestudentroom.co.uk/show....php?t=3361905
6. d and c thanks in advance nvm !

figured it out :/
Attached Images

7. someone explain how you would go about getting the range?
Attached Images

8. (Original post by studentwiz)
someone explain how you would go about getting the range?
consider range of y=e^x then consider what will happen to this range when the graph is shifted upward by k^2 units
9. Hey guys,

I've done all the real papers from 2005-2014 and also finished up the the Solomon papers. I'm hoping to finish all the Elmwood by tomorrow any suggestions for more papers that are more challenging but still relevant.

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10. (Original post by Gilo98)
consider range of y=e^x then consider what will happen to this range when the graph is shifted upward by k^2 units
1+k^2?

11. Can anyone help me with ii) ?? I'm confused..
12. (Original post by ridirahman)

Can anyone help me with ii) ?? I'm confused..
Let the length of a side be denoted r. Express V in terms of r.
You want dr/dt; you can find this using dV/dt and dV/dr. For the specific value you'll need to find r from the given volume.
13. (Original post by ridirahman)

Can anyone help me with ii) ?? I'm confused..
If the side length of the cube is x, then its volume will be x^3. We know dV/dt = 2, we can find dV/dx, and we can find dx/dt from there.
14. (Original post by Krollo)
If the side length of the cube is x, then its volume will be x^3. We know dV/dt = 2, we can find dV/dx, and we can find dx/dt from there.
Ohh okay thanks I got it now
and I have another question if you don't mind..

I've done part a) and got root over 109 cos(deta + 16.70)

How would you do part b) ?
15. (Original post by ridirahman)
Ohh okay thanks I got it now
and I have another question if you don't mind..

I've done part a) and got root over 109 cos(deta + 16.70)

How would you do part b) ?
if you sub root over 109 cos(30t + 16.7) in place of 10cos(30t) + 3sin(30t),
it becomes
H = 12 - root 109 cos (30t+16.7)
which is a transformation of the cos curve. Knowing the maximum in a normal cos curve is 1, you can work backwards to get the maximum in this one.

for part b, use your maxima and solve for t

16. Someone please help this is from the IYGB papers. I'm confused on part b what order do you do the transformations and why??

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17. (Original post by Medicjug)

Someone please help this is from the IYGB papers. I'm confused on part b what order do you do the transformations and why??

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either order so long as it is correct
(this particular transformation will not appear in an actual EDEXCEL exam after 2005)
18. useful material for the upcoming exams

http://www.thestudentroom.co.uk/show....php?t=3353445

http://www.thestudentroom.co.uk/show....php?t=3361867

http://www.thestudentroom.co.uk/show....php?t=3361905
19. (Original post by TeeEm)
either order so long as it is correct
(this particular transformation will not appear in an actual EDEXCEL exam after 2005)
If I decide to stretch it first by scale factor 1/2 do I move it by 1 unit instead of 2?

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20. (Original post by Medicjug)
If I decide to stretch it first by scale factor 1/2 do I move it by 1 unit instead of 2?

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yes

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